Question

# Two moles of an ideal monoatomic gas at 27∘C occupies a volume of V. If the gas is expanded adiabatically to the volume 23/2V, then the work done by the gas will be (γ=53,R=8.31 J/mol K)3739.5 J2627.23 J2500 J−2500 J

Solution

## The correct option is A 3739.5 JGiven, gas is monoatomic ∴γ=53 Initial Temperature (Ti)=300 K Initial volume (Vi)=V Final volume (Vf)=23/2V Work done by the gas (W)=nR(Ti−Tf)γ−1 =nRTiγ−1[1−TfTi] As, TiVγ−1i=TfVγ−1f ∴W=nRTiγ−1[1−(ViVf)γ−1] =2×8.31×300(53−1)⎡⎢ ⎢⎣1−(123/2)23⎤⎥ ⎥⎦ =3739.5 J

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