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A gas at normal temperature is suddenly compressed to one-fourth of its original volume.
If $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma=1.5$, then the increase in its temperature is
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If $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma=1.5$, then the increase in its temperature is
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Verified Answer
The correct answer is:
$273 \mathrm{~K}$
Given that, $\mathrm{V}_2=\frac{\mathrm{V}_1}{4}, \frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma=1.5$
As the process is sudden, it is an adiabatic expansion,
$$
\begin{aligned}
& \therefore \quad \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
& \therefore \quad \mathrm{T}_2=\mathrm{T}_1\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1} \\
& =\mathrm{T}_1(4)^{\gamma-1} \\
& =\mathrm{T}_1 \times(4)^{0.5} \\
& =2 \mathrm{~T}_1 \\
& \therefore \quad \mathrm{T}_2-\mathrm{T}_1=\mathrm{T}_1 \\
& \therefore \quad \mathrm{T}_2-\mathrm{T}_1=273 \mathrm{~K} \quad\left(\because \mathrm{T}_1=\text { Normal temperature }\right) \\
&
\end{aligned}
$$
As the process is sudden, it is an adiabatic expansion,
$$
\begin{aligned}
& \therefore \quad \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
& \therefore \quad \mathrm{T}_2=\mathrm{T}_1\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1} \\
& =\mathrm{T}_1(4)^{\gamma-1} \\
& =\mathrm{T}_1 \times(4)^{0.5} \\
& =2 \mathrm{~T}_1 \\
& \therefore \quad \mathrm{T}_2-\mathrm{T}_1=\mathrm{T}_1 \\
& \therefore \quad \mathrm{T}_2-\mathrm{T}_1=273 \mathrm{~K} \quad\left(\because \mathrm{T}_1=\text { Normal temperature }\right) \\
&
\end{aligned}
$$
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