A mole of oxygen, considered as a van der Waals gas, undergoes a reversible isothermal expansion at fixed temperature $T_0$ from an initial volume $V_i$ to a final volume $V_f$ . Determine the work $W_{if}$ performed on the van der Waals gas in terms of the parameters a, and b.

Numerical Application:

$T_0 = 273 K, V_i = 22.3 · 10^{−3} m^3, V_f = 3V_i, p_0 = 1.013 · 10^5 Pa, a = 0.14 Pa m^6$ and $b = 3.2 10^{−6} m^3$ .

Question

A mole of oxygen, considered as a van der Waals gas, undergoes a reversible isothermal expansion at fixed temperature $T_0$ from an initial volume $V_i$ to a final volume $V_f$ . Determine the work $W_{if}$ performed on the van der Waals gas in terms of the parameters a, and b.

Numerical Application:

$T_0 = 273 K, V_i = 22.3 · 10^{−3} m^3, V_f = 3V_i, p_0 = 1.013 · 10^5 Pa, a = 0.14 Pa m^6$ and $b = 3.2 10^{−6} m^3$ .

Numerical Application:

[latex]T_0 = 273 K, V_i = 22.3 · 10^{−3} m^3, V_f = 3V_i, p_0 = 1.013 · 10^5 Pa, a = 0.14 Pa m^6[/latex] and [latex]b = 3.2 10^{−6} m^3[/latex].

Accepted Answer

For one mole of oxygen at temperature [latex]T_0[/latex], the van derWaals equation of state (6.64) reads,

[latex] p = \frac{R T}{ u -b} - \frac{\alpha }{ u ^2} [/latex].

[latex] p = \frac{R T_0}{V -b} - \frac{\alpha }{ u ^2} [/latex].

Thus, the work is expressed as,

[latex] W_{if} = - \int_{V_i}^{V_f}{pdV} =-R T_0 \int_{V_i}^{V_f}{\frac{dV}{V - b}-\alpha }\int_{V_i}^{V_f}{\frac{dV}{V^2} } = - RT_0 \ln \Bigl(\frac{V_f -b}{V_i - b}\Bigr) -\alpha \Bigl(\frac{1}{V_f}-\frac{1}{V_i} \Bigr) -2.49.10^3 [/latex] J.

Question 6.10: A mole of oxygen, considered as a van der Waals gas, undergo...

Related Answered Questions