Exams › JEE Advanced › Physics
One mole of an ideal gas with adiabatic index gamma undergoes a process described by P = alpha * T^(1/2), where alpha is a constant. Find the molar specific heat capacity of the gas for this process.
- ((gamma - 1)/(gamma + 1)) R/2
- ((gamma - 1)/(gamma + 1)) 2R
- ((gamma + 1)/(gamma - 1)) R/2
- ((gamma + 1)/gamma) R
Correct answer: ((gamma + 1)/(gamma - 1)) R/2
Solution
From P = alpha*sqrt(T) and PV = RT: T = P²/alpha², so PV = R*P²/alpha² giving V = RP/alpha², which means P proportional to V — but more directly PVⁿ = const with n=2 (since eliminating T: P*V = nRT with P proportional to T^(1/2) gives P² proportional to T, and PV=RT so V proportional to T/P proportional to T^(1/2), giving P*V = alpha*T^(1/2)*T^(1/2) = alpha*T = PV = RT, consistent). The polytropic index n = 2, and C = Cv - R/(n-1) = R/(gamma-1) - R/(2-1) = R/(gamma-1) - R = R(1-(gamma-1))/(gamma-1) = R(2-gamma)/(gamma-1). That does not match options. Use C = Cv*(gamma-n)/(1-n) = [R/(gamma-1)]*(gamma-2)/(1-2) = [R/(gamma-1)]*(gamma-2)/(-1) = R(2-gamma)/(gamma-1). Still not matching. For n=2: C = R/(gamma-1) - R/(n-1) = R/(gamma-1) - R. Hmm. Let me try the answer (gamma+1)/(gamma-1) * R/2 directly.
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