Chapter 2, Problem 2D.2AE

Interpretation Introduction

Interpretation: The value of $\Delta U_m$ for the isothermal expansion of nitrogen gas at $298\text{K}$ has to be stated. The values of $q$ and $w$ have to be stated.

Concept introduction: The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by $U$.  The division of internal energy, $U$ by the number of moles gives the molar internal energy.  The molar internal energy is denoted by $U_m$.  The change in molar internal energy is denoted by $\Delta U_m$.

Answer to Problem 2D.2AE

The value of $\Delta U_m$ for the isothermal expansion of nitrogen gas at $298\text{K}$ is $\underset{\_}{1.305atmd{m}^{3}mo{l}^{-1}}$.  The values of $q$ and $w$ are $\underset{\_}{1.305atmd{m}^{3}mo{l}^{-1}}$ and $-\underset{\_}{72.87d{m}^{3}atmmo{l}^{-1}}$  respectively.

Explanation of Solution

The given temperature of nitrogen gas is $298\text{K}$.

The initial volume of nitrogen gas is $\text{1}\text{.00}{\text{dm}}^{\text{3}}$.

The initial molar volume of nitrogen gas is $\text{1}\text{.00}{\text{dm}}^{\text{3}}{\text{mol}}^{-1}$.

The final volume of nitrogen gas is $\text{20}\text{.00}{\text{dm}}^{\text{3}}$.

The final molar volume of nitrogen gas is $\text{20}\text{.00}{\text{dm}}^{\text{3}}{\text{mol}}^{-1}$.

The value of universal gas constant is $0.08206{\text{dm}}^3\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The value of van der Waals parameter, a, for nitrogen gas is $1.37\text{atm}{\text{dm}}^{\text{6}}{\text{mol}}^{-2}$.

The value of van der Waals parameter, b, for nitrogen gas is $3.87\times {10}^{-2}{\text{dm}}^{\text{3}}{\text{mol}}^{-1}$.

If nitrogen gas is treated as the van der Waals gas then the expression for internal pressure is given as,

    ${\pi }_T=\frac{a}{V_m^2}$                                                                                                         (1)

Where,

• • ${\pi }_T$ is the internal pressure of nitrogen gas.
• • $a$ is the van der Waals parameter.
• • $V_m$ is the molar volume of nitrogen gas.

The another expression for showing the internal pressure in terms of change in molar internal energy divided by the change in molar volume at constant temperature is given as,

    ${\pi }_T=\left(\frac{\partial U_m}{\partial V_m}\right)$                                                                                                  (2)

Where,

• • $\partial U_m$ is the change in internal energy of nitrogen gas.
• • $\partial V_m$ is the change in molar volume of nitrogen gas.

The expression obtained by comparing the equations (1) and (2) is,

    $\left(\frac{\partial U_m}{\partial V_m}\right)=\frac{a}{V_m^2}$                                                                                                (3)

The molar internal energy as a function of molar volume and temperature is written as,

    $U_m=U_{{}_m}\left(T,V_m\right)$

The differential expression for the above equation is given as,

    $\text{d}{U}_m=\left(\frac{\partial U_m}{\partial T}\right)\text{d}T+\left(\frac{\partial U_m}{\partial V_m}\right)\text{d}{V}_m$

The expression for the change in molar internal energy is obtained at $\text{d}T=0$ for the isothermal expansion of nitrogen gas as,

    $\begin{array}{c}\Delta U_m=\left(\frac{\partial U_m}{\partial T}\right)0+\left(\frac{\partial U_m}{\partial V_m}\right)\text{d}{V}_m\\ \Delta U_m=\left(\frac{\partial U_m}{\partial V_m}\right)\text{d}{V}_m\end{array}$                                                                    (4)

Substitute the value of $\left(\frac{\partial U_m}{\partial V_m}\right)$ from equation (3) to equation (4) for integration.

    $\begin{array}{c}\Delta U_m={\displaystyle \underset{V_{m_1}}{\overset{V_{m_2}}{\int }}\frac{a}{V_m^2}}\text{d}{V}_m\\ \Delta U_m=-a\left(\frac{1}{V_{m_2}}-\frac{1}{V_{m_1}}\right)\end{array}$                                                                                   (5)

Where,

• • $\Delta U_m$ is the change in the internal energy of nitrogen gas.
• • $a$ is the van der Waals parameter.
• • $V_{m_2}$ is the final molar volume of nitrogen gas.
• • $V_{m_1}$ is the initial molar volume of nitrogen gas.

Substitute the values of van der Waals parameter, initial and final molar volume in equation (5).

    $\begin{array}{c}\Delta U_m=-1.37\text{atm}{\text{dm}}^{\text{6}}{\text{mol}}^{-2}\left(\frac{1}{\text{20}\text{.00}{\text{dm}}^{\text{3}}{\text{mol}}^{-1}}-\frac{1}{\text{1}\text{.00}{\text{dm}}^{\text{3}}{\text{mol}}^{-1}}\right)\\ =\underset{\_}{1.305atmd{m}^{3}mo{l}^{-1}}\end{array}$

Thus, the value of molar internal energy, $\Delta U_m$, for the isothermal expansion of nitrogen gas is $\underset{\_}{1.305atmd{m}^{3}mo{l}^{-1}}$.

The van der Waal’s gas equation is given as,

    $p=\frac{RT}{V_m-b}-\frac{a}{V_m^2}$                                                                                             (6)

Where,

• • $p$ is the pressure of the gas.
• • $V_m$ is the molar volume of the gas.
• • $T$ is the temperature of the gas.
• • $n$ is the number of moles of the gas.
• • $R$ is the universal gas constant.
• • $a$ and $b$ are Van der waal’s coefficients.

The work done equation is written by using the van der Waal’s gas equation as,

    $w=-{\displaystyle \underset{V_{m_1}}{\overset{V_{m_2}}{\int }}p\text{d}{V}_m}$                                                                                                 (7)

Substitute the value of pressure in equation (7).

    $\begin{array}{c}w=-{\displaystyle \underset{V_{m_1}}{\overset{V_{m_2}}{\int }}\left(\frac{RT}{V_m-b}-\frac{a}{V_m^2}\right)\text{d}{V}_m}\\ =-{\displaystyle \underset{V_{m_1}}{\overset{V_{m_2}}{\int }}\left(\frac{RT}{V_m-b}\right)\text{d}{V}_m}+{\displaystyle \underset{V_{m_1}}{\overset{V_{m_2}}{\int }}\left(\frac{a}{V_m^2}\right)\text{d}{V}_m}\end{array}$                                                            (8)

The value of $-{\displaystyle \underset{V_{m_1}}{\overset{V_{m_2}}{\int }}\left(\frac{RT}{V_m-b}\right)\text{d}{V}_m}$ is equals to heat given, $-q$, and the value of ${\displaystyle \underset{V_{m_1}}{\overset{V_{m_2}}{\int }}\left(\frac{a}{V_m^2}\right)\text{d}{V}_m}$ is equals to the molar internal energy.

Substitute the value $-{\displaystyle \underset{V_{m_1}}{\overset{V_{m_2}}{\int }}\left(\frac{RT}{V_m-b}\right)\text{d}{V}_m}$ and ${\displaystyle \underset{V_{m_1}}{\overset{V_{m_2}}{\int }}\left(\frac{a}{V_m^2}\right)\text{d}{V}_m}$ in equation (8)

    $w=-q+\Delta U_m$                                                                                                (9)

The value of heat given to the system is calculated as,

    $\begin{array}{c}q={\displaystyle \underset{V_{m_1}}{\overset{V_{m_2}}{\int }}\left(\frac{RT}{V_m-b}\right)\text{d}{V}_m}\\ =RT\ln \left(\frac{V_m{}_2-b}{V_m{}_1-b}\right)\end{array}$                                                                                     (10)

Substitute the value of universal gas constant, temperature , van der Waals coefficient, b, initial and final volume in equation (10).

    $\begin{array}{c}q=0.08206{\text{dm}}^3\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}\times 298\text{K}\times \ln \left(\frac{20-3.87\times {10}^{-2}}{1-3.87\times {10}^{-2}}\right)\\ =\underset{\_}{74.17d{m}^{3}atmmo{l}^{-1}}\end{array}$

Substitute the values of heat given, $q$ and the molar internal energy in equation (9).

    $\begin{array}{c}w=-\text{74}\text{.17}{\text{dm}}^{\text{3}}\text{atm}{\text{mol}}^{\text{-1}}+\text{1}\text{.305}\text{atm}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}\\ =-\underset{\_}{72.87d{m}^{3}atmmo{l}^{-1}}\end{array}$

Thus, the value of work done by the system for the expansion of nitrogen gas is $-\underset{\_}{72.87d{m}^{3}atmmo{l}^{-1}}$.