An ideal gas of mass m and temperature T1 undergoes a reversible isothermal process from an initial pressure P1 to final pressure P2. The heat loss during the process is Q. The entropy change ΔS of the gas is
A
mRIn(P2P1)−QT1
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B
mRIn(P1P2)
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C
mRIn(P2P1)
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D
zero
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Solution
The correct option is BmRIn(P1P2) We know that dS=δQT
Intergrating between states 1 and 2 gives ∫21ΔS=∫21δQT S2−S1=1T∫21δQ∵T=C =1T×Q1−2=mRTInV2V2T mRInV2V1=mRInp1p2
where p1p2=V2V1 for isothermal process
Alternatively:
We know that the entropy change in terms of temperautre and pressure rations. S2−S1=mcpInT2T1−S2−S1=mRInp2p1
where T1=T2 for isothermal process ∴S2−S−1=−mRInp2p1=mRInp1p2