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Chapter 4 The Gaseous State NO 2
26

Chapter 4 The Gaseous State

Apr 03, 2022

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Page 1: Chapter 4 The Gaseous State

Chapter 4 The Gaseous State

NO2

Page 2: Chapter 4 The Gaseous State

AIR

Page 3: Chapter 4 The Gaseous State

Chemistry of Gases

SO3 .. corrosive gas

SO2…burning of sulfur in air

Page 4: Chapter 4 The Gaseous State

Pressure & Temperature

For 76cm of Hg, P = ρgh = 1.01325 x 105 kg m-1s-2

1 atm ≡ 1.01325 x 105 Pa

Page 5: Chapter 4 The Gaseous State

1662 Boyle “The Spring of the Air and its Effects”PV = C (fixed T & amount of gas)

~1760 CharleThe definition of the Temperature

All gases expand with increasing temperature by the same extent.t = c [(V/V0)-1] for all gases

1802 Gay-Lussac reproted c of 267 oClater refined to be c = 273.15

V = V0[1+(t/273.15oC)]

Kelvin T = 273.15 + t(Celsius)

Page 6: Chapter 4 The Gaseous State

Boyle’s Law

• The stirling engine, a heat engine invented by a Scottish minister, has been considered for use in automobile engines because of itsefficiency. In such an engine, a gas goes through a four-step cycle of (1) expansion at constant T, (2) cooling at constant V, (3) compression at constant T to its original volume, and (4) heating at constant V to its original temperature. Suppose the gas start at a pressure of 1.23 atm and the volume of the gas changes from 0.350 L to 1.31 L during its expansion at constant T. Calculated the pressure of the gas at the ene of this step in the cycle.

Page 7: Chapter 4 The Gaseous State

Charels’s Law

• A gas originally at a temperature of 26.5 °C is cooled at constant pressure. Its volume decreases from 5.40 L to 5.26 L. Determine its new temperature in degrees Celsius.

Page 8: Chapter 4 The Gaseous State

The Ideal Gas Law

PV = nRTR is a universal gas constant

R = 0.082058 L atm mol-1 K-1

=8.3145 J mol-1K-1

Page 9: Chapter 4 The Gaseous State

The Ideal Gas Law

• Iodine heptafluoride (IF7) can be made at elevated temperatures by the reaction

I2(g) + 7 F2(g) 2 IF7(g)Suppose 63.6 L of gaseous product is made by this

reaction at 300°C and a pressure of 0.459 atm. Calculated the volume this gas will occupy if heated to 400°C at a pressure of 0.980 atm.

Page 10: Chapter 4 The Gaseous State

The ideal gas law

• The pressure of a poisonous gas inside a sealed container is 1.47 atm at 20°C. If the brometic pressure is 0.96 atm, to what temperature (in degree Celsius) must the container and its contents be cooled so that the container can be opened with no risk of gas spurting out?

Page 11: Chapter 4 The Gaseous State

Mixtures of Gases

Partial pressurePA = nA [RT/V]

XA = nA/ntotPA = XAPtot

NOTE Ex. 4.6 (p.107)

Page 12: Chapter 4 The Gaseous State

The Kinetic Theory of GasesEmpirical ideal gas law can be explained by the kinetic theory of gases

Assumptions1. Pure gas consists of identical molecules separated by great distances2. Constantly moving in random directions with a speed distribution3. No forces between collisions – straight lines with constant velocities4. The collisions are elastic

Page 13: Chapter 4 The Gaseous State

Momentum change per collision

Δpx,mol = m(-vx)-mvx = -2 mvx = -Δpx,wall

Time between collisions to the original faceΔt = [2l/vx]The momentum transferred per second[Δpx,wall/ Δt] = (mvx

2)/l

For N molecules of mass m,F = (mvx1

2)/l + (mvx22)/l +….. = [Nm/l] vx

2

Page 14: Chapter 4 The Gaseous State

P = F/A = [Nm/Al]vx2

PV = Nmvx2

Since u2 = vx2 + vy

2 + vz2 = 3vx

2

PV = ⅓Nmu2, u2 is the mean-square speedFrom PV=nRT, ⅓Nmu2 = nRT, ⅓N0mu2 = RTKinetic energy per mole = ½N0mu2 = (3/2)RTOru2 = [3RT]/M, molar mass

Another look at T

Page 15: Chapter 4 The Gaseous State

Distribution of Molecular SpeedsRoot-mean square speed

Urms = (u2)½ = (3RT/M)½

Thermal equilibrium is a prerequisite for the definition ofTemperature

Page 16: Chapter 4 The Gaseous State
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Effusion1846 Graham’s law of effusion

Page 22: Chapter 4 The Gaseous State

Frequency of Molecular Collisions – Mean Free path - Diffusion

Page 23: Chapter 4 The Gaseous State

Real Gases: Intermolecular Forces

Van der Waals Equation

[P+a(n2/V2)][V‐nb] = nRT

Page 24: Chapter 4 The Gaseous State
Page 25: Chapter 4 The Gaseous State
Page 26: Chapter 4 The Gaseous State

Intermolecular Forces

Lennard-Jones Potential

VLJ(R) =4ε[(σ/R)12 – (σ/R)6]