Question:
Two samples A and B of gas initially at the same pressure and temperature. They are compressed from volume \(V\) to \(\frac{V}{2}\) (A isothermally and B adiabatically). The relation between pressure of the gas A (\(P_A\)) and pressure of gas B (\(P_B\)) is
Two samples A and B of gas initially at the same pressure and temperature. They are compressed from volume \(V\) to \(\frac{V}{2}\) (A isothermally and B adiabatically). The relation between pressure of the gas A (\(P_A\)) and pressure of gas B (\(P_B\)) is
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Adiabatic compression heats the gas, so its pressure rises more.
Updated On: Oct 1, 2026
- \(P_A < P_B\)
- \(P_A > P_B\)
- \(P_A = P_B\)
- \(P_A = 2P_B\)
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The Correct Option is A
Solution and Explanation
Step 1: Isothermal (A)
\(P_AV=P_A\cdot\frac V2\) gives \(P_A=2P\).
Step 2: Adiabatic (B)
\(PV^\gamma=P_B\left(\frac V2\right)^\gamma\) gives \(P_B=2^\gamma P\).
Step 3: Compare
Since \(\gamma>1\), \(2^\gamma>2\), so \(P_B>P_A\). Option (A).
Final Answer:
\(P_A<P_B\), option (A). \[ \boxed{\text{(A)}} \]
\(P_AV=P_A\cdot\frac V2\) gives \(P_A=2P\).
Step 2: Adiabatic (B)
\(PV^\gamma=P_B\left(\frac V2\right)^\gamma\) gives \(P_B=2^\gamma P\).
Step 3: Compare
Since \(\gamma>1\), \(2^\gamma>2\), so \(P_B>P_A\). Option (A).
Final Answer:
\(P_A<P_B\), option (A). \[ \boxed{\text{(A)}} \]
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