Discussion
Methane has compressibility factor value of 0.9 at reduced pressure of 1.0 and reduced temperature of 1.5. For propane, the critical temperature and pressure are 369.8 K and 42.48 bar, respectively. Take R = 8.314 kJ/kmol.K. Applying the principle of corresponding states, the molar volume of propane (in m3/kmol) at the same reduced pressure and temperature is
Answer: D
The molar volume of propane is 0.977 m3/kmol.
By the principle of corresponding states, gases at the same reduced pressure and temperature have the same compressibility factor, so propane also has Z = 0.9. Its actual conditions are
\[ \begin{aligned} T &= 1.5 \times 369.8 = 554.7\ \text{K} \\ P &= 1.0 \times 42.48\ \text{bar} = 4248\ \text{kPa} \end{aligned} \]
\[ \begin{aligned} \bar{v} &= \frac{ZRT}{P} = \frac{0.9 \times 8.314 \times 554.7}{4248} \\ &= 0.977\ \text{m}^{3}\text{/kmol} \end{aligned} \]
Treating propane as an ideal gas (Z = 1) would give 1.086, and 0.355, 0.526 and 0.791 do not follow from Z = 0.9 at these conditions.
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