Question:
A monoatomic gas \(\left(\gamma=\frac{5}{3}\right)\) at a pressure of \(4\) atm is compressed adiabatically so that its temperature rises from \(27^\circ\text{C}\) to \(327^\circ\text{C}\). The pressure of the gas in its final state is
A monoatomic gas \(\left(\gamma=\frac{5}{3}\right)\) at a pressure of \(4\) atm is compressed adiabatically so that its temperature rises from \(27^\circ\text{C}\) to \(327^\circ\text{C}\). The pressure of the gas in its final state is
Show Hint
For an adiabatic process,
\[
\frac{P_2}{P_1}
=
\left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}}.
\]
For a monoatomic gas,
\[
\gamma=\frac{5}{3},
\]
so
\[
\frac{\gamma}{\gamma-1}=\frac{5}{2}.
\]
Updated On: Jun 26, 2026
- \(2^{\frac{5}{3}}\ \text{atm}\)
- \(2^{\frac{10}{3}}\ \text{atm}\)
- \(2^{\frac{5}{2}}\ \text{atm}\)
- \(2^{\frac{9}{2}}\ \text{atm}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the given data.
Initial pressure, \[ P_1=4\ \text{atm}. \] For a monoatomic gas, \[ \gamma=\frac{5}{3}. \] Initial temperature, \[ T_1=27^\circ\text{C}=300\ \text{K}. \] Final temperature, \[ T_2=327^\circ\text{C}=600\ \text{K}. \] Therefore, \[ \frac{T_2}{T_1} = \frac{600}{300} = 2. \]
Step 2: Use the adiabatic relation between temperature and pressure.
For an adiabatic process, \[ TV^{\gamma-1}=\text{constant} \] and \[ PV^\gamma=\text{constant}. \] Eliminating \(V\), we obtain \[ T^\gamma P^{\,1-\gamma} = \text{constant}. \] Hence, \[ \left(\frac{T_2}{T_1}\right)^\gamma = \left(\frac{P_2}{P_1}\right)^{\gamma-1}. \] Therefore, \[ \frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}}. \]
Step 3: Substitute the values.
Since \[ \gamma=\frac{5}{3}, \] \[ \gamma-1=\frac{2}{3}. \] Thus, \[ \frac{\gamma}{\gamma-1} = \frac{\frac{5}{3}}{\frac{2}{3}} = \frac{5}{2}. \] Hence, \[ \frac{P_2}{P_1} = 2^{\frac{5}{2}}. \] Therefore, \[ P_2 = 4\times 2^{\frac{5}{2}}. \] Since \[ 4=2^2, \] \[ P_2 = 2^2\times 2^{\frac{5}{2}}. \] \[ P_2 = 2^{\frac{9}{2}}\ \text{atm}. \]
Step 4: Final conclusion.
Hence, the final pressure of the gas is \[ \boxed{2^{\frac{9}{2}}\ \text{atm}} \] Therefore, the correct option is \[ \boxed{(4)} \]
Initial pressure, \[ P_1=4\ \text{atm}. \] For a monoatomic gas, \[ \gamma=\frac{5}{3}. \] Initial temperature, \[ T_1=27^\circ\text{C}=300\ \text{K}. \] Final temperature, \[ T_2=327^\circ\text{C}=600\ \text{K}. \] Therefore, \[ \frac{T_2}{T_1} = \frac{600}{300} = 2. \]
Step 2: Use the adiabatic relation between temperature and pressure.
For an adiabatic process, \[ TV^{\gamma-1}=\text{constant} \] and \[ PV^\gamma=\text{constant}. \] Eliminating \(V\), we obtain \[ T^\gamma P^{\,1-\gamma} = \text{constant}. \] Hence, \[ \left(\frac{T_2}{T_1}\right)^\gamma = \left(\frac{P_2}{P_1}\right)^{\gamma-1}. \] Therefore, \[ \frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}}. \]
Step 3: Substitute the values.
Since \[ \gamma=\frac{5}{3}, \] \[ \gamma-1=\frac{2}{3}. \] Thus, \[ \frac{\gamma}{\gamma-1} = \frac{\frac{5}{3}}{\frac{2}{3}} = \frac{5}{2}. \] Hence, \[ \frac{P_2}{P_1} = 2^{\frac{5}{2}}. \] Therefore, \[ P_2 = 4\times 2^{\frac{5}{2}}. \] Since \[ 4=2^2, \] \[ P_2 = 2^2\times 2^{\frac{5}{2}}. \] \[ P_2 = 2^{\frac{9}{2}}\ \text{atm}. \]
Step 4: Final conclusion.
Hence, the final pressure of the gas is \[ \boxed{2^{\frac{9}{2}}\ \text{atm}} \] Therefore, the correct option is \[ \boxed{(4)} \]
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