Lesson A

, Lesson A, Page 1 - What is Thermodynamics?

In 1698, Thomas Savery invented the steam engine.The steam engine converted heat into mechanical power.

The machine was used to pump water out of coal mines, but the principles behind its operationwere not well-understood.

In 1712, Thomas Newcomen built an improved steam engine, but it was expensive to build and not very reliable.

James Watt invented the first inexpensive, reliable steam engine in 1765.  Watt’s steam engine was more than six times more efficient than Newcomen’s !

Watt’s steam engine catalyzed the industrial revolution. In 1798, Count Rumford noticed that canons became hot as they were bored out. Rumsford showed that more heat was released in the boring process than could have been

originally contained in the metal. He concluded that the mechanical boring process was producing heat.  This was a major

departure from the caloric theory that was widely accepted at the time. In 1824, Carnot introduced ideal gas cycle analysis in his work “Reflection on the Motive Power of

Fire” In the 1840’s Mayer, Joule and Helmholtz developed the idea that energy could not be created or

destroyed.  Energy is conserved. This principle is now known as the 1st Law of Thermodynamics.  We will learn more about it in

chapters 4 and 5. Rudolf Clausius stated that no cycle can transfer heat from a low temperature region to a high

temperature region with no other interaction with its surroundings. This principle is known as the 2nd Law of Thermodynamics and we will learn a great deal more

about it in chapters 6 and 7. The contribution of these scientists and many others led to the development of the broad field of

thermodynamics that we know today. So, what areas of thermodynamics will we study in this program ?  Flip the page and see.

Jump to New pageShow / Hide Notes

© B-Cubed, 2003, 2005, 2006. All rights reserved.

Timeline of the Early Developments of Thermodynamics Newcomen invented an improved steam engine1712

1798Count Rumford began canon-boring experiments (dealing with the conversion of work into heat)

1840sMayer (1842), Joule (1847) and Helmholtz (1847) independently arrived at the conservation of

energy principle

Carnot published "Reflections on the Motive Power of Fire"1824

Clausius formulated the Second Law of Thermodynamics 1850

Thermodynamics Fundamentals

Work..... Energy..... Heat..... System..... Boundary..... Surroundings..... State..... Property.....

Process..... Reversible / Irreversible processes..... Ideal Gas..... Perfect Gas..... Semi Perfect Gas..... Thermal Efficiency.....

Introduction

This page identifies the factors that are fundamental to understanding thermodynamics.  

Work

Work is defined simply as the product of force (SI units = newtons) and the distance moved in the direction of the force (SI units = metres).  If a block is moved against a constant frictional force of 1 N through a distance of 1 m then 1 N.m (= 1 Joule), of work has been exerted in moving the block.....

Energy

Energy at is simplest level is the capacity of a system to do work.   A compressed spring has potential energy because, when released it can exert, a force over a distance until the spring has is no longer compressed.

A water in motion possesses kinetic energy which can be used to do work using a water turbine..

Potential energy is the form of energy which is stored e.g a body has potential energy due to its position in a gravitation field.  Kinetic energy is the form of energy a body possesses because of its motion.

Energy can have many different forms including

Thermal energy: sub-microscopic particles in motion. (form of kinetic energy)

Mechanical energy: macroscopic objects in motion. (form of kinetic energy)

Electrical energy: movement of electrons through a conductor.(form of kinetic energy)

Sound energy: compression/expansion of spaces between molecules. (form of kinetic energy)

Chemical potential energy: position of electrons relative to atomic nuclei in bound atoms. (form of potential energy)

Gravitational energy: position of an object in a gravitational field. (form of potential energy)

Electrostatic energy: relative position of charged particles.(form of potential energy)

The total stored energy of a body, substance or system is a property and is given the symbol E and is generally used not as an absolute value but as measure of change between different states ΔE.  The change of internal energy can be obtained from measurements of heat and work.

The stored energy of a body, substance or system which is independent of electricity, atomic, sound, magnetism, surface tension, motion and gravity is called the internal energy is given the symbol U.

Notes on the internal energy and the total stored energy and its relevance to the first law are provided on webpage Thermodynamic Laws

Energy is intangible ..It is not practical to measure the total stored energy of a body, substance or system.  Its presence can only be recognised by its effects on its surrounding or connected materials or mechanisms.  A rock located at ground level in a normal environment does not seem to possess any energy but it has sufficient potential energy to destroy a creature standing below it in a deep hole: it has sufficient kinetic energy in the form of heat to melt ice: it may have

sufficient chemical energy to heat a house ( if it is coal): it may have sufficient nuclear energy to destroy a city.....

Heat

Heat is the energy form which is transferred by a difference in temperature.   If a body at 50 deg C is positioned in a fluid at 10 deg C.   The fluid temperature rises as heat is transferred from the body.   (At the microscopic level the energy possessed as a result of its temperature is kinetic energy of the molecules and atoms.)

System

In thermodynamics a system is a 3D region in space under study.   A system can be an isolated, adiabetic, closed system or an open system and it is surrounded by a defined boundary.. The outside of the boundary is called the surroundings

For isolated system matter and energy do not cross the boundary..it is not influenced in any way by the part of space which is external to the system boundaries.

For adiabetic systems matter and heat do not cross the boundary..(energy in the form of work can cross the boundary)

For a closed system a fixed amount of matter which is enclosed by a boundary.   Only heat and work can be transferred across the boundary.

For an open system matter, heat and work flow across the boundary.

A simple system is a system that does not contain any internal adiabatic, rigid and impermeable boundaries and is not acted upon by external forces.

A composite System is a system that is composed of two or more simple systems.

Closed System

Boundary

The boundary encloses a system and may be real (physical) or imaginary.

A real boundary may be fixed e.g. a gas in a bottle, or movable e.g a gas in a cylinder which is compressed or expanded by movement of a piston.

An imaginary boundary is one not based on a physical control surface e.g. an engine could be analysed as a system contained within an imaginary boundary.

Imaginary and Real Boundaries

Control Surface

An open system is often called a control volume and its boundary and most specifically the boundary under scrutiny is called the control surface.  Mass may flow across a control surface.  A cylinder piston in motion may be a control surface.

Surroundings

Everything outside a system boundary is called the surroundings.  Normally the term surroundings is restricted to those outside the system that in some way interact with the system or affect the behaviour of the system.

Thermodynamic State of a System

The condition of the system as characterized by the values of its state properties.

Stable equilibrium is a state in which the system is not capable of finite spontaneous change to another state without a finite change in the state of the surroundings.... Many types of equilibrium must be fulfilled -- thermal, mechanical, phase (material) and chemical.

State postulate: The equilibrium state of a simple closed system can be completely characterized by two independently variable properties and the masses of the species contained within the system.

Property ..A characteristic of a system

In the context of a thermodynamic system a "state property" identifies a condition of the system (or substance) at an equilibrium state which is independent of the path of the process by which the state has been reached.  The change in the pressure (P) or the temperature (T) of a system between two equilibrium states is the same for all paths and therefore the pressure and temperature are properties.  All properties of equilibrium states e.g. P,V,m and T are characteristics of the activities of large quantities of molecules..

Extensive properties are those properties with values representative of the sum of parts are called extensive properties e.g. m (mass), V (volume), (H) total enthalpy.

Intensive properties are those with values not representative of the sum of parts are called intensive properties e.g. the temperature

(T). Pressure is also an intensive property

Specific properties are extensive properties per unit mass.   Specific properties are intensive.

Some Typical Thermodynamic (extensive) properties of fluids include..

(V) Volume (m3) (m)Mass (kg) (H)Enthalpy (J) (U) Internal Energy (J) (S) Entropy (J)

Some Typical Thermodynamic (intensive) properties of fluids include..

T temperature [K] ρ density [kg/m3] cp specific heat at constant pressure

[J/kg·K] cv specific heat at constant volume

[J/kg·K] μ dynamic viscosity [N/m2s] ν kinematic viscosity [m2/s] k thermal conductivity [W/m·K] α thermal diffusivity [m2/s] β volumetric thermal expansion

coefficient [K-1] h specific enthalpy [J/kg] s specific entropy [J/kg·K] G gibbs free energy [J/kg]

A primitive property can in principle be specified by describing an simple operation or test to which the system is subjected. e.g using

mechanical measurements ... pressure, volume, thermometric temperature (T) and heat capacity.

A derived property is a property that is mathematically defined in terms of primitive properties.

Process

A thermodynamic process may be defined as the progress of a thermodynamic system proceeding from an initial state to a final state.   The series of states the substance or system experiences as it progresses through the process is called the path of the process.

Typically, a thermodynamic process can be characterised, according to what system property e.g.temperature, pressure, or volume, etc., are held fixed.   Furthermore, it is useful to group these properities into pairs, in which the variable held constant is one member of the pair.   The six most common thermodynamic processes are shown below:

An isobaric process occurs at constant pressure...ΔP= 0

An isochoric process, or isometric process, occurs at constant volume...ΔV= 0

An isothermal process occurs at a constant temperature...ΔT= 0

An isentropic process occurs at a constant entropy...ΔS= 0 ( Adiabetic and reversible)

An isenthalpic process occurs at a constant enthalpy...Δh= 0

An adiabatic process occurs without loss or gain of heat...ΔQ= 0

A cyclic process occurs with same initial and final states

A polytropic process has the relationship pV k = constant

There are a number of thermodynamic process types encountered by engineers including non-flow, steady flow, semi-flow and unsteady flow.   These are described as follows:

Non flow processes are those involving no flow of matter across the system boundaries

Steady flow processes involve fluid entering and leaving the system control volume these flows do not change with time and the internal energy of the control is also fixed in the time period under consideration

A Semi-flow process involves fluid flow into a control volume which may be rigid charging a gas bottle or flexible -blowing up a balloon

An unsteady flow process is one with a variable internal energy- i.e changing liquid level in a boiler

Reversible /Irreversible Process

If the substance or system passes through a continuous series of equilibrium states in progressing through the process as it receives or rejects energy it is referred to as a reversible process.  The path of this theoretical process is generally shown on diagrams as a full line.  If the process is reversed, in the thermodynamic sense, it would leave no trace of itself.

In the real world there are no reversible processes..All processes are irreversible and are shown on diagrams as broken lines...Factors which make a process irreversible include friction, unrestrained gas expansion, heat transfer across finite temperature difference, mixing,chemical reactions etc. etc...

Ideal Gas

An ideal gas is one that behaves according according to the assumptions that the volume of a gas molecules can be discounted and that molecules do not attract each other.   Various relationships have been arrived at for and ideal gas including Charles Law, Boyle's Law and Avogradros Law. ref. Ideal Gas .. An ideal gas conforms to the ideal gas law

Pv = nRT

P = Pressure = (Pa) N/m 2 v = Volume = m 3 T = Absolute Pressure = deg Kelvin R = Universal Gas Constant = 8,314 J /mole.K = 8

314 J /kmole.K n = Number of moles

The laws and rules for ideal gases are only reasonably accurate for gases at low pressures and moderately high temperatures...At pressures around 1 bara or less the ideal gases are generally reasonably accurate for real gases.

Perfect Gas

A perfect gas is an ideal gas for which the values of the specific heats cp and cv are assumed constant.   ref. Properties This is an idealisation of the behaviour of real gases at low pressures e.g. oxygen and nitrogen at atmospheric pressure and ambient temperature.

Semi-Perfect Gas

Semi perfect gases are those subject to a wide variation of temperature such that it cannot be assumed that the specific heats are constan.   Semi Perfect gases are ideal gases for which the values of the specific heats cp and cv are allowed to

vary as function of T alone.   ref. Properties.

Thermal Efficiency

The thermal efficiency is a measure of the success of a thermodynamic process.   It is generally expressed in simple terms as the ratio of the energy desired and the energy expended.

Typical efficiencies include

Automobile IC engines ...12-15% Gas Turbine ...12-16% Steam Power Plant ...38-41% Solar Cell ...15% Fuel Cell ...40-60% Electric Motors ...90%

Intensive and extensive properties

In the physical sciences, an intensive property (also called a bulk property, intensive quantity, or intensive variable), is a physical property of a system that does not depend on the system size or the amount of material in the system: it is scale invariant.

By contrast, an extensive property (also extensive quantity, extensive variable, or extensive parameter) of a system is directly proportional to the system size or the amount of material in the system.

For example, density is an intensive property as it does not depend on the property, while mass and volume are extensive properties. Note that the ratio of two extensive properties that scale in the same way is scale-invariant, and hence an intensive property.

Table of Contents

1 Intensive properties

 1.1

Combined intensive properties

 1.2

Joining systems

 1.3

Examples

2 Extensive properties

 2.1

Combined extensive properties

 2.2

Examples

3 Related extensive and intensive properties

4 Distinction from perceptions

5 See also

6 References

Intensive properties

An intensive property is a physical quantity whose value does not depend on the amount of the substance for which it is measured. For example, the temperature of a system is the same as the temperature of any part of it. If the system is divided, the temperature of each subsystem is identical. The same applies to the density of a homogeneous system, if divided in half, the mass and the volume change in the identical ratio and the density remains unchanged.

According to the state postulate, for a sufficiently simple system, only two independent intensive variables are needed to fully specify the entire state of a system. Other intensive properties can be derived from the two known values.

Some intensive properties, such as viscosity, are empirical macroscopic quantities and are not relevant to extremely small systems.

Combined intensive properties

There are four properties in any thermodynamic system, two intensive ones and two extensive ones.

If a set of parameters, {ai}, are intensive properties and another set, {Aj}, are extensive properties, then the function F({ai},{Aj}) is an intensive property if for all α,

It follows, for example, that the ratio of two extensive properties is an intensive property - density (intensive) is equal to mass (extensive) divided by volume (extensive).

Joining systems

Let there be a system or piece of substance a of amount ma and another piece of substance b of amount mb which can be combined without interaction. [For example, lead and tin combine without interaction, but common salt dissolves in water and the properties of the resulting solution are not a simple combination of the properties of its constituents.] Let V be an intensive variable. The value of variable V corresponding to the first substance is Va, and the value of V corresponding to the second substance is Vb. If the two pieces a and b are put together, forming a piece of substance "a+b" of amount ma+b = ma+mb, then the value of their intensive variable V is:

which is a weighted mean. Further, if Va = Vb then Va + b = Va = Vb, i.e. the intensive variable is independent of the amount. Note that this property holds only as long as other variables on which the intensive variable depends stay constant.

In a thermodynamic system composed of two monatomic ideal gases, a and b, if the two gases are mixed, the final temperature T is

a weighted mean where Ni is the number of particles in gas i, and Ti is the corresponding temperature.

Note that you have to measure the amounts in the same unit that was used to calculate the intensive property from the extensive property. So when you interpolate density, you have to measure the properties in volume, as density is mass per volume. The formula makes no sense when you measure the properties in mass (kg).

Examples

Examples of intensive properties include:

temperature

chemical potential

density

specific gravity

viscosity

velocity

electrical resistivity

spectral absorption maxima (in solution)

specific energy

specific heat capacity

hardness

melting point and boiling point

pressure

ductility

elasticity

malleability

magnetization

concentration

Extensive properties

An extensive property is a physical quantity whose value is proportional to the size of the system it describes. Such a property can be expressed as the sum of the properties for the separate subsystems that compose the entire system.[citation needed]

Extensive properties are the counterparts of intensive properties, which are intrinsic to a particular subsystem and remain constant regardless of size. Dividing one type of extensive property by a different type of extensive property will in general give an intensive value. For example, mass (extensive) divided by volume (extensive) gives density (intensive).

Combined extensive properties

If a set of parameters {ai} are intensive properties and another set {Aj} are extensive properties, then the function F({ai},{Aj}) is an extensive property if for all α,

Thus, extensive properties are homogeneous functions (of degree 1) with respect to {Aj}. It follows from Euler's homogeneous function theorem that

where the partial derivative is taken with all parameters constant except Aj. The converse is also true - any function which obeys the above relationship will be extensive.[citation needed]

Examples

Examples of extensive properties include [citation needed]:

entropy

enthalpy

energy

mass

particle number

stiffness

volume

Related extensive and intensive properties

Thermodynamics

[show]Branches

Classical · Statistical · ChemicalEquilibrium / Non-equilibriumThermofluids

[show]Laws

Zeroth · First · Second · Third

[show]Systems

State:EquilibriumInstruments (heat bath, reservoir)Phase of matter

Processes:Isobaric · Isochoric · IsothermalAdiabatic · Isentropic · IsenthalpicQuasistatic · PolytropicReversibility · Endoreversibility · Irreversibility

Cycles:Heat engines · Heat pumpsThermal efficiency

[hide]System properties

Property diagramsIntensive and extensive properties

State functions:Temperature / Entropy (intro.) †Pressure / Volume (specific) †Chemical potential / Particle number †(† Conjugate variables)Reduced properties

Process functions:Work · Heat

[show]Material properties

Specific heat capacity 

c =T ∂S

N∂T

Compressibility β = −

1 ∂V

V∂p

Thermal expansion α = 1 ∂V

V∂T

Property database

[show]Equations

Carnot's theoremClausius theoremFundamental relationIdeal gas lawMaxwell relations

Table of thermodynamic equations

[show]Potentials

Free energy · Free entropy

Internal energy U(S,V)

Enthalpy H(S,p) = U + pV

Helmholtz free energy

A(T,V) = U − TS

Gibbs free energy G(T,p) = H − TS

[show]History and Culture

Philosophy:Entropy and time · Entropy and lifeBrownian ratchet · Maxwell's demonHeat death paradox · Loschmidt's

paradoxSynergetics

History:General · Heat · Entropy · Gas lawsPerpetual motionTheories:Caloric theory · Vis viva · Theory of heatMechanical equivalent of heat · Motive powerPublications:"An Experimental Enquiry Concerning ... Heat""On the Equilibrium of Heterogeneous Substances""Reflections on the Motive Power of Fire"

Timelines of:Thermodynamics · Heat engines

Art:Maxwell's thermodynamic surface

[show]Scientists

Daniel BernoulliSadi CarnotBenoît Paul Émile ClapeyronRudolf ClausiusHermann von HelmholtzJulius Robert von MayerWilliam RankineJohn Smeaton

Georg Ernst StahlBenjamin ThompsonWilliam ThomsonJohn James Waterston

v • d

See also: List of thermodynamic properties

Although not true for all physical properties, there are a number of properties which have corresponding extensive and intensive analogs, many of which are thermodynamic properties. Examples of such extensive thermodynamic properties, which are dependent on the size of the thermodynamic system in question, include volume (V), internal energy (U), enthalpy (H), entropy (S), Gibbs free energy (G), Helmholtz free energy (A), and heat capacities (Cv and Cp) (in the sense of thermal mass). Note that the main symbols of these extensive thermodynamic properties shown here are capital letters. Except for volume (V), these extensive properties are dependent on the amount of material (substance) in the thermodynamic system in question.

For homogeneous substances, these extensive thermodynamic properties each have analogous intensive thermodynamic properties, which can be expressed on a per mass basis, and the corresponding intensive property symbols would be the lower case letters of the corresponding extensive property. Examples of intensive thermodynamic properties, which are independent on the size of the thermodynamic system in question and are analogous to the extensive ones mentioned above, include specific volume (v), specific internal energy (u), specific enthalpy (h), specific entropy (s), specific Gibbs free energy (g), specific Helmholtz free energy (a), and specific heat capacities (cv and cp, sometimes simply called specific heats). These intensive thermodynamic properties are effectively material properties which are valid at a point in a thermodynamic system or at a point in space at a certain time. These intensive properties are dependent on the conditions at that point such as temperature, pressure, and material composition, but are not considered dependent on the size of a thermodynamic system or on

the amount of material in the system. See the table below. Specific volume is volume per mass, the reciprocal of density which equals mass per volume.

Corresponding extensive and intensive thermodynamic properties

Extensiveproperty

Symbol

SI unitsIntensiveproperty**

Symbol

SI units

Volume V m3 or l* Specific volume*** vm3/kg or l*/kg

Internal energy U J Specific internal energy u J/kg

Entropy S J/K Specific entropy s J/(kg·K)

Enthalpy H J Specific enthalpy h J/kg

Gibbs free energy G JSpecific Gibbs free energy

g J/kg

Heat capacityat constant volume

CV J/KSpecific heat capacityat constant volume

cv J/(kg·K)

Heat capacityat constant pressure

CP J/KSpecific heat capacityat constant pressure

cP J/(kg·K)

* l = liter, J = joule

** specific properties, expressed on a per mass basis

*** Specific volume is the reciprocal of density.

If a molecular weight can be assigned for the substance, or the number of moles in the system can be determined, then each of these thermodynamic properties can be expressed on a per mole basis. These intensive properties could be named after the analogous extensive properties but with the word "molar" preceding them; thus molar volume, molar internal energy, molar enthalpy, molar entropy, etc. Although the same small letters can be used as in the analogous specific properties indicating they are intensive, sometimes the corresponding capital letters have been used (and understood to be on a per mole basis), and there seems to be no universally agreed upon symbol convention for these molar properties. A well known molar volume is that of

an ideal gas at STP (Standard Temperature and Pressure); this molar volume = 22.41 liters per mole.

Distinction from perceptions

Certain perceptions are often described (or even "measured") as if they are intensive or extensive physical properties, but in fact perceptions are fundamentally different from physical properties. For example, the colour of a solution is not a physical property. A solution of potassium permanganate may appear pink, various shades of purple, or black, depending upon the concentration of the solution and the length of the optical path through it. The colour of a given sample as perceived by an observer (i.e., the degree of 'pinkness' or 'purpleness') cannot be measured, only ranked in comparison with other coloured solutions by a panel of observers. Attempts to quantify a perception always involve an observer response, and biological variability is an intrinsic part of the process for many perceived properties. A given volume of permanganate solution of a given concentration has physical properties related to the colour: the optical absorption spectrum is an extensive property, and the positions of the absorption maxima (which are relatively independent of concentration) are intensive properties. A given absorption spectrum, for a certain observer, will always be perceived as the same colour; but there may be several different absorption spectra which are perceived as the same colour: there is no precise one-to-one correspondence between absorption spectrum and colour even for the same observer.

The confusion between perception and physical properties is increased by the existence of numeric scales for many perceived qualities. However, this is not 'measurement' in the same sense as in physics and chemistry. A numerical value for a perception is, directly or indirectly, the expected response of a group of observers when perceiving the specified physical event.

Examples of perceptions related to an intensive physical property:

Temperature: in this case all observers will agree which is the hotter of two objects.

Loudness of sound; the related physical property is sound pressure level. Observers may disagree about the relative loudness of sounds with different acoustic spectra.

Hue of a solution; the related physical property is the position of the spectral absorption maximum (or maxima).

Examples of perceptions related to an extensive physical property:

Color of a solution: The related physical property is the transmission or absorption spectrum

THERMODYNAMICS (PART-I) IIT-JEE INTRODUCTION & FIRST LAW OF THERMODYNAMICS 1) Which of the following statement is true about thermodynamics

a) Thermodynamics will help in predicting whether a physical or chemical change is possible under given conditions. b) Thermodynamics only deals with the initial and final states of the system and is not helpful in evolving the mechanism of the process. c) The rate of a reaction can be evolved from thermodynamics. d) The mechanism of a reaction can be evolved from thermodynamics. 1) a only 2) a & b 3) a,b & c 4) a & d

2) Choose the incorrect statement

1) Open systems can exchange both energy and matter with its surroundings.2) Closed systems can only exchange energy and do not exchange matter with its surroundings.3) Isolated systems can exchange energy and matter with its surroundings.4) None.

3)Which of the following is not an intensive property?

1) Entropy 2) Density 3) Temperature 4) Pressure

4) Which of the following is an extensive property?

1) Entropy 2) Enthalpy 3) Volume 4) All

5) Extensive properties, out of a) boiling point, b) viscosity, c) pH, d) emf of a cell and e) molar heat capacity; are

1) a, b & c 2) b & e 3) c & e 4) None

6) Which of the following is not a state function?

1) Internal energy 2) Enthalpy 3) Work 4) Free energy

7)Which of the following is a state function and not an intensive property?

1) Temperature 2) Volume 3) Pressure 4) None

8) In an adiabatic process,

1) dT = 0 2) dU = 0 3) 0q 4) All Note: In an adiabatic process, there is no exchange of heat between system and surroundings. Hence dq = 0 and dU = w. However there is change in temperature and internal energy during this process.

9) In which of the following processes, dT =0? 1) Adiabatic process 2) Isobaric process 3) Isothermal process 4) Isochoric process 10) In an isochoric process, 1) dT = 0 2) dP = 0 3) dV = 0 4) dH = 0 Note: in isochoric (dV=0) and isobaric (dP=0) processes, the pressure-volume work is always zero.

11) In a cyclic process, 1) dU = 0 2) dH = 0 3) dT = 0 4) All 12) If the systems A and B are in thermal equilibrium with system C, then system A is also in thermal equilibrium with system B. In thermodynamics, this statement is known as 1) 1st law 2) 2nd law 3) 3rd law 4) zeroeth law 13) Thermos flask is an example of 1) Closed system 2) Open system 3) Isolated system 4) All 14) The correct statement about isothermal process is 1) dT = 0 2) dU = 0 3) q = -w 4) All 15) The statement which is not true according to 1st law of thermodynamics is 1) Energy can neither be created nor destroyed although it can be changed from one form to another. 2) The total energy of the universe is constant. 3) The change in the internal energy of a closed system is equal to the sum of heat lost or absorbed by the system and work either done by the system or done on the system 4) The change in internal energy of a system is zero in all the processes

16) The work done during the free expansion of an ideal gas is equal to 1) Pext 2) dV 3) dU 4) zero 17) Which of the following is true about isothermal free expansion of a gas? 1) dU = 0 2) dT = 0 3) PextdV = 0 4) All 18) The work done during isothermal irreversible change will be given by 1) -Pext(Vf-Vi) 2) 2.303logfVnRTV

3) logfiVnRTV

4) zero Note: In the irreversible process, the work is done against constant external pressure which differs largely from the internal pressure.

19) The work done during an isothermal reversible change will be given by 1) -Pext(Vf-Vi) 2) 2.303logfVnRTV

3) logfiVnRTV

4) zero Note: In case of reversible process, the internal pressure is almost equal to the external pressure. This internal pressure is given by nRT / V.

20)Two litres of an ideal gas at a pressure of 10 atm expands isothermally into vacuum until its total volume becomes 100 litres.The amount of heat absorbed in the expansion is 1) 2.303 J 2) 100 J 3) 90 J 4) None Note: As the process is isothermal, q=-w=PdV; And as the expansion is done against zero pressure(Pext=0), the work done as well as the heat absorbed will be zero.

21) Which of the following statements is true

1) The work done in reversible expansion is less than the work done in irreversible expansion.

2) The work done in reversible expansion is equal to the work done in irreversible expansion.

3) The work done in reversible expansion is greater than the work done in irreversible expansion.4) All. Note: But in case of compression, |wrev| < |wirrev| 22)Ten litres of an ideal gas at a pressure of 10 atm expands isothermally against a constant pressure of 1 atm. until its total volume becomes 100 litres.The work done in the expansion is 1) -90 L-atm 2) -10 L-atm 3) -23.03 L-atm 4) -230.3 L-atm Hint: This is the case of irreversible expansion.

23)Ten litres of an ideal gas at a pressure of 10 atm expands reversibly under isothermal conditions until its total volume becomes 100 litres.The work done in the expansion is 1) -90 L-atm 2) -10 L-atm 3) -230.3 L-atm 4) -100 L-atm Note: The work done is maximum in case of reversible expansion. (negative sign only indicates the work is done by the system)

24)The amount of heat absorbed, when 2.5 L of an ideal gas is compressed isothermally under constant atmospheric pressure until the volume becomes 0.5 L is 1) 2 J 2) -202.65 J 3) +202.6 J 4) -2 J Note: 1) During isothermal compression, heat is liberated by the system and hence negative with respect to it. 2) 1 L-atm = 101.325 J

25) 2 moles of an ideal gas at STP is expanded isothermally in infinitisimally small steps until the volume is doubled. The amount of heat absorbed during this process is 1) 1.37 kJ 2) 8.3 kJ 3) -1.37 kJ 4) None 26) 1 liter of an ideal gas, initially at a pressure of 10 atm, is allowed to expand at constant temperature

to 10 liters by reducing the external pressure to 1 atm in a single step. The work done during this process is 1) 10 L-atm 2) -100 L-atm 3) -9 L-atm 4) 0 L-atm 27) The area of which of the following graph shows PV work done during reversible expansion of ideal gas at constant temperature?

28) 25 L of an ideal gas is compressed isothermally under constant atmospheric pressure until the vol- ume becomes 5 L. The change in internal energy is 1) 20 J 2) -2026.5 J 3) +2026 J 4) 0 J 29) There is a balloon of given volume, V1, containing a gas at temperature, T1. When the balloon is placed in a colder room at temperature, T2, the balloon’s temperature starts to drop. What are the signs of the system’s q, w, and E for this process? 1) +q, +w, +E 2) -q, -w, -E 3) -q, -w, +E 4) -q, +w, -E 30) 11.2 L of a hydrogen gas at 273 K temperature and 1 atm of pressure in a sealed rigid container is heated to double its temperature. The change in internal energy dU will be equal to 1) +w 2) +q 3) -w 4) -q Note: As the process is isochoric, the change in internal energy is only due to exchange of heat

dU =qv

31) An ideal gas at 10 atm pressure and occupying 0.1 L is expanded to 1.1 L by supplying 101.325 J of heat against

constant atmospheric pressure irreversibly. The change in the temperature of the gas during this process is 1) 10 K 2) 1.1 K 3) 1K 4) No change Hint: dU = q+w = 101.325 J + (-101.325 J) = 0

32) A gas is allowed to expand at constant temperature from a volume of 1.0 L to 10.1 L against an external pressure of 0.50 atm. If the gas absorbs 250 J of heat from the surroundings, what are the values of q, w, and E respectively? 1) 250 J, -461 J & -211 J 2) -250 J, -461 J & -711 J 3) 250 J, -461 J & -711 J 4) 250 J, -4.55 J & 245 J 33)In a process, 800 J of heat is absorbed by a system and 350 J of work is done by the system. The change in internal energy for the process is 1) 350 J 2) 450 J 3) -450 J 4) -350 J 34) When is dHsys= dEsys? 1) When qv= qp.

2) In reactions involving only liquids and solids. 3) In reactions running under a vacuum (P=0). 4) All 35)Which of the following reactions could do work of expansion on the surroundings? 1) 2CO(g) + O2(g) 2CO2(g)

2) Fe2O3(s) + 2Al(s) Al2O3(s) + 2Fe(s) 3) CH4(g) + 2O2(g) CO2(g) + 2H2O(g) 4) 2N2O(g)

2N2(g) + O2(g) 36)Suppose a gas in a piston is expanded at constant pressure and the temperature goes down. Which of the following correctly describes the signs for the work, the heat for the system and the energy change of the system?

1) Work is positive, q is negative, and dU is negative.

2) Work is negative, q is may be positive or negative, and dU is negative.

3) Work is positive, q is may be positive or negative, and dU is negative.

4) Work is positive, q is positive, and dU is negative.37) In order to have 0 E for a process, which of the following conditions must be obeyed ?

a) q > 0 b) w > 0 c) q + w > 0 d) q > 0 such that |q| > |w| (where |q| and |w| are absolute values) The correct options are 1) a,b,c & d 2) b &d 3) a & d 4) c & d ENTHALPY & HEAT CAPACITY 1) The amount of heat exchanged between system and surroundings under constant pressure is called 1) Entropy 2) Enthalpy 3) Internal energy 4) Free energy Note: The amount of heat exchanged at constant pressure i.e., p

q H U P V

2) Enthalpy(H) of a system can be represented by 1) U + PV 2) U-PV 3) q+w 4) dU + PdV 3) Consider the following statements about enthalpy.

a) It is an intensive property

b) It is a state function.

c) It is an extensive property.

d) It is a path function.The correct statements are 1) a & d 2) b & c

3) a & c 4) b & d 4) Hydrogen chloride gas readily dissolves in water, releasing 75.3 kJ/mol of heat in the process. If one mole of HCl at 298 K and 1 atm pressure occupies 24.5 liters, findU for the system when one mole of HCl dissolves in water under these conditions. 1) +2.48 kJ 2) -75.3 kJ 3) -72.82 kJ 4) +75.3 kJ Hint: The volume of the liquid is negligible. The process is isothermal and isobaric.

Note: -72.82 kJ > -75.3 kJ; So the change in internal energy(dU) will be greater than the change in enthalpy(dH)i.e., dU > dH. But if absolute

values are taken, 75.3 kJ > 72.3 kJ i.e., the heat liberated will be greater than decrement in internal energy. It is because, the compression of the

gas which increases the internal energy. Also remember, the magnitude of work done on the system is less than heat liberated during the

process.

5) One mole of an ideal gas is dissolved in a solvent by liberating 0.27 kJ of heat. What is the change in internal energy when one mole of gas is dissolved completely in this solvent at 273 K and 1 atm pressure. 1) -2 kJ 2) -0.27 kJ 3) +2 kJ 4) +0.53 kJ Note: Now +2 kJ > 0.27 kJ. Therefore dU > dH. This is true even when the absolute values are taken. Why? In this case the value of work done on the system during compression is greater than the amount of heat lost.

6)

If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar

and 100°C is 41kJ mol-1. The internal energy change, when 1 mol of water is vaporised at 1 bar pressure and

100°C will be 1) 37.9 kJ mol–1 2) 41 kJ mol–1 3) 47.9 kJ mol–1 4) 18.3 kJ mol–1 Reaction: H2(liquid) H2(gas) Formula: g

Where n = no. of moles of gaseous products - no. of moles of gaseous reactants g

U H n RT

7) Which of the following is not an intensive property? 1) Molar heat capacity 2) Specific heat capacity 3) Heat capacity 4) All 8) The amount of heat required to raise the temperature of 1.00 g of aluminium by 1oC is called its 1) Enthalpy 2) Heat capacity 3) Specific heat 4) Molar heat

9) The molar heat capacity for NaCl is 50.50 J mol-1 K-1. What is the specific heat? 1) 0.8640 J g-1 K-1 2) 50.5 J g-1 K-1 3) 8.640 J g-1 K-1 4) 4.184 J g-1 K-1 10)The amount of heat necessary to raise the temperature of 60.0 g of aluminium from 15oC to 55oC is. (Molar heat capacity ofAl is 24 J mol–1K–1) 1) 2.133 J 2) 1.066 J 3) 2.133 kJ 4) 2.4 kJ Formula: m m q=n.C . T Where n = no. of moles C = molar heat capacity T = raise or lowering of temperature

11) The heat capacity of methyl alcohol (MW = 32.05 g/mol) is 80.3 J mol-1 K-1. The quantity of heat that will be evolved when the temperature of 2610 g of methyl alcohol falls from 22oC to 2oC 1) 1.5 x 102 kJ 2) 1.3 x 102 kJ 3) 1.7 x 102 kJ 4) 7 kJ

12) The amount of heat absorbed by one mole of an ideal gas in an isochoric process to raise the temperature from

1.1oC to 11.1oC is 120 kJ/mol. The Cv and Cp values of the gas in kJ mol-1K-1 will be 1) 12 kJ & 3.7 kJ 2) 10 kJ & 12 kJ 3) 12 kJ & 20.3 kJ 4) 20.3 kJ & 12 kJ Use the formulae: Formula: m m q=n.C . T Where n = no. of moles C = molar heat capacity T = raise or lowering of temperature

13) The difference in Cp and Cv values for liquids and solids will be equal to 1) R 2) nR 3) 2R 4) 0 14) Choose the incorrect statement 1) The molar heat capacities of the metallic elements are almost identical. This is the basis of theLaw of Dulong and Petit, which served as an important tool for estimating the atomic weights of some elements. 2) The intermolecular hydrogen bonding in water and alcohols results in anomalously high heat capacities for these liquids; the same is true for ice, compared to other solids. 3) The heat capacity values for graphite and diamond are very high as the solids that are more

“ordered” tend to have larger heat capacities. 4) None. Note: The molar heat capacities of metals are almost equal to 3R.

15) The molar heat capacity values of noble gases at contant volume are almost equal to 1)32R 2)52R 3)72R 4) R Note: In case of noble gases (which are mono-atomic) only three translational degrees of freedom are possible and each of these contribute12 Rto heat capacity and hence the Cv = 32R . The Cp value will be given by Cv + R = 52 R

16) The theoretical molar heat capacities of diatomic molecules at constant volume and at fairly high temperatures is almost equal to 1)32R 2) R 3)72R 4)52R Note: Diatomic molecules have 3 translational degrees of freedom (contribution of 3R/2), 2 rotational degrees of freedom (contribution of 1R/2 from each) and 1 vibrational degree of freedom ( contribution of R). Hence Cv = 72R (at high temp.) But at low temperatures (eg., room temperature), the vibrational degree of freedom can be neglected and hence Cv = 52R In the same way, for poly-atomic molecules, the contributions are as follows From three translational degrees of freedom --1 1 1 3 2 2 2 2 R R R R

From three rotational degrees of freedom --1 1 1 3 2 2 2 2 R R R R

From 3N-6 vibrational modes -- (3N-6)R (where N = number of atoms in poly-atomic molecule)

17)Calculate the enthalpy change on freezing of 1.0 mol of water at10.0oC to ice at –10.0oC. Given fusH = 6.03 kJ mol–1 at 0°C; C p [H2O(l)] = 75.3 J mol–1 K–1 ;Cp [H2O(s)] = 36.8 J mol–1 K–1

1) 7.151 kJ 2) -6.03 kJ 3) 3.63 kJ 4) -7.151 kJ

18) If 1000 calories are added to 20 g of ice at -10oC, the final temperature will be (Specific heat of ice = 0.48 cal/g-K and Heat of fusion of ice = 80 cal/g.) 1) 380 K 2) 0 K 3) 273 K 4) 283.2 K Note: As the heat supplied is not sufficient to melt the ice completely, finally there will be a mixture of ice and water at 0oC

19) A coffee-cup calorimeter is calibrated by adding 1840 J of heat to the water in the calorimeter and measuring a 1.72oC rise in temperature. When some

NH4Cl(s) is added to the same water in the calorimeter, the temperature falls by 1.04oC. The enthalpy change due to the dissolving of NH4Cl(s)

is 1) -1112 J 2) +1250 J 3) +3040 J 4) +1112 J 20) Which of the following statements is true?

1) q = dH at constant P; q = dE at constant T 2) q = dH at constant T; q = dE at constant V 3) q = dH at

constant V; q = dE at constant P 4) q = dH at constant P; q = dE at constant V

21) An ice cube at 0oC weighing 9.0 g is dropped into an insulated vessel containing 72 g of water at 50oC. What is the

final temperature of water after the ice has melted and a constant temperature has been reached? The latent heat

of fusion of ice is 6.01 kJ/mol and the molar heat capacity of H2O is 75.4 J mol-1K-1. 1) 36oC 2) 40oC 3) 44oC 4) 32oC 22) What is o

U

when one mole of liquid water vaporises at 100oC if the heat of vaporisation o

vapH

of water at 100 oC is 40.66 kJ.mol-1

? 1) 40.66 kJ.mol-1 2) 24.66 kJ.mol-1 3) 36.73 kJ.mol-1 4) -40.66 kJ.mol-1

23) The value of Helium gas is equal to 1) 1.67 cal 2) 1.4 cal 3) 6 cal 4) 10 cal 5 Note: 1.67 3 3 For mono atomic gases 3 2 5 pv v p v

C cal

C R C cal C C R cal

24) Temperature of one mole of Neon gas is increased by 1oC, hence, increase in internal energy is 1) 5 cal 2) 3 cal 3) 9 cal 4) 2 cal 25) Enthalpy change for a reaction does not depend upon

1) the physical states of reactants and products.

2) use of different reactants for the same product.

3) the nature of intermediate reaction steps.4) the differences in initial or final temperatures of involved substances.

26) The specific heat of water is 4.18 J g-1 K-1 and that of stainless steel is 0.51 J g-1 K-1. The heat that must be

supplied to a 750.0 g stainless steel vessel containing 800.0 g of water to raise its tempera- ture from 20.0oC to

the boiling point of water 1) 6.98 kJ 2) 29.8 kJ 3) 69.8 kJ 4) 298 kJ 27) The temperature of a substance represents

1) the total heat content of a the particles in a substance

2) the speed of the fastest particles in the substance

3) the speed of the slowest particles in the substance4) the average kinetic energy of the particles in a system. 28) A bomb calorimeter was calibrated by burning a sample of benzoic acid (C6H5COOH) which has a known heat of reaction, H° = –3227 kJ/mol. When 1.22 g of benzoic acid is burned in the calorimeter, the temperature is increased by 0.75 °C. The heat capacity of the calorimeter and its contents will be 1) 32 kJ K-1 2) 4.3 kJ K-1 3) 83 kJ K-1 4) 43 kJ K-1

29) The heat of combustion of methyl alcohol, CH3OH, is -715 kJ mol-1. When 2.85 g of CH3OH was burned in a

bomb calorimeter, the temperature of the calorimeter changed from 24.05 °C to 29.19 °C. The heat capacity of

the calorimeter is 1) 12.4 kJ/°C 2) 124 kJ/°C 3) 12.4 J/°C 4) 1.24 kJ/°C THERMOCHEMISTRY 1)

Which statement isinc orrec t about endothermic reactions? 1) the system absorbs energy from its surroundings 2) the enthalpy of products is lower than the enthalpy of the reactants 3) the thermal kinetic energy of the surroundings will decrease 4) the enthalpy change will have a positive value 2) Which of the following is the standard state of carbon at STP? 1) C as CO2 (g) 2) C as graphite (s) 3) C as CH4 (g) 4) C as diamond (s) 3) Which of the following statements is/are true about an exothermic reaction?

I) the energy absorbed in bond breaking is more than the energy released in bond formation

II) the system absorbs energy

III) the potential energy of the reactants is less than the potential energy of the productsIV) the thermal kinetic energy of the surroundings will increase 1) I and III only 2) II and III only 3) II and IV only 4) IV only 4) How much heat is absorbed when 3.00 grams of SiO2 react with excess carbon according to the reaction below? o

rH

for the reaction is +624.7 kJ. SiO2(s) + 3C(s)

SiC(s) + 2CO(g)

1) 208 kJ 2) 5.06 kJ 3) 10.4 kJ 4) 31.2 kJ 5) H of which of the following reactions is equal to the standard enthalpy of formation( o

fH

) of NH3? 1) 2NH3(g) 3H2(g) + N2(g) 2) NH3(g) 3/2H2(g) + 1/2N2(g) 3) 3H(g) + N(g) NH3(g) 4) 3/2H2(g) + 1/2N2(g) NH3(g) Note: The standard enthalpy of formation, o

fH

, is the amount of heat either liberated or absorbed during the formation of one mole of a compound from its elements under standard conditions.

6) Given: H2(g) + Br2(l) 2 HBr(g) ; o

rH

= –72.8 kJ Calculate the amount of energy absorbed or released when 15.0 g of HBr (g) is formed. 1) 6.75 kJ released 2) 13.5 kJ released 3) 4.85 kJ absorbed 4) 607 kJ absorbed 7) Which of the following has non zero standard enthalpy of formation at 25oC? 1) Cl2(g) 2)O2(g) 3) Na(g) 4) F2(g) 8) The standard heat of formation of NH3 is (Given N2(g) + 3H2 (g) 2NH3 (g) ; o

rH

= -91 kJ ) 1) +91 kJ 2) + 45.5 kJ 3) -19 kJ 4) -45.5 kJ 9) The enthalpy change for the reaction of 50 mL of ethylene with 50 mL of H2 at 1.5 atm pressure is H = -0.31 kJ. The change in internal energy will be

1) 0.66 kJ.mol-1 2) -0.3024 kJ.mol-1 3) +0.3024 kJ.mol-1 4) -0.66 kJ.mol-1

10) In a constant-volume bomb calorimeter an unknown compound reacted with excess oxygen to give carbon dioxide

and water. The temperature of 2 kg of water in the calorimeter rose from 12.72oC to 20.72oC. The heat capacity

of the calorimeter is 2.02 kJ.K-1 and the specific heat of water is 4.184 J/goC. The heat given off by the

combustion reaction under these conditions is: 1) 41,600 J 2) -41.6 kJ 3) -83.2 kJ 4) -33.5 kJ Note: The heat given off indicates the internal energy change, as the reaction is occuring at constant volume, and not is not equal to the enthalpy change.

11) In a constant - volume bomb calorimeter, 4g of methane is burned in excess of oxygen. The tem- perature of 0.5

Kg of water in the calorimeter rose from 12oC to22oC The heat capacity of calorim- eter is 20.1 kJ.K-1. The

enthalpy of combustion of methane under standard conditions will be 1) 223.1 kJ.mol-1 2) 891.6 kJ.mol-1 3) 888 kJ.mol-1 4) 20.92 kJ.mol-1 Note: Enthalpy of combustion is the amount of heat liberated when one mole of substance is completely burned in excess of oxygen at standard conditions (constant temperature and pressure). In this case the amount of heat liberated is equal toU . This value should be converted toH

by using the formula. g

H U nRT

12) B5H9 burns in air according to the following reaction. 2B5H9(g) + 12O2(g) 5B2O3(s) + 9H2O(l)

What is the molar heat of reaction for the combustion of B5H9 if the reaction between 0.1 g of B5H9 and excess

oxygen in a bomb calorimeter raises the temperature of the 852 g of water surrounding the calorimeter by

1.57oC? [At wts: B = 10.81 amu; H = 1.008 amu; O = 16.00 amu; the heat capacity of water is 4.184 J / goC] 1) 5.60 x 103 J/mol 2) 9.14 x 102 kJ/mol 3) 3.54 x 103 kJ/mol 4) 4.46 x 103 kJ/mol 13) In the reaction, CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ; o

rH

= -890.4 kJ At 1.0 atm and 273 K , how much work is involved per mole of CH4(g) with the volume change that occurs upon reaction. 1) -4.5 kJ 2) -2.2 kJ 3) 2.2 kJ 4) 4.54 kJ Note: The work is done on the system and is equal to 4.54 kJ. Here g

n =-ve and hence,H

E i.e., -890.4 kJ< - 885.8 kJ.

But if the absolute values are considered, the decrease in internal energy is less than heat liberated. This is because Some amount of decrease in internal energy, due to loss of heat, is compensated by compression work done on the system.

14) When burned in oxygen, 10.0 g of phosphorus generated enough heat to raise the temperature of 2950 g of water from 8.0oC to 28.0oC. The heat of formation of P4O10 from P4(s) and O2(g) is 1) -30.6 kJ mol-1 2) -306 kJ mol-1 3) -3060 kJ mol-1 4)-6120 kJ mol-1 15) Using the following data: I) N2(g) + 3O2(g) + H2(g) 2HNO3(aq) ;H = -414.8 kJ II) N2O5(g) + H2O(g) 2HNO3(aq) ;H = 218.4 kJ III) 2H2O(g) 2H2(g) + O2(g) ;H = 483.6 kJ What is theH for the reaction: 2N2O5(g) 2N2(g) + 5O2(g)

1) 90.8 kJ 2) -876.4 kJ 3) 782.8 kJ 4)1750 kJ 16. Which thermochemical equation corresponds to the enthalpy diagram shown above?

1) 2 H2 (g) + O2 (g) 2 H2O (g) + 486.3 kJ

2) 2 H2 (g) + O2 (g) + 486.3 kJ 2 H2O (g)

3) 2 H2 (g) + O2 (g) 2 H2O (g) – 486.3 kJ

4) H2 (g) + 1/2 O2 (g) – 486.3 kJ H2O (g)17. The standard heat of combustion (Hº ) of ethanol is – 1372 kJ/mol ethanol. How much heat is released when a 20.0 g sample burns? 1) 68.6 kJ 2) 29.8 kJ 3) 3.16 x 103 kJ 4) 595 kJ 18. Which is the correct enthalpy level diagram for the following reaction: CaO(s) + 3C(s) + 462.3 kJ CaC2(s) + CO(g)

19. Given the following thermochemical equation 2 C2H3Br (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (g) + 2 HBr (g) Hº = 1150 kJ What is the change in energy accompanying the production of 1.5 moles of CO2(g)? 1) 1725 kJ 2) 766.7 kJ 3) 431.3 kJ 4) 9.799 kJ

20. When 1.75 g of CaCl2 dissolves in 125 g of water in a coffee-cup calorimeter, the temperature increases by

2.44ºC. What is the heat change per mole of CaCl2 dissolved? Assume that all the heat is absorbed by the

water. 1)11.3 kJ/mol of CaCl2 2)1.13 kJ/mol of CaCl2 3)729 kJ/mol of CaCl2 4)80.9 kJ/mol of CaCl2

21. 5.5 g of sodium hydroxide is dissolved in 175 mL of water. Using a coffee-cup calorimeter, the temperature change

of the water is measured to be +2.10C. The specific heat capacity of water is 4.184

J/gC. What is the thermochemical equation for this process? 1) NaOH(s) NaOH(aq) 1.54 kJ 2) NaOH(s)1.54 kJNaOH(aq) 3) NaOH(s) NaOH(aq) 11.2 kJ 4) NaOH(s) 11.2 kJNaOH(aq)

22. A mass of 100.0 g of dilute hydrochloric acid is placed in a coffee cup calorimeter. The tempera-

ture of the solution is recorded to be 17.5C. A piece of magnesium ribbon of mass 0.601 g is

placed in the solution and the final temperature is recorded to be 39.6C. Calculate the molarenthalpy change for this reaction : Mg(s)2 HCl(a q)MgCl2(aq)H2( g) 1) + 15.4 kJ/mol of Mg 2) - 263 kJ/mol of Mg 3) - 374 kJ/mol of Mg 4) + 5.56 kJ/mol of Mg 23. Which process isn ot endothermic? 1) H2O(s) H2O(l) 2) 2H2O(g) 2H2(g) + O2(g) 3) H2O(g) H2O(l) 4) Al2O3 + 2Fe(l) 2Al + Fe2O3 24. Calculate theHo for the following reaction: 2ClF3(g) + 2NH3(g) N2(g) + 6HF(g) + Cl2(g) Given this information : ClF3(g)f Ho -261.0 kJ/mol HF(g) f Ho -271.1 kJ/mol NH3(g)f Ho -46.11 kJ/mol 1) -2241 kJ 2) - 1013 kJ 3) -578.2 kJ 4) 578.2 kJ 25. Given the thermochemical equation 2Al(s) + 3/2O2(g) Al2O3(s) ; Ho = -95.6 kJ determine Ho for the following reaction 2Al2O3(s) 4Al(s) + 3O2(g) 1) -95.6 kJ 2) +95.6 kJ 3) -47.8 kJ 4) +191.2 kJ 26. For the reaction: NH4NO3(s) N2(g) + 2H2O(g) + 1/2O2(g) Ho = -1.50 kJ/g NH4NO3 (molecular mass = 80.05 amu). If 0.105 g NH4NO3 decompose in

a bomb calorimeter with a heat capacity of 5.510 J/oC initially at 21.00oC, the final temperature of the calorimeter and its contents will be numerically (in K): 1) 223

2) 281 3) 323 4) 381 27. Calculate the molar enthalpy of combustion of propylene, C3H6(g), when it reacts with O2(g) to give CO2(g) and H2O(l). [ o

fH

(C3H6(g)) = +20.4 kJ/mol, o

fH

(CO2(g)) = -393.5 kJ/mole, o

fH

(H2O(l)) = -285.8 kJ/mole] 1) +699.7 kJ 2) -658.6 kJ 3) -1926.3 kJ 4) -2018 kJ 28. Using the following data: N2(g) + 3O2(g) + H2(g) 2HNO3(aq) H = -414.8 kJ N2O5(g) + H2O(g) 2HNO3(aq) H = 218.4 kJ 2H2O(g) 2H2(g) + O2(g) H = 483.6 kJ Determine H for the reaction: 2N2O5(g) 2N2(g) + 5O2(g) 1)149.6 kJ 2) 90.8 kJ 3) -876.4 kJ 4) 782.8 kJ 29. Calculate the standard molar enthalpy of formation of CO2(g) in the reaction: C(s) + O2(g) CO2(g) given the following standard enthalpy changes: 2C(s) + O2(g) 2CO(g) Ho = -221.0 kJ 2CO(g) + O2(g) 2CO2(g) Ho = -566.0 kJ 1) -393.5 kJ 2) +393.5 kJ 3) +787.0 kJ 4) -787.0 kJ 30. Calculate theHo for the reaction: S(s) + O2(g) SO2(g) Given the following data: S(s) + 3/2O2(g) SO3(g) Ho = -395.2 kJ/mol 2SO2(g) + O2 2SO3(g) Ho = -198.2 kJ/mol 1) 592.2 kJ/mol 2) -197.0 kJ/mol 3) -296.1 kJ/mol 4) -593 kJ/mol 31. Given the following data: 3/2O2(g) + 2B(s) B2O3(s) Ho = -1264 kJ/mol

O3(g) + 2B(s) B2O3(s) Ho = -1406 kJ/mol The change in enthalpy for the reaction converting oxygen (O2(g)) to ozone (O3(g)) at 298 K and 1 atm will be 3/2O2(g) O3(g) 1) -1406 kJ 2) -1264 kJ 3) -2670 kJ 4) +142 kJ 32. Given the following data N2(g) + O2(g) 2NO(g) H = 180.8 kJ 1/2N2(g) + O2(g) NO2(g) H = 33.9 kJ calculate the enthalpy change for the following reaction NO(g) + 1/2O2(g) NO2(g) 1) 214.7 kJ 2) 146.9 kJ 3) -56.5 kJ 4) 56.5 kJ 33. Calculate the enthalpy change, Ho, for the combustion of benzene, C6H6, given the following C6H6(l) + 15/2O2(g) 6CO2(g) + 3H2O(l) o

fH

values in kJ/mol are as follows: C6H6(l) = 49.0; CO2(g) = -393.5; H2O(l) = -285.8 1) -3169.4 kJ 2) 3267.4 kJ 3) -728.3 kJ 4) -3267.4 kJ 34. Calculate Ho for the reaction: Na2O(s) + SO3 (g) Na2SO4(s) given the following: (1) Na(s) + H2O(l) NaOH(s) + 1/2 H2(g) Ho = – 146 kJ (2) Na2SO4 (s) + H2O(l) 2NaOH(s) + SO3(g) Ho = + 418 kJ (3) 2Na2O (s) + 2H2(g) 4Na(s) + 2H2O(l) Ho = + 259 kJ 1) + 823 kJ 2) – 581 kJ 3) – 435 kJ 4) + 531 kJ 35. Given equations (1) and (2), calculate the enthalpy change for equation (3). (1) Pb(s) + PbO2(s) + 2 SO3(g) 2 PbSO4(s) Ho = -775 kJ (2) SO3(g) + H2O(l) H2SO4(aq) Ho = -133 kJ (3) Pb(s)PbO2(s)2 H2SO4(aq) 2 PbSO4(s)2 H2O(l) 1) – 908 kJ 2) – 642 kJ 3) – 509 kJ 4) + 642 kJ 36. Given that : S(s) + O2(g) SO2(g) Hº = – 296.8 kJ/mol 2 SO3(g) 2 SO2(g) + O2(g) Hº = + 197.8 kJ/mol Determine the enthalpy change of the reaction: 2 S(s) + 3 O2(g) 2 SO3(g)

1) – 99 kJ 2) – 495 kJ 3) + 495 kJ 4) – 791.4 kJ 37. Use the following reactions to calculateH° for: 4 HCl (g) + O2(g) 2 Cl2(g) + 2 H2O(g) H2(g) + Cl2 (g) 2 HCl(g) H° = – 185.0 kJ 2 H2(g) + O2(g) 2 H2O(g) H° = – 483.7 kJ 1) +113.7 kJ 2) –298.7 kJ 3) +298.7 kJ 4) –113.7 kJ 38. Given equations (I), (II), and (III), calculate the standard enthalpy of formation of acetylene, C2H2, as shown in equation (IV). (I) C (s) + O2 (g) CO2 (g) Hº = – 393.5 kJ (II) H2 (g) + ½ O2 (g) H2O (l) Hº = – 285.8 kJ (III) 2 C2H2 (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (l) Hº = – 2598.8 kJ (IV) 2 C (s) + H2 (g) C2H2 (g) 1) 253.6 kJ 2) 226.6 kJ 3) 1801 kJ 4) -3278 kJ

39) A volume of 50.0 mL of 0.5 M NaOH was added to 20.0 mL of 0.5 M H2SO4 in a calorimeter whose heat capacity

is 39.0 J/K. The temperature of the resulting solution increased by 3.6 °C. The standard enthalpy of

neutralization of H2SO4(aq) with NaOH(aq) will be 1) 13.7 kJ mol-1 2) 53.7 kJ mol-1 3) 5.37 kJ mol-1 4) 23.7 kJ mol-1 40) The standard enthalpy of neutralization of acetic acid with sodium hydroxide will be 1) =53.7 kJ mol-1 2) <53.7 kJ mol-1 3) >53.7 kJ mol-1 4) None Hint: When weak acids or bases participate in the neutralisation, some amount of heat is required for their complete dissociation. Hence the amount of heat liberated during neutralisation will be less than 53.7 kJ mol-1.

41) The heat of neutralisation of a strong acid and a strong alkali is 57.0 kJ mol-1. The heat released when 0.5 mole of HNO3 solution is mixed with 0.2 mole of KOH is 1) 57.0 kJ 2) 11.4 kJ 3) 28.5 kJ 4) 34.9 kJ 42) If the bond dissociation energies of XY, X2 and Y2 are in the ratio of 1:2:0.5 and standard heat of formation of XY is -100 kJ mol-1, then the bond dissociation energy of Y2 will be 1) 400 kJ mol-1 2) 100 kJ mol-1 3) 200 kJ mol-1 4) 800 kJ mol-1 Note: Key to the questions and updates, if any, can be downloaded from

http://groups.google.com/group/adichemadi