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Q.

Two moles of helium gas γ=5/3 at 27C is expanded at constant pressures until its volume is doubled. Then it undergoes an adiabatic change until the temperatures returns to its initial value. The work done during adiabatic process is _____ (universal gas constant = 8.3Jmol1K1 )

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a

7470J

b

4780J

c

4770J

d

7070J

answer is A.

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Detailed Solution

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Helium gas constant γ=5/3

Temperature,  T=27ºC

T=273+27=300K

Initial Volume,V1=V,Final volumeV2=2V,

At constant pressure,P=constant,

V1/T1=V2/T2,T2=2T1,T2 = 600 K

Work done,W= {nR(T2T1)}/(γ1), =(2×8.3×30023)  =7470 Joule

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