Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Textbook Question
Chapter 1, Problem 1.51E

Numerically evaluate for one mole of methane acting as a van der Waals gas at (a) T = 298 K and V = 25.0 L and (b) T = 1000 K and V = 250.0 L. Comment on which set of conditions yields a number closer to that predicted by the ideal gas law.

p V T , n

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

pVT,n is to be evaluated for one mole of methane acting as a van der Waals gas at T = 298 K and V = 25.0 L.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Answer to Problem 1.51E

pVT,n  for one mole of methane acting as a van der Waals gas at T = 298 K and V = 25.0 L is calculated as -0.0395 atm/L

Explanation of Solution

The ideal gas equation can be represented as;

PV = nRT …(1)

Notably, the van Waals equation improves the ideal gas law by adding two significant terms in the ideal gas equation: one term is to account for the volume of the gas molecules and another term is introduced for the attractive forces between them. The non-ideal gas equation represented as;

p+ an2V2 V-nb=nRT…(2)

In the above equation,

p+ an2V2 Correction term introduced for molecular attraction

V-nb correction term introduced for volume of molecules

a’ and ‘b’ are called as van der Waals constants

Rearranging equation (2) we get pressure p of real gas as,

p= nRTV-nb- an2V2…(3)

On differentiation with respect to volume V, at constant T, n we get

pV= -nRTV-nb2- 2an2V3…(4)

Given for methane,

Number of moles = n = 1 mole

Temperature of gas = T = 298 K

Volume of gas = V = 25.0 L

Value of constant ‘a’ for methane = 2.253atmL2/mol2

Value of constant ‘b’ for methane = 0.0428 L/mol

Substituting the values in equation (4), we get,

pVT,n= -1.0 mol x 0.0820 L. atmK. mol x 298 K25 L-1.0 mol x 0.0428 Lmol2+ 2 x 2.253 atm. L2mol2 x 1.0 mol225.0 L3

                = -24.525 L. atm 624.998 L2+ 4.50 atm . L215625 L3

pVT,n= -0.0395atmL

Besides, differentiating the equation (1) for ideal gas with respect to volume V, we get

pVT,n= -nRTV2 …(5)

Substituting the given parameters for methane in equation (5), we get for ideal gas

pVT,n= -1.0 mol x 0.0820 L. atmK. mol x 298 K25 L2

pVT,n= -0.0395 atmL

Conclusion

pVT,n for one mole of methane acting as a van der Waals gas at T = 298 K and V =

25.0 L is calculated as -0.0395 atm/L

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

pVT,n is to be evaluated for one mole of methane acting as a van der Waals gas at

T = 1000 K and V = 250.0 L.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Answer to Problem 1.51E

pVT,n for one mole of methane acting as a van der Waals gas at T = 1000 K and V =

250.0 L is calculated as -0.0013 atm/L

Explanation of Solution

The ideal gas equation can be represented as;

PV = nRT … (1)

The non-ideal gas equation represented as;

p+ an2V2 V-nb=nRT … (2)

In the above equation,

p+ an2V2 Correction term introduced for molecular attraction

V-nb correction term introduced for volume of molecules

a’ and ‘b’ are called as van der Waals constants.

Rearranging equation (2) we get pressure P of real gas as,

p= nRTV-nb- an2V2 … (3)

On differentiation with respect to volume V, at constant T, n we get

pV= -nRTV-nb2- 2an2V3 …(4)

Given for methane,

Number of moles = n = 1 mole

Temperature of gas = T = 1000 K

Volume of gas = V = 250.0 L

Value of constant ‘a’ for methane = 2.253atmL2/mol2

Value of constant ‘b’ for methane = 0.0428 L/mol

Substituting the values in equation (4), we get,

pVT,n= -1.0 mol x 0.0820 L. atmK. mol x 1000 K250 L-1.0 mol x 0.0428 Lmol2+ 2 x 2.253 atm. L2mol2 x 1.0 mol2250 L3

                = -82.3 L. atm 62499.99 L2+ 4.50 atm . L215625 L3

pVT,n= -0.0013 atmL

Besides, differentiating the equation (1) for ideal gas with respect to volume V, we get

pVT,n= -nRTV2 …(5)

Substituting the given parameters for methane in equation (5), we get for ideal gas,

pVT,n= -1.0 mol x 0.0820 L. atmK. mol x 1000 K250 L2

pVT,n= -0.0013atmL

Non-ideal gas value is in close proximity to ideal gas values.

Conclusion

pVT,n for one mole of methane acting as a van der Waals gas at T = 1000 K and V = 250.0 L is calculated as -0.0013 atm/L. The given parameters are in close proximity to ideal gas equation.

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