Question:
A thermodynamic system goes from states
\[
\text{i)}\quad (P,V)\ \rightarrow\ (P,2V)
\]
\[
\text{ii)}\quad (P_1,V)\ \rightarrow\ (2P_1,V)
\]
Then the works done in these two cases are
A thermodynamic system goes from states
\[
\text{i)}\quad (P,V)\ \rightarrow\ (P,2V)
\]
\[
\text{ii)}\quad (P_1,V)\ \rightarrow\ (2P_1,V)
\]
Then the works done in these two cases are
Show Hint
Remember:
\[
\text{Isobaric process: }
W=P\Delta V
\]
and
\[
\text{Isochoric process: }
W=0.
\]
No volume change means no work done by the gas.
Updated On: Jul 29, 2026
- \[ \text{i) Zero} \qquad \text{ii) Zero} \]
- \[ \text{i) Zero} \qquad \text{ii) }P_1V \]
- \[ \text{i) }PV \qquad \text{ii) }P_1V \]
- \[ \text{i) }PV \qquad \text{ii) Zero} \]
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
Work done in a thermodynamic process is
\[
W=\int P\,dV.
\]
Thus, work is done only when there is a change in volume.
Step 1: Find the work done in case (i). The system changes from \[ (P,V) \rightarrow (P,2V). \] Pressure remains constant. Hence the process is isobaric. \[ W=P(V_f-V_i). \] \[ W=P(2V-V). \] \[ W=PV. \] Therefore, \[ \boxed{W_1=PV}. \]
Step 2: Find the work done in case (ii). The system changes from \[ (P_1,V) \rightarrow (2P_1,V). \] Volume remains constant. Hence the process is isochoric. For constant volume, \[ dV=0. \] Therefore, \[ W=\int P\,dV=0. \] Thus, \[ \boxed{W_2=0}. \]
Step 3: Write the final result. \[ \boxed{ W_1=PV, \qquad W_2=0 } \] Hence, \[ \boxed{\text{Answer = (D)}} \]
Step 1: Find the work done in case (i). The system changes from \[ (P,V) \rightarrow (P,2V). \] Pressure remains constant. Hence the process is isobaric. \[ W=P(V_f-V_i). \] \[ W=P(2V-V). \] \[ W=PV. \] Therefore, \[ \boxed{W_1=PV}. \]
Step 2: Find the work done in case (ii). The system changes from \[ (P_1,V) \rightarrow (2P_1,V). \] Volume remains constant. Hence the process is isochoric. For constant volume, \[ dV=0. \] Therefore, \[ W=\int P\,dV=0. \] Thus, \[ \boxed{W_2=0}. \]
Step 3: Write the final result. \[ \boxed{ W_1=PV, \qquad W_2=0 } \] Hence, \[ \boxed{\text{Answer = (D)}} \]
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