An ideal gas at $27^{\circ}\mathrm{C}$ is compressed adiabatically to $1/32$ of its original volume.

Question

An ideal gas at $27^{\circ}\mathrm{C}$ is compressed adiabatically to $1/32$ of its original volume. The rise in temperature is (Take $\gamma = 1.4$)

Professor PC Thomas Classes · Answer

(b) : The initial temperature is $T_1 = 27^{\circ}\mathrm{C} = 300 \mathrm{~K}$. For an adiabatic process, $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$. Here, $\gamma - 1 = 1.4 - 1 = 0.4 = 2/5$. Thus, $T_2 = T_1 (V_1/V_2)^{2/5} = 300 	imes (32)^{2/5}$. Since $32 = 2^5$, we have $T_2 = 300 	imes (2^5)^{2/5} = 300 	imes 2^2 = 300 	imes 4 = 1200 \mathrm{~K}$. The rise in temperature is $\Delta T = T_2 - T_1 = 1200 - 300 = 900 \mathrm{~K}$.