Discussion
A gas obeys the following equation of state:
\[ P(\bar{v} - b) = RT + \frac{aP^{2}}{T}, \]
where \(\bar{v}\) is molar volume, and a, b are constants with values \(a = 10^{-5}\) J.K/Pa2.kmol and \(b = 8 \times 10^{-2}\) m3/kmol. Take Cp = 30 kJ/kmol.K. At 10 bar and 500 K, the value of the Joule-Thomson coefficient (in K/Pa) is
Answer: B
The coefficient is \(-4\times10^{-6}\) K/Pa.
The Joule-Thomson coefficient follows from the equation of state through
\[ \begin{aligned} \mu &= \frac{1}{C_{p}}\Big[T\left(\frac{\partial\bar{v}}{\partial T}\right)_{P} \\ &\quad - \bar{v}\Big] \end{aligned} \]
Solve the equation of state for the molar volume and differentiate at constant pressure:
\[ \begin{aligned} \bar{v} &= b + \frac{RT}{P} + \frac{aP}{T} \\ \left(\frac{\partial\bar{v}}{\partial T}\right)_{P} &= \frac{R}{P} - \frac{aP}{T^{2}} \end{aligned} \]
\[ \begin{aligned} T\left(\frac{\partial\bar{v}}{\partial T}\right)_{P} &= \frac{RT}{P} - \frac{aP}{T} \\ T\left(\frac{\partial\bar{v}}{\partial T}\right)_{P} - \bar{v} &= -b - \frac{2aP}{T} \end{aligned} \]
With \(a = 10^{-5}\), \(P = 10^{6}\) Pa, \(T = 500\) K and \(b = 0.08\) m3/kmol:
\[ \begin{aligned} \mu &= -\frac{0.08 + 0.04}{30000} \\ &= -4\times10^{-6} \text{ K/Pa} \end{aligned} \]
The coefficient is negative, so this gas warms on throttling. The two positive options have the wrong sign, and dropping the factor 2 on the \(aP/T\) term would give \(-3.3\times10^{-6}\), which is not offered.
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