A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial…

A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_i$ $T_i$ (in kelvin) and the final temperature is $a T_i$, the value of $a$ is

Solution

In adiabatic process, $ \begin{aligned} & T V^{\gamma-1}=\text { constant } \\ & \therefore T_i V_i^{0.4}=T_f V_f^{0.4} \\ & \text { (as } \gamma=1.4 \text { for diatomic gas) } \\ & \text { or } \quad T_i V_i^{0.4}=\left(\alpha T_i\right)\left(\frac{V_i}{32}\right)^{0.4} \\ & \text { or } \quad \alpha(32)^{0.4}=4 \\ & \end{aligned} $ $\therefore$ The correct answer is 4

Asked in: JEE Advanced 2010 (Paper 2)

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