JEE Main 2014PhysicsThermodynamicsMediumMCQ

JEE Main 2014 — Thermodynamics Question with Solution

JEE Main 2014 (09 Apr Online)

Question

The equation of state for a gas is given by PV = nRT + α V , where n is the number of moles and α   is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are T0 and P0 respectively. The work done by the gas when its temperature doubles isobarically will be :

Choose an option

Show full solutionCorrect option: C
Correct answer
C P 0 T 0 R P 0 - α

Step-by-step explanation

PV = nRT + α V

P 0 dV dT = R + α dV dT

P 0 - α dV dT = R

∴    dV = RdT P 0 - α

∴    W = ∫ P 0 dV = ∫ T 0 2 T 0 P 0 R P 0 - α dT

= P 0 T 0 R P 0 - α

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About this question

This is a previous-year question from JEE Main 2014, covering the Thermodynamics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.