Consider an ideal gas whose entropy is given by S=(n)/(2) bar(r) 5R l ...
Solution:
Given,
Entropy (S) = (n/2)σ5R ln(U/n2R) + σnR ln(V/n)
where n = number of moles, R = universal gas constant, U = internal energy, V = volume, and σ = constant.
To find: Specific heat at constant volume (Cv)
We know that,
Cv = (∂U/∂T)v
where U = internal energy, T = temperature, and v = constant volume.
We need to express entropy in terms of temperature and volume to find the specific heat.
Step 1: Express entropy in terms of temperature and volume
Using the relation,
dS = (∂S/∂T)v dT + (∂S/∂V)T dV
We get,
(∂S/∂T)v = (n/2)σ5R (1/U)(∂U/∂T)v
(∂S/∂V)T = σnR (1/V)
Substituting the given values, we get,
(∂S/∂T)v = (n/2)σ5R (1/U)Cv
(∂S/∂V)T = σnR (1/V)
Integrating both sides, we get,
S = (n/2)σ5R ln(U/Cv^(5/2)) + σnR ln(V)
Step 2: Find the internal energy
Differentiating the entropy equation with respect to temperature, we get,
(∂S/∂T)v = (n/2)σ5R (1/U)(∂U/∂T)v
Substituting the given entropy equation, we get,
(∂S/∂T)v = (n/2)σ5R (1/U)Cv
Therefore,
Cv = (∂U/∂T)v = (∂T/∂U)v^(-1)
Using the given entropy equation,
S = (n/2)σ5R ln(U/Cv^(5/2)) + σnR ln(V)
Differentiating with respect to U, we get,
(∂S/∂U)v = (n/2)σ5R (Cv^(5/2)/U)
Substituting the above values, we get,
Cv = (5/2)nR
Therefore, the specific heat at constant volume is (Cv) = (5/2)nR.
Answer: Option (a) (5/2)nR