Work Done During Isochoric and Isobaric Process - Thermodynamics

Work Done During Isochoric Process

We know, work done ( dW = P dV).

In an isochoric process, volume (V) is constant, so (dV = 0). Hence, work done during an isochoric process is zero.

According to the First Law of Thermodynamics:

$\Delta Q = \Delta U + \Delta W$

Since ( $\Delta W = 0$), so ( $\Delta Q = \Delta U$). Thus, the heat absorbed by the system (or gas) is used to change the internal energy and hence the temperature of the system (or gas).

Work Done During Isobaric Process

Work done:

$dW = P dV = P(V_{2} - V_{1})$

In an isobaric process, (P) is constant. Now,

$P V_{1} = \mu R T_{1}$ and $P V_{2} = \mu R T_{2}$

$dW = P(V_{2} - V_{1}) = \mu R (T_{2} - T_{1})$

According to the First Law of Thermodynamics:

$\Delta Q = \Delta U + \Delta W$

Therefore, heat absorbed b the system is partly used to change its internal energy and partly used to do work.