JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2021
- Hard
- Numerical
In a certain thermodynamical process, the pressure of a gas depends on its volume as \(kV ^{3}\). The work done when the temperature changes from \(100^{\circ} C\) to \(300^{\circ} C\) will be .......... \(nR\), where \(n\) denotes number of moles of a gas.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(50\)
Step-by-step Solution
Detailed explanation
\(P=k V^{3}\)
\(T_{i}=100^{\circ} C \)
\(T _{ f }=300^{\circ} C\)
\(\Delta T =300-100\)
\(\Delta T =200^{\circ} C\)
\(P = kV ^{3}\)
now \(PV = nRT\)
\(\therefore kV ^{4}= nRT\)
now \(4 kV ^{3} dV = nRdT\)
\(\therefore PdV = nRdT / 4\)
\(\therefore\) Work \(=\int PdV =\int \frac{ nRdT }{4}=\frac{ nR }{4} \Delta T\)
\(=\frac{200}{4} \times nR =50 nR\)
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