A reversible cyclic process for an ideal gas is shown below. Here, P, V, and T are pressure, volume and temperature, respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.
Temperature (T) The correct option(s) is (are)
U BC and w ab = P 2 (V 2 -V 1 )
H AC
H CA < 
U CA and q AC = 
U BC
H AC and 
H CA > 
U CA
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(b, c)
H AC + 
H CB + 
H BA =0
H BA = 0 (Temperature is constant)
H AC = 
H BC ...(1)
We know that q p = 
H (In path BC, P = constant)
Hence q BC = 
H BC
From Eq. (1)
q BC = 
H AC
q BC =-P 2 (V 1 -V 2 ) = P 2 (V 2 -V 1 )
H CA =nC p (T 1 -T 2 ) = -nC P (T 2 -T 1 )
U CA =nC v (T 1 -T 2 ) = -nC v (T 2 -T 1 )
As, C P > C v
So, 
H CA < 
U CA
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