Correct Answer – 9
Gas is diatomic as `gamma = 1.4`, thus, `C_(V) = (5)/(2)R` and `C_(P) = (7)/(2)R`, Given, `Delta H = 85J` at constant pressure
Also, `Delta T (85)/(nC_(P)) = (140 xx 2)/(7 n xx R) = (40)/(n)`
Now,
`W = – nRT Delta T = – n xx 2 xx (40)/(n) = – 80 J`
Also, `q_(p) = DeltaH = Delta U + (-W)`
`Delta U = Delta H + W = 85 – 80`
`Delta U = 5J`

Suppose the sample contains n moles. Also,
suppose the volume changes from `V^(1) to V^(2)` and the
temperature changes from `T^(1) to T^(2)`.
The heat supplied is
`DeltaQ = nCp(T^(2) -T^(1))`.
(a) The change in internal energy is `DeltaU = nCv (T^(2) – T^(1)) = Cv / Cp nCP (T^(2) – T^(1))`
`= Cv / Cp DeltaQ = 140 J / 1.4 = 100 J`.
(b) The work done by the gas is
`DeltaW = DeltaQ – DeltaU
140 J – 100 J = 40 J`.

Correct Answer – A
`dQ=n C_(p)dT`
`dU=nC_(V)dT`
`dW=dQ-dU=n(C_(p)-C_(V))dT`
`(dW)/(dQ)=(nC+_(p))-(C_(V)dT)/(nC_(p)dT)=(C_(p))/(C_(p))-(C_(V))/(C_(P))`
`(dW)/(100)=1-(C_(V))/(C_(p))=1-(1)(1.4)=(0.4)/(1.4)=(4)/(14)=(2)/(7)`
`dW=100xx(2)(7)=28.57 J`