Question:
A gas is expanded from an initial state to a final state along a path on a \(P\)-\(V\) diagram. The path consists of (i) an isothermal expansion of work \(50\;J\), (ii) an adiabatic expansion and (iii) an isothermal expansion of work \(20\;J\). If the internal energy of gas is changed by \(-30\;J\), then the work done by gas during adiabatic expansion is
A gas is expanded from an initial state to a final state along a path on a \(P\)-\(V\) diagram. The path consists of (i) an isothermal expansion of work \(50\;J\), (ii) an adiabatic expansion and (iii) an isothermal expansion of work \(20\;J\). If the internal energy of gas is changed by \(-30\;J\), then the work done by gas during adiabatic expansion is
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For an isothermal process of an ideal gas, \(\Delta U=0\). For an adiabatic process, \(Q=0\), so by first law \(\Delta U=-W\).
Updated On: Jun 22, 2026
- \(40\;J\)
- \(100\;J\)
- \(30\;J\)
- \(20\;J\)
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The Correct Option is C
Solution and Explanation
Step 1: Understand internal energy change in isothermal process.
For an ideal gas in an isothermal process, temperature remains constant.
Therefore, change in internal energy is \[ \Delta U=0 \] So, for both isothermal expansions, \[ \Delta U_1=0 \] and \[ \Delta U_3=0 \]
Step 2: Relate total internal energy change to adiabatic process.
The total change in internal energy is given as \[ \Delta U_{\text{total}}=-30\;J \] Since only the adiabatic process contributes to the change in internal energy, \[ \Delta U_{\text{adiabatic}}=-30\;J \]
Step 3: Apply first law for adiabatic expansion.
For an adiabatic process, \[ Q=0 \] Using first law of thermodynamics, \[ \Delta U=Q-W \] So, \[ \Delta U=0-W \] \[ \Delta U=-W \] Given, \[ \Delta U=-30\;J \] Therefore, \[ -30=-W \] \[ W=30\;J \]
Step 4: Final conclusion.
Hence, the work done by the gas during adiabatic expansion is \[ \boxed{30\;J} \]
For an ideal gas in an isothermal process, temperature remains constant.
Therefore, change in internal energy is \[ \Delta U=0 \] So, for both isothermal expansions, \[ \Delta U_1=0 \] and \[ \Delta U_3=0 \]
Step 2: Relate total internal energy change to adiabatic process.
The total change in internal energy is given as \[ \Delta U_{\text{total}}=-30\;J \] Since only the adiabatic process contributes to the change in internal energy, \[ \Delta U_{\text{adiabatic}}=-30\;J \]
Step 3: Apply first law for adiabatic expansion.
For an adiabatic process, \[ Q=0 \] Using first law of thermodynamics, \[ \Delta U=Q-W \] So, \[ \Delta U=0-W \] \[ \Delta U=-W \] Given, \[ \Delta U=-30\;J \] Therefore, \[ -30=-W \] \[ W=30\;J \]
Step 4: Final conclusion.
Hence, the work done by the gas during adiabatic expansion is \[ \boxed{30\;J} \]
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