Answer to Question #89306 in Molecular Physics | Thermodynamics for Shivam Nishad

Question

Derive the expression for the work done by one mole of van der Waals gas during the isothermal expansion from volume V1, to V2 at temperature T.

Answer 1

Van der Waals equation of state for one mole of a real gas is

$$\left( P+\frac{a}{V_{m}^{2}} \right)\left( {{V}_{m}}-b \right)=RT$$

where V_m is the molar volume of the gas, R is the universal gas constant, T is temperature, P is pressure, a and b are the Van der Waals' constants

Find $P\left( {{V}_{m}} \right)$

$$P\left( {{V}_{m}} \right)=\frac{RT}{{{V}_{m}}-b}-\frac{a}{V_{m}^{2}}$$

Work done by one mole of gas during the isothermal expansion from volume V₁, to V₂ at temperature T is

$$W=\int\limits_{{{V}_{_{1}}}}^{{{V}_{2}}}{P\left( {{V}_{m}} \right)d{{V}_{m}}}=\int\limits_{{{V}_{_{1}}}}^{{{V}_{2}}}{\left( \frac{RT}{{{V}_{m}}-b}-\frac{a}{V_{m}^{2}} \right)d{{V}_{m}}}$$

$$W=\left. \left( RT\ln \left( {{V}_{m}}-b \right)+\frac{a}{{{V}_{m}}} \right) \right|_{{{V}_{1}}}^{{{V}_{2}}}$$

 

$$W=RT\left( \ln \left( {{V}_{2}}-b \right)-\ln \left( {{V}_{1}}-b \right) \right)+\frac{a}{{{V}_{2}}}-\frac{a}{{{V}_{1}}}$$

$$W=RT\ln \frac{{{V}_{2}}-b}{{{V}_{1}}-b}+a\frac{{{V}_{1}}-{{V}_{2}}}{{{V}_{1}}{{V}_{2}}}$$