91Ó°ÊÓ

The values of work done along path a to b and b to c are given by

$$\begin{gathered} {\text{W}}_{ab}=5\text{J} \\ {\text{W}}_{bc}=4\text{J} \end{gathered}$$

According to the First Law of Thermodynamics, energy can only be changed from one form to another and cannot be generated or destroyed. We can find the change in the internal energy using the first law of thermodynamics along the paths ab and bc. From this, we can find the change in the internal energy of the gas when it is taken along the straight path from a to c.

Formula:

According to first law of thermodynamics, the change in internal energy is given by

$\mathrm{Δ}\text{E}=\text{Q}−\text{W}$ …(¾±)

where heat energy Q is taken positive if it is added to the system and work done W is taken as positive if it is done by the gas.

Change intheinternal energy along path a-b is

$\mathrm{Δ}{\text{E}}_{ab}=0$

(Since, the process is isothermal)

Since the process b-c is adiabatic,${\text{Q}}_{bc}=0$.

The change intheinternal energy along path b-c using equation (i) is found to be

$\begin{array}{c}\mathrm{Δ}{\text{E}}_{bc}=−{\text{W}}_{bc}\\ =−4\text{J}\end{array}$

Since the change intheinternal energy is a state function, thus the change in internal energy along path a to c is as follows:

$\begin{array}{c}\mathrm{Δ}{\text{E}}_{ac}=\mathrm{Δ}{\text{E}}_{ab}+\mathrm{Δ}{\text{E}}_{bc}\\ =0\text{ J}−4\text{ J}\\ =−4\text{J}\end{array}$

Therefore, the change in the internal energy of the gas when it is taken along the straight path from a to c is$−4\text{J}$.