The volume of an ideal gas increases 8 times and temperature becomes $(1 / 4)^{ ext {th }}$ of init

Question

The volume of an ideal gas increases 8 times and temperature becomes $(1 / 4)^{	ext {th }}$ of initial temperature during a reversible change. If there is no exchange of heat in this process $(\Delta \mathrm{Q}=0)$ then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

Professor PC Thomas Classes · Answer

For an adiabatic process ($\Delta Q = 0$), $T V^{\gamma-1} = 	ext{constant}$. We are given $V_2 = 8V_1$ and $T_2 = \frac{1}{4}T_1$. Therefore, $T_1 V_1^{\gamma-1} = \left(\frac{T_1}{4}ight)(8V_1)^{\gamma-1}$. This simplifies to $4 = 8^{\gamma-1}$, which means $2^2 = (2^3)^{\gamma-1} = 2^{3\gamma-3}$. Equating the exponents: $2 = 3\gamma - 3 \Rightarrow 3\gamma = 5 \Rightarrow \gamma = \frac{5}{3}$. A heat capacity ratio of $\frac{5}{3}$ corresponds to a monoatomic gas. Among the options, Helium (He) is monoatomic.