InterviewSolution
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InterviewSolution
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For the case of an ideal gas find the equation of the process (in the variables `T, V`) in which the molar heat capacity varies as : (a) `C = C_V + alpha T` , (b) `C = C_V + beta V `, ( c) `C = C_v + ap` , where `alpha, beta` and `a` are constants. |
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Answer» Heat capacity is given by `C = C_V + (RT)/(V) (dV)/(dT)` (a) Given `C = C_V + alpha T` So, `C_V + alpha T = C_V + (RT)/(V)(dV)/(dT)`or, `(alpha)/(R) dT = (dV)/(V)` Integrating both sides, we get `(alpha)/(R)T = 1n C_0 = 1n VC_0 , C_0` is a constant. Or, `V. C_0 = e^(alpha T//R)` or `V . e^(alpha T//R) = (1)/(C_0) = constant` (b) `C = C_V + beta V` and `C = C_V + (RT)/(V) (dV)/(dT)` so, `C_V (RT)/(V) (dV)/(dT) = C_V + beta V` or, `(RT)/(V)(dV)/(dT) = beta V` or, `(dV)/(V^2) = (beta)/(R) (dT)/(T)` or, `V^-2 = (dT)/(T)` Integrating both sides, we get `(R)/(beta) (V^-1)/(beta - 1) = 1n T + 1n C_0 = 1n T.C_0` So, `1n T.C_0 = -(R)/(betaV) T. C_0 = e^(-R//beta V) ` or, `T e^(-R//beta V) = (1)/(C_0) = constant` ( c) `C = C_V + ap` and `C = C_V + (RT)/(V) (dV)/(dT)` So, `C_V + ap = C_V + (RT)/(V) (dV)/(dT)` so, `ap = (RT)/(V) (dV)/(dT)` or, `a (RT)/(V) = (RT)/(V) (dV)/(dT)` (as `p = (RT)/(V)` for one mole of gas) or, `(dV)/(dT) =a` ro, `dV = adT`or `dT = (dV)/(a)` So, `T = (V)/(a) + constant` or `V - aT = constant`. |
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