Question

Explain the thermodynamics of the isobaric process.  

Answer 1

1. A thermodynamic process that is carried out at constant pressure i.e., Δp = 0 is called the isobaric process. 
2. For an isobaric process, none of the quantities ΔU, Q, and W is zero. 
3. The temperature of the system changes, i.e., ΔT ≠ 0.
4. The energy exchanged is used to do work as well as to change internal energy causing an increase in temperature. Thus, Q = ΔU + W. 
5. As work is done volume changes during the process.
6. Heat exchanged in case of an isobaric process:
   a. Consider an ideal gas undergoing volume expansion at constant pressure.
   b. If V_i and T_i are its volume and temperature in the initial state of a system and V_f and T_f are its final volume and temperature respectively, the work done in the expansion is given by,
   W = pdV = p(V_f – V_i) = nR(T_f – T_i) ….(1)
   c. Also, the change in the internal energy of a system is given by, ΔU = nC_VΔT = nC_V(T_f – T_i) ….(2)
   Where, C_V is the specific heat at constant volume, and ΔT = (T_f – T_i) is the change in its temperature during the isobaric process.   
   d. According to the first law of thermodynamics, the heat exchanged is given by, Q = ΔU + W
   Using equations (1) and (2) we get,
   Q = nC_V(T_f – T_i) + nR(T_f – T_i)
   ∴ Q = (nC_V + nR) (T_f – T_i)
   ∴ Q = nC_p(T_f – T_i)   .............(∵ C_p = C_V + R)
   Where, C_p is the specific heat at constant pressure.
7. The p-V diagram for an isobaric process is called isobar. It is shown in the figure below.