Question 6 of 6: Isothermal Compression of Non-Ideal Ethylene

Question 6: Isothermal Compression of Non-Ideal Ethylene (Part B)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $n = 10\ \text{mol}$; $T = 298.15\ \text{K}$; initial $(P_1, V_1) = (21.7\ \text{bar},\ 1000\ \text{cm}^3/\text{mol})$; final $V_2 = 100\ \text{cm}^3/\text{mol}$; ethylene $T_c = 282.3\ \text{K}$, $P_c = 50.4\ \text{bar}$.

Find. (a) the final pressure $P_2$; (b) the work of compression.

Approach. Identify the EOS by back-substituting the stated initial state (van der Waals reproduces 21.7 bar exactly), then evaluate $P_2$ and integrate the reversible isothermal work with the vdW $P(V)$.

1. van der Waals constants and EOS check. With $R = 83.14\ \text{cm}^3\!\cdot\!\text{bar/(mol}\cdot\text{K)}$: $$a = \frac{27R^2T_c^2}{64P_c} = 4.611\times10^{6}\ \tfrac{\text{cm}^6\cdot\text{bar}}{\text{mol}^2}, \qquad b = \frac{RT_c}{8P_c} = 58.21\ \tfrac{\text{cm}^3}{\text{mol}}$$ $$P_1 = \frac{RT}{V_1-b} - \frac{a}{V_1^2} = \frac{24788}{941.79} - 4.611 = 26.32 - 4.61 = 21.7\ \text{bar}\ \checkmark$$
2. (a) Final pressure. At $V_2 = 100\ \text{cm}^3/\text{mol}$: $$P_2 = \frac{24788}{100-58.21} - \frac{4.611\times10^{6}}{100^2} = 593.2 - 461.1 = \boxed{132\ \text{bar}}$$
3. (b) Reversible isothermal work. Work done on the gas is $-\!\int_{V_1}^{V_2} P\,dV$; integrating the vdW $P(V)$: $$W_{\text{on}} = -\left[RT\ln\frac{V_2-b}{V_1-b} + a\!\left(\frac{1}{V_2}-\frac{1}{V_1}\right)\right] = -\big[24788\ln(0.04437) + 4.611\times10^{6}(0.009)\big]$$ $$W_{\text{on}} = -[-77234 + 41500] = 35734\ \tfrac{\text{cm}^3\cdot\text{bar}}{\text{mol}} = 3573\ \text{J/mol}, \qquad W = 10\,(3573) = \boxed{35.7\ \text{kJ}}$$ The attractive term $a(1/V_2-1/V_1)$ returns 41.5 kJ/mol on compression, so the required work is well below the ideal-gas estimate.