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A polyatomic gas follows a law $T^2 V^\alpha=$ constant. Find $\alpha$ for which the heat exchange of gas in the process becomes zero.
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Verified Answer
The correct answer is:
$\alpha=\frac{2}{3}$
The process in which there is no heat exchange of gas, it is called adiabatic process.
In this process, $T V^{\gamma-1}=k$ [constant $]$
Comparing Eqs. (i) and (ii), we get
$$
\alpha=2(\gamma-1)
$$
Using $\gamma=\frac{4}{3}$, for polyatomic gas
$$
\alpha=2\left(\frac{4}{3}-1\right)=\frac{2}{3}
$$
In this process, $T V^{\gamma-1}=k$ [constant $]$
Comparing Eqs. (i) and (ii), we get
$$
\alpha=2(\gamma-1)
$$
Using $\gamma=\frac{4}{3}$, for polyatomic gas
$$
\alpha=2\left(\frac{4}{3}-1\right)=\frac{2}{3}
$$
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