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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.

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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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