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.

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}$

Interpretation Introduction

(a)

Interpretation:

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}$ 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

${\left(\partial p|\partial V\right)}_{T,n }$ 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;

$\left(p+ \frac{a{n}^2}{V^2}\right) \left(V-nb\right)=nRT$…(2)

In the above equation,

$\left(p+ \frac{a{n}^2}{V^2}\right)$ Correction term introduced for molecular attraction

$\left(V-nb\right)$ 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,

$p= \frac{nRT}{\left(V-nb\right)}- \frac{a{n}^2}{V^2}$…(3)

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

$\left(\frac{\partial p}{\partial V}\right)= \frac{-nRT}{{\left(V-nb\right)}^2}- \frac{2a{n}^2}{V^3}$…(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,

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}= \frac{-1.0\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{x}\mathrm{ }0.0820\mathrm{ }\frac{\mathrm{L}.\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}}{\mathrm{K}.\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}\mathrm{ }\mathrm{x}\mathrm{ }298\mathrm{ }\mathrm{K}}{{\left(25\mathrm{ }\mathrm{L}-1.0\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{x}\mathrm{ }0.0428\mathrm{ }\frac{\mathrm{L}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)}^2}+\mathrm{ }\frac{2\mathrm{ }\mathrm{x}\mathrm{ }2.253\mathrm{ }\frac{\mathrm{a}\mathrm{t}\mathrm{m}.\mathrm{ }{\mathrm{L}}^2}{{\mathrm{m}\mathrm{o}\mathrm{l}}^2}\mathrm{ }\mathrm{x}\mathrm{ }1.0\mathrm{ }{\mathrm{m}\mathrm{o}\mathrm{l}}^2}{{\left(25.0\mathrm{ }\mathrm{L}\right)}^3}$

$=\mathrm{ }\frac{-24.525\mathrm{ }\mathrm{L}.\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{ }}{{624.998\mathrm{ }\mathrm{L}}^2}+\mathrm{ }\frac{4.50\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{ }.\mathrm{ }{\mathrm{L}}^2}{{15625\mathrm{ }\mathrm{L}}^3}$

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}= -0.0395\frac{atm}{L}$

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

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}= -\frac{nRT}{V^2}$ …(5)

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

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}=\mathrm{ }\frac{-1.0\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{x}\mathrm{ }0.0820\mathrm{ }\frac{\mathrm{L}.\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}}{\mathrm{K}.\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}\mathrm{ }\mathrm{x}\mathrm{ }298\mathrm{ }\mathrm{K}}{{\left(25\mathrm{ }\mathrm{L}\right)}^2}$

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}=\mathrm{ }-0.0395\mathrm{ }\frac{atm}{L}$

Conclusion

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}$ 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

Interpretation Introduction

(b)

Interpretation:

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}$ 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

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}$ 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;

$\left(p+ \frac{a{n}^2}{V^2}\right) \left(V-nb\right)=nRT$ … (2)

In the above equation,

$\left(p+ \frac{a{n}^2}{V^2}\right)$ Correction term introduced for molecular attraction

$\left(V-nb\right)$ 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,

$p= \frac{nRT}{\left(V-nb\right)}- \frac{a{n}^2}{V^2}$ … (3)

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

$\left(\frac{\partial p}{\partial V}\right)= \frac{-nRT}{{\left(V-nb\right)}^2}- \frac{2a{n}^2}{V^3}$ …(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,

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}= \frac{-1.0\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{x}\mathrm{ }0.0820\mathrm{ }\frac{\mathrm{L}.\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}}{\mathrm{K}.\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}\mathrm{ }\mathrm{x}\mathrm{ }1000\mathrm{ }\mathrm{K}}{{\left(250\mathrm{ }\mathrm{L}-1.0\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{x}\mathrm{ }0.0428\mathrm{ }\frac{\mathrm{L}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)}^2}+\mathrm{ }\frac{2\mathrm{ }\mathrm{x}\mathrm{ }2.253\mathrm{ }\frac{\mathrm{a}\mathrm{t}\mathrm{m}.\mathrm{ }{\mathrm{L}}^2}{{\mathrm{m}\mathrm{o}\mathrm{l}}^2}\mathrm{ }\mathrm{x}\mathrm{ }1.0\mathrm{ }{\mathrm{m}\mathrm{o}\mathrm{l}}^2}{{\left(250\mathrm{ }\mathrm{L}\right)}^3}$

$=\mathrm{ }\frac{-82.3\mathrm{ }\mathrm{L}.\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{ }}{{62499.99\mathrm{ }\mathrm{L}}^2}+\mathrm{ }\frac{4.50\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{ }.\mathrm{ }{\mathrm{L}}^2}{{15625\mathrm{ }\mathrm{L}}^3}$

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}= -0.0013\mathrm{ }\frac{\mathrm{a}\mathrm{t}\mathrm{m}}{\mathrm{L}}$

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

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}= -\frac{nRT}{V^2}$ …(5)

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

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}=\mathrm{ }\frac{-1.0\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{x}\mathrm{ }0.0820\mathrm{ }\frac{\mathrm{L}.\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}}{\mathrm{K}.\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}\mathrm{ }\mathrm{x}\mathrm{ }1000\mathrm{ }\mathrm{K}}{{\left(250\mathrm{ }\mathrm{L}\right)}^2}$

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}= -0.0013\frac{\mathrm{a}\mathrm{t}\mathrm{m}}{\mathrm{L}}$

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

Conclusion

${\left(\frac{\partial p}{\partial V}\right)}_{T,n}$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.