A sample of gas undergoes a transition from an initial state ato a final state bby three different paths (processes), as shown in the $\mathit{p}\mathbf{-}\mathbf{}\mathit{V}$diagram in Fig. 18-57, where localid="1662963822546" ${\mathit{V}}_{\mathbf{b}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}{\mathit{V}}_{\mathbf{i}}$. The energy transferred to the gas as heat in process 1 is localid="1662963830858" $\mathbf{10}{\mathit{p}}_{\mathbf{i}}{\mathit{V}}_{\mathbf{i}}$.In terms oflocalid="1662963841315" ${\mathit{p}}_{\mathbf{i}}{\mathit{V}}_{\mathbf{i}}$, what are (a) the energy transferred to the gas as heat in process 2 and (b) the change in internal energy that the gas undergoes in process 3?

The universe's energy is constant, according to the first law of thermodynamics. It cannot be created or destroyed, but it can be exchanged between the system and its surroundings. The law essentially deals with how work and heat transport cause changes in energy states. Work done is the area under the P- V curve; we can use this fact to find the work done. Then, by using the first law of thermodynamics in which heat is the addition of change in internal energy and work done, we can find the required quantities.

Formula:

According to first law of thermodynamics, $Q=W+∆\mathrm{U}$ …(i)

where Q is heat transferred, W is work done, and $∆\mathrm{U}$is change in internal energy.

The work done by the system, $W=p_i\left(V_b-V_i\right)$. …(ii)

a.

Now, heat transferred in process 1 is $Q_1=10P_i{V}_i$, and the work done in process 1 is the area under P-V curve. Using equation (ii), it is found to be

Now, change in internal energy in process 1,using equation (i) and the above values, can be found to be

$$\begin{gathered} W_1=p_i\left(5V_i-V_i\right) \\ =4p_{i}V \end{gathered}$$

…(iii)

Work done in process 2 is as follows:

$$\begin{gathered} W_2=p_i\left(V_b-V_i\right)+0.5\times \left(\frac{3p_i}{2}-p_i\right)\times \left(V_b-V_i\right) \\ =p_i\left(5V_i-V_i\right)+0.5\times \left(0.5p_i\right)\times \left(5V_i-V_i\right) \\ =4p_i{V}_i+\frac{4}{4}{p}_i{V}_i \\ =5p_i{V}_i \end{gathered}$$

So, the heat transferred in process 2,using the above values and equation (i),is

(Note: The internal energy of a system depends on the end points.)

$$\begin{gathered} Q_2=5p_i{V}_i+6p_i{V}_i \\ =11p_i{V}_i \end{gathered}$$

Hence, the heat transferred in the process 2 is $11p_i{V}_i$.

Process 3 starts at a and ends at b; so,the change in internal energy is as follows:

$$\begin{gathered} \Delta U_3=U_b-U_a \\ =6p_i{V}_i\text{(from equation (iii))} \end{gathered}$$

Hence, the value of the change in internal energy in process 3 is.