The second virial coefficient of methane can be approximate by the empirical equation: where , , and — States of Matter and Gaseous State Chemistry Question
Question
The second virial coefficient of methane can be approximate by the empirical equation: $B = a + b \cdot e^{-c/T^2}$ where $a = -0.2 \text{ bar}^{-1}$, $b = 0.22 \text{ bar}^{-1}$, and $c = 950 \text{ K}^2$. What is the value of the Boyle temperature of methane (in K)? ($\ln 1.1 = 0.095$)
💡 Solution & Explanation
Boyle temperature $T_B$ is the temperature where the second virial coefficient $B = 0$. So, $0 = a + b \cdot e^{-c/T_B^2} \implies e^{-c/T_B^2} = -a/b$. Substituting given values: $e^{-950/T_B^2} = -(-0.2) / 0.22 = 0.2 / 0.22 = 1/1.1$. Taking natural log on both sides: $-950/T_B^2 = \ln(1/1.1) = -\ln(1.1) = -0.095$. $950 / T_B^2 = 0.095 \implies T_B^2 = 950 / 0.095 = 10000$. Taking the square root gives $T_B = 100\text{ K}$. Formatted as a four digit integer, it is 0100. Therefore, correct answer is 0100.