InterviewSolution
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InterviewSolution
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One mole of an ideal gas with heat capacity at constant pressure `C_p` undergoes the process `T = T_0 + alpha V`, where `T_0` and `alpha` are constants. Find : (a) heat capacity of the gas as a function of its volume , (b) the amount of heat transferred to the gas, if its volume increased from `V_1` to `V_2`. |
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Answer» (a) Heat capacity is given by `C = C_V + (RT)/(V)(dV)/(dT) ` (see solution of `2.52`) We have `T = T_0 + alpha V` or, `V = (T)/(alpha) - (T_0)/(alpha)` After differentiating, we get, `(dV)/(dT) = (1)/(alpha)` Hence `C = C_V + (RT)/(V).(1)/(alpha) = (R)/(gamma - 1) +(R(T_0 + alpha V))/(V) .(1)/(alpha)` =`(R)/(gamma - 1)+ R((T_0)/(alpha V) + 1) = (gamma R)/(gamma - 1)+(RT_0)/(alpha V) = C_V + (RT)/(alpha V) = V_p + (RT_0)/(alpha V)` (b) Given `T = T_0 + alpha V` As `T = (pV)/( R)` for one mole of gas `p = (R)/(V) (T_0 + alpha V) = (RT)/(V) = alpha R` Now `A = int_(V_1)^(V_2) pdV = int_(V_1)^(V_2) ((RT_0)/(v) + alpha R) dV` (for one mole) =`RT_0 1n(V_2)/(V_1) + alpha (V_2 - V_1)` `Delta U = C_v (T_2 - T_1)` =`C_V[T_0 + alpha V_2 - T_0 alpha V_1] = alpha C_V (V_2 - V_1)` By the first law of thermodynamics `Q = Delta U + A` =`(alpha R)/(gamma - 1) (V_2 - V_1) + RT_0 1n (V_2)/(V_1) + alpha R (V_2 - V_1)` =`alpha R(V_2 - V_1)[1 + (1)/(gamma - 1)] + RT_0 1n (V_2)/(V_1)` . =`alpha C_p(V_2 - V_1) + RT_0 1n (V_2)/(V_1)` =`alpha C_p (V_2 - V_1) + RT_0 1n (V_2)/(V_1)`. |
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