InterviewSolution
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InterviewSolution
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An ideal gas has a molar heat capacity `C_v` at constant volume. Find the molar heat capacity of this gas as a function of its volume `V`, if the gas undergoes the following process : (a) `T = T_0 e^(alpha v)` , (b) `p = p_0 e^(alpha v)`, where `T_0, p_0`, and `alpha` are constants. |
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Answer» (a) By the first law of thermodynamics `dQ = dU + dA = vC_VdT + pd V` Molar specfic heat according to definition `C = (dQ)/(v dT) = (C_V dT + pdV)/(vdT)` =`(v C_V dT + (vRT)/(V) dV)/(v dT) = C_V + (RT)/(V) (dV)/(dT)`, We have `T = T_0 e^(alpha V)` After differentiating, we get `dT = alpha T_0 e^(alpha V). dV` So, `(dV)/(DT) = (1)/(alpha T_0 e^(alpha V))`, Hence `C= = C_V + (RT)/(V).(1)/(alpha T_0 e^(alpha V)) = C_V + (RT_0 e^(alpha V))/(alpha VT_0 e^(alpha v)) = C_v + (R)/(alpha V)` (b) Process is `p = p_0 e^(alpha V)` `P = (RT)/(V) = p_0 e^(alpha V)` or, `T = (p_0)/(R) e^(alpha v). V` So, `C = C_V + (RT)/(V) (dV)/(dT) = C_V + p_0 e^(alpha V).(R)/(p_0 e^(alpha V)(1 + alpha V)) = C_V + (R)/(1 + alpha V)`. |
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