Thermal Physics - Isothermal

Suppose one mole of an ideal gas undergoes the reversible cycle ABCA shown in the P-V diagram above, where AB is an isotherm. The molar heat capacities are C_p at constant pressure and C_v at constant volume. The net heat added to the gas during the cycle is equal to

A. RT_h (V₂/V₁)
B. −C_p(T_h − T_c)
C. C_p(T_h − T_c)
D. RT_h ln (V₂/V₁) − C_p(T_h − T_c)
E. RT_h ln (V₂/V₁) − R(T_h − T_c)

(GR9677 #15)

Solution:

AB Isotherm → T Constant = T_h

Ideal Gas: PV = nRT 
P = nRT / V

n = 1 mole,

W_AB = _V1∫^V₂ P dV = RT_h V1∫^V₂ (1/V) dV  = RT_h ln (V₂/V₁)

BC Isobaric → P constant = P₂

W_BC = _V2∫^V₁ P dV = P₂ (V₁− V₂) =  P₂V₁ −  P₂V₂

From the diagram:

P₂V₁ = nRT_c 

P₂V₂ = nRT_h

W_BC = nR(T_c − T_h)  = R(T_c − T_h)

W_CA = 0  since V constant

Total W = W_AB + W_BC = RT_h ln (V₂/V₁) + R(T_c − T_h)

or

W = RT_h ln (V₂/V₁) − R(T_h − T_c)

Answer: E