91Ó°ÊÓ

P-V diagram representing four processes is given.

We can infer by analyzing the p-v graph that the work done is the area under the path. So by comparing the area under each path, we can rank the four paths according to work done by the gas. From the formula for change in internal energy, we can infer that internal energy depends on the temperature change. By comparing the change in temperature from the graph, we can find the ranking of paths 1, 2, and 3 according to the change in the internal energy of the gas.

Formulae:

Work done by the gas at constant pressure, $W={∫}_{v_i}^{v_f}pdv$ …(¾±)

Change in internal energy at constant volume, $\mathrm{Δ}E=n{C}_V\mathrm{Δ}T$ …(¾±¾±)

The work done can be written as the area under the curve.

From analyzing the P-V graph and considering equation (i), we can see thatthearea under the first path is greater as compared totheother path, and the area under the path four is less thantheother path.

From that, we can conclude that the rank of the path according to work done is

$1>2>3>4.$

Change in internal energy for the gasdepends on change in temperature.

From P-V curve:

Thus, the change in temperature for path 1 is as follows:

$\begin{array}{c}\mathrm{Δ}{T}_1=700 \mathrm{K}-500 \mathrm{K}\\ =200\mathrm{K}\end{array}$ …(¾±¾±¾±)

Thus, the change in temperature for path 2 is as follows:

$\begin{array}{c}\mathrm{Δ}{T}_2=500 \mathrm{K}-500 \mathrm{K}\\ =0\mathrm{K}\end{array}$ …(¾±±¹)

Thus, the change in temperature for path 3 is as follows:

$\begin{array}{c}\mathrm{Δ}{T}_2=400 \mathrm{K}-500 \mathrm{K}\\ =-100\mathrm{K}\end{array}$ …(±¹)

From equations (iii), (iv), and (v), we can write that $\mathrm{Δ}{T}_1>\mathrm{Δ}{T}_2>\mathrm{Δ}{T}_3$.

So, from the above value of temperature and equation (ii), we get that role="math" localid="1662973411994" $∆E_{int,1}â‰\yen ∆E_{int,2}â‰\yen ∆E_{int,3}.$

So, the ranking of paths 1, 2, and 3 according to internal energy is $1>2>3.$