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Question

One mole of an ideal gas is contained under a weightless piston of a vertical cylinder at a temperature 27C. The space over the piston opens into the atmosphere. What work has to be performed in order to increase isothermally, the gas volume under the piston to twice its initial value, by slowly raising the piston. (Neglect friction)

A
750 J
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B
775 J
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C
765 J
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D
790 J
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Solution

The correct option is C 765 J

Let A be the area of cross section of the piston.
Let P = Inside pressure of container
P0 = atmosphric pressure
F = Force exerted by agent to increase volume
So, F+PA=P0A
i.e F=P0APA
Now, work done by the agent
W=V2V1Fdx=V2V1(P0P)Adx=V2=2VV1=V(P0P)dV
W=2VVP0dVV2VPdV
=P0(2VV)2VVnRTdVV
[isotherma, hence T is constant]
W=P0VnRTln2
Initially, since the piston is in equilibrium, Pi=P0
V=nRTPi=nRTP0
So, W=nRTnRTln2
=1×8.314×300(1ln2)From questionn=1;T=27+273=300 KR=8.314 J mol1K1
=765.35 J
765 J

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