91Ó°ÊÓ

Energy cannot be generated or destroyed; according to the First Law of Thermodynamics, it can only be changed from one form to another.

According to the first law of thermodynamics,

$\mathit{Q}\mathbf{=}\mathit{W}\mathbf{+}\mathbf{∆}\mathit{U}$

Here, Q is the heat, W is the work done, and $∆U$ is the change in internal energy.

The work done will be the area under the p-V curve.

$$\begin{gathered} W=A_{triangle}+A_{rec\tan gle} \\ =\frac{hb}{2}+wl \\ =\frac{1}{2}\left(0.01-0.002\right)\left(3.5×{10}^5-{10}^5\right)+\left(0.01-0.002\right)\left({10}^5-0\right) \\ =1800J \end{gathered}$$

In the ac path, the pressure is kept constant. So,

$$\begin{gathered} p∆V=nR∆T \\ n∆T=\frac{p∆V}{R} \\ {C}_v=\frac{3}{2}R \end{gathered}$$

The change in internal energy can be found by

$$\begin{gathered} ∆U=n{C}_v∆T \\ ∆U=\frac{3}{2}R×\frac{p∆V}{R} \\ =1.5p∆V \\ =1.5×{10}^5×\left(0.01-0.002\right) \\ =1200J \end{gathered}$$

The heat involved will be

$$\begin{gathered} Q=W+∆U \\ =1800+1200 \\ =3000J \end{gathered}$$

In the ac path, the pressure is kept constant. So,

$$\begin{gathered} p∆V=nR∆T \\ n∆T=\frac{p∆V}{R} \\ {C}_v=\frac{3}{2}R \end{gathered}$$

The change in internal energy can be found by

$$\begin{gathered} ∆U=n{C}_v∆T \\ ∆U=\frac{3}{2}R×\frac{p∆V}{R} \\ =1.5p∆V \\ =1.5×{10}^5×\left(0.01-0.002\right) \\ =1200J \end{gathered}$$

The pressure is kept constant in the process. So, work done will be

$$\begin{gathered} W=p∆V \\ ={10}^5\left(0.01-0.002\right) \\ =800J \end{gathered}$$

The heat involved is given by

$$\begin{gathered} Q=W+∆U \\ =800+1200 \\ =2000J \end{gathered}$$

Heat in part-(a) is more than that of part-b because in the part-a, additional work is done due to change in the volume while the pressure is increasing, but in part-b, the pressure is kept constant.