Chem 163B first midterm review - University of California ...chen.chemistry.ucsc.edu/163Breview01TAs.pdfEXERCISE •P 2.17 A vessel containing 1.50 mole of an ideal gas with 2 Ü=1.00

CHEM 163B THERMODYNAMICS

1ST MIDTERM REVIEWJia Lu (Gabby) [email protected] Hour: Monday 3:00-4:00pm @ PSB 145Gabe Mednick [email protected] Hour: Wednesday 1:00-2:00pm @ PSB 145

TRUE / FALSE

• Real gas can be approximated to be ideal gas at high temperature and high pressure.• A can of soda is an isolated system.• If the pressure of a real gas is calculated (using van der Waals equation) to be larger

than ideal gas (using PV=nRT), the interaction between this real gas molecules is dominated by attraction.

• For systems undergo isothermal expansion from the same initial state and to reach the same final state, the work done by reversible process is always larger than irreversible process.

• Ideal gas undergoes expansion into vacuum environment, work is done by the system.

• P and T are intensive parameters / independent of the size of a given system.

TRUE / FALSE

• Real gas can be approximated to be ideal gas at high temperature and high pressure. (F : low pressure)

• A can of soda is an isolated system. (F : closed)• If the pressure of a real gas is calculated (using van der Waals equation) to be larger

than ideal gas (using PV=nRT), the interaction between this real gas molecules is dominated by attraction. (F : pressure will decrease if attraction is dominant)

• For systems undergo expansion from the same initial state and to reach the same final state, the work done by reversible process is always larger than irreversible process. (True)

• Ideal gas undergoes expansion into vacuum environment, work is done by the system. (F : P=0 for vacuum, no work is done)

• P and T are intensive parameters / independent of the size of a given system. (True)

TRUE / FALSE

• First Law of Thermodynamics states that: if two systems are separately in equilibrium with a third system, these two systems are in equilibrium with each other.

• For adiabatic processes, there is no heat transfer (∆ 0).• For isenthalpic process, internal energy is kept constant.• For ideal gas undergoes isothermal processes, ∆ and ∆ are constant.• Hess’s Law defines that: reaction enthalpy is the sum of enthalpy of all the reaction

steps.• If 0, repulsive part of the potential is dominant.

TRUE / FALSE

• The First Law of Thermodynamics states that: if two systems are separately in equilibrium with a third system, these two systems are in equilibrium with each other. (F : Zeroth Law)

• For adiabatic processes, there is no heat transfer (∆ 0). (True)• For isenthalpic process, internal energy is kept constant. (F : constant enthalpy)• For ideal gas undergoes isothermal processes, ∆ and ∆ are constant. (True : ∆∆ 0)

• Hess’s Law defines that: reaction enthalpy is the sum of enthalpy of all the reaction steps. (True)

• If 0, repulsive part of the potential is dominant. (F: If 0, attractive)

EXERCISE• 1 mole of monatomic ideal gas undergoes an adiabatic irreversible compression,

from =1 atm, =300 K, = 24.62 L, against a constant external pressure =2 atm to a final volume =17.2332 L; i.e the external pressure is instantaneously raised to 2 atm and the gas is compressed. For this process, calculate

, , ∆ , ≡ , ∆ .

EXERCISE• 1 mole of monatomic ideal gas undergoes an adiabatic irreversible

compression, from =1 atm, =300 K, = 24.62 L, against a constant external pressure =2 atm to a final volume =17.2332 L; i.e the external pressure is instantaneously raised to 2 atm and the gas is compressed. For this process, calculate , , ∆ , ≡ , ∆ .

• Adiabatic compression, q=0

• ∆ 1.50

• Monoatomic ideal gas,

• ∆ ∆ 120 , ∆ 420 or T

• ∆ ∆ ∆ ∆ 2.49

Cannot use

because T is not constant

EXERCISE

• P 2.17 A vessel containing 1.50 mole of an ideal gas with =1.00 bar and , = 5R/2 is in thermal contact with a water bath. Treat the vessel, gas, and water bath as being in thermal equilibrium, initially at 298K, and as separated by adiabatic walls from the rest of the universe. The vessel, gas, and water bath have an average heat capacity of = 2450 J/K. The gas is compressed reversibly to = 20.5 bar. What is the temperature of the system after thermal equilibrium has been established?

EXERCISE

• P 2.17 A vessel containing 1.50 mole of an ideal gas with =1.00 bar and , = 5R/2 is in thermal contact with a water bath. Treat the vessel, gas, and water bath as being in thermal equilibrium, initially at 298K, and as separated by adiabatic walls from the rest of the universe. The vessel, gas, and water bath have an average heat capacity of = 2450 J/K. The gas is compressed reversibly to = 20.5 bar. What is the temperature of the system after thermal equilibrium has been established? Isothermal reversible compression:

• ln ln 11.2

Adiabatic walls: 0, ∆ ∆

• ∆ ∆ 3.05, 301

EXERCISE

• The oxidation of and reaction with water droplets in the air is one of the processes related to acid rain. For the following reaction carried out at 298K (assume gasses are ideal):

12 →

• What is ∆ ° at 298K[∆ , ,° 814.0 , ∆ , ,

° 285.8 , and ∆ , ,

° 296.8 ?

• What is ∆ ° at 298K?• Assuming that heat capacities are independent of temperature, what is ∆ ° at 398K

[ ̅ , , 29.4.

, ̅ , , 39.9.

, ̅ , , 75.3.

, and ̅ , ,138.9

.] ?

EXERCISE

• The oxidation of and reaction with water droplets in the air is one of the processes related to acid rain. For the following reaction carried out at 298K(assume gasses are ideal):

12 →

• What is ∆ ° at 298K[∆ , ,° 814.0 , ∆ , ,

° 285.8 , and ∆ , ,

° 296.8 ?

• What is ∆ ° at 298K?• Assuming that heat capacities are independent of temperature, what is ∆ ° at 398K

[ ̅ , , 29.4.

, ̅ , , 39.9.

, ̅ , , 75.3.

, and ̅ , ,138.9

.] ?

∆ ° ∆ ,° ∆ ,

° 231.3

∆ ° ∆ ° ∆ 227.6

∆ ° ∆ ° ∆ 230.4

EXERCISE

• For a van der Waals gas,

• What is ?

• Knowing , which holds for all substances (with 0; dn 0 , when dV=0, ∆ ̅ ∆ . However, for ideal gas, ∆ ̅ ∆ , even if volume is not constant (i.e. even when dV 0 in the equation above.) why?

• For an ideal gas, explain:

EXERCISE

• For a van der Waals gas,

• What is ?

• Knowing , which holds for all substances (with 0; dn 0 , when dV=0, ∆ ̅ ∆ . However, for ideal gas, ∆ ̅ ∆ , even if volume is not constant (i.e. even when dV 0 in the equation above.) why?

• For an ideal gas, explain:

For ideal gas, U only depend on T. If U is constant, T is constant.

For ideal gas (point mass & no interactions), U only depend on T, 0Also need ̅ to be constant,

∆ ̅ dT n ̅ ∆T

EXERCISE

• Joule-Thompson cooling of a gas occurs when the pressure of the gas drops as it goes through a porous plug separating a high pressure chamber from one at 1 atm. The process is carried out at constant enthalpy. The appropriate measure of the

cooling is given by , show that or ̅

EXERCISE

• Joule-Thompson cooling of a gas occurs when the pressure of the gas drops as it goes through a porous plug separating a high pressure chamber from one at 1 atm. The process is carried out at constant enthalpy. The appropriate measure of the

cooling is given by , show that or ̅

0

1

EXERCISE

• P2.28 A 3.50 mole sample of an ideal gas with , 3 /2 is expanded adiabatically against a constant external pressure of 1.45 bar. The initial temperature and pressure are 310 and 15.2 . The final pressure is

1.45 . Calculate , , ∆ ∆ for the process.

EXERCISE

• P2.28 A 3.50 mole sample of an ideal gas with , 3 /2 is expanded adiabatically against a constant external pressure of 1.45 bar. The initial temperature and pressure are 310 and 15.2 . The final pressure is

1.45 . Calculate , , ∆ ∆ for the process.

Using or

No, γ is for reversible adiabatic processes only!

EXERCISE

• P2.28 A 3.50 mole sample of an ideal gas with , 3 /2 is expanded adiabatically against a constant external pressure of 1.45 bar. The initial temperature and pressure are 310 and 15.2 . The final pressure is

1.45 . Calculate , , ∆ ∆ for the process.

• Adiabatic expansion, q 0, ∆• ,

• Ideal gas, PV=nRT

• ,

•,

,

∆ , ∆∆ ∆ ∆