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A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is $\mathrm{TV}^{\mathrm{x}}=$ constant, then $\mathrm{x}$ is:
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Verified Answer
The correct answer is:
$\frac{2}{5}$
Equation of adiabatic change is
$$
\mathrm{TV}^{\gamma-1}=\text { constant }
$$
Put $\gamma=\frac{7}{5}$, we get: $\gamma-1=\frac{7}{5}-1$
$\therefore \mathrm{x}=\frac{2}{5}$
$$
\mathrm{TV}^{\gamma-1}=\text { constant }
$$
Put $\gamma=\frac{7}{5}$, we get: $\gamma-1=\frac{7}{5}-1$
$\therefore \mathrm{x}=\frac{2}{5}$
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