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