MHT CET · Physics · Kinetic Theory of Gases
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
- A \(273 \mathrm{~K}\)
- B \(373 \mathrm{~K}\)
- C \(473 \mathrm{~K}\)
- D \(573 \mathrm{~K}\)
Answer & Solution
Correct Answer
(A) \(273 \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
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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