MHT CET · Physics · Thermal Properties of Matter
A monoatomic ideal gas initially at temperature ' \(\mathrm{T}_1\) ' is enclosed in a cylinder fitted with massless, frictionless piston. By releasing the piston suddenly the gas is allowed to expand to adiabatically to a temperature ' \(\mathrm{T}_2\) '. If ' \(\mathrm{L}_1\) ' and ' \(\mathrm{L}_2\) ' are the lengths of the gas columns before and after expansion respectively, then \(\frac{T_2}{T_1}\) is
- A \(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\)
- B \(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\)
- C \(\left(\frac{L_1}{L_2}\right)^{2 / 3}\)
- D \(\left(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right)^{2 / 3}\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{L_1}{L_2}\right)^{2 / 3}\)
Step-by-step Solution
Detailed explanation
For an adiabatic process
\(
\begin{aligned}
& \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
\therefore \quad & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}
\end{aligned}
\)
For a monoatomic gas, \(r=\frac{5}{3}\)
\(
\begin{aligned}
& \Rightarrow \gamma-1=\frac{5}{3}-1=\frac{2}{3} \\
& \mathrm{~V}_1=\mathrm{AL}_1 \text { and } \mathrm{V}_2=\mathrm{AL}_2 \\
\therefore \quad & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left[\frac{\mathrm{AL}_1}{\mathrm{AL}_2}\right]^{2 / 3}=\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3} \\
& \gamma-1=\frac{2}{3}
\end{aligned}
\)
For an adiabatic process,
\(
\begin{aligned}
& \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\frac{2}{3}} \\
\mathrm{~V} & \propto \mathrm{L} \\
\therefore \quad \frac{\mathrm{T}_2}{\mathrm{~T}_1} & =\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{\frac{2}{3}}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
\therefore \quad & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}
\end{aligned}
\)
For a monoatomic gas, \(r=\frac{5}{3}\)
\(
\begin{aligned}
& \Rightarrow \gamma-1=\frac{5}{3}-1=\frac{2}{3} \\
& \mathrm{~V}_1=\mathrm{AL}_1 \text { and } \mathrm{V}_2=\mathrm{AL}_2 \\
\therefore \quad & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left[\frac{\mathrm{AL}_1}{\mathrm{AL}_2}\right]^{2 / 3}=\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3} \\
& \gamma-1=\frac{2}{3}
\end{aligned}
\)
For an adiabatic process,
\(
\begin{aligned}
& \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\frac{2}{3}} \\
\mathrm{~V} & \propto \mathrm{L} \\
\therefore \quad \frac{\mathrm{T}_2}{\mathrm{~T}_1} & =\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{\frac{2}{3}}
\end{aligned}
\)
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