Thermodynamics

JEE Main level, numerical answer, medium

48 of 190

Question 48 of 190

In a certain thermodynamical process, the pressure of a gas depends on its volume as kV3k{V}^{3}. The work done when the temperature changes from 100∘C100^{\circ}C to 300∘C300^{\circ}C will be xnRxnR where nn denotes number of moles of a gas find xx ;

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Answer: 50

P=kV3P=k{V}^{3}
{T}_{i}=100^{\circ}C&{T}_{f}=300^{\circ}C
ΔT=300−100\Delta T=300-100
ΔT=200∘C\Delta T=200^{\circ}C
P=kV3P=k{V}^{3}
now PV=nRTPV=nRT
∴kV4=nRT∴k{V}^{4}=nRT
now 4kV3dV=nRdT4k{V}^{3}dV=nRdT
∴PdV=nRdT/4∴PdV=nRdT/4
∴∴ Work =∫PdV=∫nRdT4=nR4ΔT=\int PdV=\int \dfrac{nRdT}{4}=\dfrac{nR}{4}\Delta T
=2004×nR=50nR=\dfrac{200}{4}\times nR=50nR

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