A sample of argon gas at $1 ext{ atm}$ pressure and $27^\circ ext{C}$ expands reversibly and adiab

Question

A sample of argon gas at $1	ext{ atm}$ pressure and $27^\circ	ext{C}$ expands reversibly and adiabatically from $1.25	ext{ dm}^3$ to $2.50	ext{ dm}^3$. Calculate the magnitude of the enthalpy change ($|\Delta H|$ in $	ext{J}$) for this process. (Given: $C_{V,m}$ for argon is $12.48	ext{ J K}^{-1}	ext{ mol}^{-1}$, $R = 8.314	ext{ J K}^{-1}	ext{ mol}^{-1}$, and $2^{2/3} = 1.587$. Report your answer to the nearest integer)

Professor PC Thomas Classes · Answer

For monoatomic argon gas, $C_{V,m} = 1.5R \approx 12.48	ext{ J K}^{-1}	ext{ mol}^{-1}$, so $\gamma = C_{P,m}/C_{V,m} = 5/3$. For a reversible adiabatic expansion, $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1} \implies T_2 = T_1 (V_1/V_2)^{2/3} = 300 	imes (1.25/2.50)^{2/3} = 300 	imes (0.5)^{2/3} = 300 / 1.587 = 189.04	ext{ K}$. The number of moles of gas is $n = P_1V_1 / (RT_1) = (101325	ext{ Pa} 	imes 1.25 	imes 10^{-3}	ext{ m}^3) / (8.314 	imes 300) = 126.66 / 2494.2 = 0.05078	ext{ mol}$. The enthalpy change is given by $\Delta H = n C_{P,m} (T_2 - T_1) = n (\gamma C_{V,m}) (T_2 - T_1) = 0.05078 	imes (5/3 	imes 12.48) 	imes (189.04 - 300) = 0.05078 	imes 20.8 	imes (-110.96) = -117.18	ext{ J}$. Thus, the magnitude $|\Delta H| \approx 117	ext{ J}$.