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Question

Five moles of an ideal gas at 27oC are allowed to expand isothermally from an initial pressure of 10.0 atm to a final pressure of 4.0 atm against a constant external pressure of 1.0 atm. Thus, work done will be:

A
1870.6J
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B
1850.6J
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C
18.47J
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D
+18.47J
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Solution

The correct option is A 1870.6J
Solution:- (A) 1870.6J
As we know that,
W=Pext.ΔV
W=Pext.(VfVi)
From ideal gas equation,
PV=nRT
V=nRTP
Therefore,
W=Pext.(nRTPfnRTPi)
W=Pext.nRT(1Pi1Pf).....(1)
Given:-
External pressure (Pext.)=1atm
Initial pressure (Pi)=4atm
Final pressure (Pf)=10atm
No. of moles (n)=5 moles
Temperature (T)=27=(27+273)=300K
Gas constant (R)=8.314J/molK
Substituting all these values in eqn(1), we have
W=1×5×8.314×300×(14110)
W=12471×320=1870.6J
Hence the work done will be 1870.6J.

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