Two moles of an ideal gas is contained in a cylinder fitted with a frictionless movalbe piston, exposed to the atmosphere, at an initial temperature `T_(0)`. The gas is slowly heated so that its volume becomes fout times the initial value. The work done by gas is

Options

A. Zero
B. `2RT_(0)`
C. `4RT_(0)`
D. `6RT_(0)`

Answer

D

Explanation

Given that gas is slowly heated , which means it reains in eqilibrium (more or less) with the atmosphere, i.e . The process takes place at constant pressure.

From the equation of ideal gas law , pV= nRT
For infinitesimal change , `pdV=nRdTrArrpDeltaV=nRDeltaT`
Also , `PDeltaV` = Work done by the gas = `DeltaW`
`therefore" " DeltaW=nRDeltaT`
`Also " " DeltaVpropDeltaT`
`therefore" " DeltaTpropDeltaVpropV_(2)-V_(1)`
`Given, " "V_(2)=4V_(1)`
`therefore" " DeltaTprop4V_(1)-V_(1)prop3V_(1)prop3T_(0)`
Also given m = 2 moles
The expression for work becomes
`DeltaW=nRDeltaat=2R3T_(0)=6RT_(0)`