MHT CET · Physics · Thermodynamics
A monoatomic ideal gas, initially at temperature \(T_1\) is enclosed in a cylinder fitted with frictionless piston. The gas is allowed to expand adiabatically to a temperature \(\mathrm{T}_2\) by releasing the piston suddenly. \(L_1\) and \(L_2\) are the lengths of the gas columns before and after the expansion respectively. The ratio \(T_2 / T_1\) is
- A \(\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3}\)
- B \(\left[\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right]^{2 / 3}\)
- C \(\left[\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right]^{1 / 2}\)
- D \(\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{1 / 2}\)
Answer & Solution
Correct Answer
(A) \(\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3}\)
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, \(\gamma=\frac{5}{3}\)
\(\begin{array}{ll}
& \Rightarrow \gamma-1=\frac{5}{3}-1=\frac{2}{3} \\
& \text {Volume }=\text { Area } \times \text { Length, } \\
\therefore \quad & \mathrm{V}_1=\mathrm{AL}_1 \text { and } \mathrm{V}_2=\mathrm{AL}_2 \\
\therefore \quad & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left[\frac{\mathrm{AL}_1}{\mathrm{AL}]_2}\right]^{2 / 3}=\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3} \\
\therefore \quad & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{\frac{2}{3}}
\end{array}\)
\(\begin{aligned}
& \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
\therefore \quad & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}
\end{aligned}\)
For a monoatomic gas, \(\gamma=\frac{5}{3}\)
\(\begin{array}{ll}
& \Rightarrow \gamma-1=\frac{5}{3}-1=\frac{2}{3} \\
& \text {Volume }=\text { Area } \times \text { Length, } \\
\therefore \quad & \mathrm{V}_1=\mathrm{AL}_1 \text { and } \mathrm{V}_2=\mathrm{AL}_2 \\
\therefore \quad & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left[\frac{\mathrm{AL}_1}{\mathrm{AL}]_2}\right]^{2 / 3}=\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3} \\
\therefore \quad & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{\frac{2}{3}}
\end{array}\)
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