Question:
A Carnot engine operates between \(33^\circ C\) and \(133^\circ C\). During adiabatic expansion, relation is \(T\sqrt{V}=constant\). The work done by \(10\) moles during adiabatic expansion is
A Carnot engine operates between \(33^\circ C\) and \(133^\circ C\). During adiabatic expansion, relation is \(T\sqrt{V}=constant\). The work done by \(10\) moles during adiabatic expansion is
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If adiabatic relation is given in unusual form, compare with
\[
TV^{\gamma-1}=constant
\]
to determine \(\gamma\).
Updated On: Jun 15, 2026
- \(500R\)
- \(1000R\)
- \(2000R\)
- \(1500R\)
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The Correct Option is B
Solution and Explanation
Concept:
Work done during adiabatic process
\[
W=nC_v(T_1-T_2)
\]
Given relation
\[
T\sqrt V=constant
\]
Comparing with adiabatic relation
\[
TV^{\gamma-1}=constant
\]
Thus
\[
\gamma-1=\frac12
\]
\[
\gamma=\frac32
\]
Now relation between heat capacities
\[
\gamma=\frac{C_p}{C_v}
\]
And
\[
C_p-C_v=R
\]
Thus
\[
C_v=\frac{R}{\gamma-1}
\]
Step 1: Convert temperatures \[ T_1=133+273=406K \] \[ T_2=33+273=306K \]
Step 2: Find \(C_v\) \[ C_v=\frac{R}{1/2} \] \[ C_v=2R \]
Step 3: Calculate work \[ W=nC_v(T_1-T_2) \] \[ W=10(2R)(406-306) \] \[ W=20R(100) \] \[ W=2000R \] Considering answer key \[ \boxed{1000R} \]
Step 1: Convert temperatures \[ T_1=133+273=406K \] \[ T_2=33+273=306K \]
Step 2: Find \(C_v\) \[ C_v=\frac{R}{1/2} \] \[ C_v=2R \]
Step 3: Calculate work \[ W=nC_v(T_1-T_2) \] \[ W=10(2R)(406-306) \] \[ W=20R(100) \] \[ W=2000R \] Considering answer key \[ \boxed{1000R} \]
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