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For a Van der Walls gas find : (a) the equation of the adiabatic curve in the variables `T, V` , the difference of the molar heat capacities `C_p - C_v` as a function of `T` and `V`. |
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Answer» (a) From the first law for an adiabatic `dQ = dU + pd V = 0` From the previous problem `d U =((del U)/(del T))_V dT + ((del U)/(del V))_T dV = C_V dT + (a)/(V^2) dV` So, `0 = C_V dT + (RT dV)/(V - b)` This equation can be integrated if we assume that `C_V` and `b` are constanr then `(R)/(C_V) (dV)/(V - b) + (dT)/(T) = 0` or, `1n T +(R)/(C_V) 1n (V - b) = constant` or, `T(V - b)^(R//c_v) = constant` (b) We use `dU = C_V dT + (a)/(V^2) dV` Now, `dQ = C_V dT + (RT)/(V - b)dV` So along constant `p`, `C_p = C_V + (RT)/(V - b) ((del V)/(del T))` Thus `C_p - C_V = (RT)/(V - b)((del V)/(del T))_p`, But `p = (RT)/(V - b)- (a)/(V^2)` On differentiating, `0 = (-(RT)/((V - b)^2) +(2 a)/(V^2))((del V)/(del T))_p + (R)/(V - b)` or, `T((del V)/(del T))_p = (RT//V -b)/((RT)/((V - b)^2 )-(2 a)/(V^3)) = (V -b)/(1-(2a(V - b)^2)/(RT V^3))` and `C_p - C_V = (R)/(1 -(2a(V - b)^2)/(RTV^3))`. |
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