Question

(a)Rankthe four paths of Fig $\mathbf{19}\mathbf{-}\mathbf{16}$according to the work done by the gas, greatest first. (b) Rank paths $\mathbf{1}\mathbf{,}\mathbf{}\mathbf{2}$and $\mathbf{3}$according to the change in the internal energy of the gas, most positive first and most negative last.

Short answer

1. The ranking of four paths according to work done by the gas is$1>2>3>4.$
2. The ranking of paths 1,2, and 3 according to the change in the internal energy of the gas, most positive first and most negative last is$1>2>3.$

Step-by-step solution

Stating the given data

P-V diagram representing four processes is given.

Understanding the concept of internal energy and work done

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, $W={\int }_{v_i}^{v_f}pdv$ …(i)

Change in internal energy at constant volume, $\Delta E=n{C}_V\Delta T$ …(ii)

(a) Calculation of the rank of the paths according to the work done

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

$1>2>3>4.$

(b) Calculation of the rank of the four paths according to change in internal energy 

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:

$\begin{array}{c}\Delta T_1=700 \mathrm{K}-500 \mathrm{K}\\ =200\mathrm{K}\end{array}$ …(iii)

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

$\begin{array}{c}\Delta T_2=500 \mathrm{K}-500 \mathrm{K}\\ =0\mathrm{K}\end{array}$ …(iv)

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

$\begin{array}{c}\Delta T_2=400 \mathrm{K}-500 \mathrm{K}\\ =-100\mathrm{K}\end{array}$ …(v)

From equations (iii), (iv), and (v), we can write that $\Delta T_1>\Delta T_2>\Delta T_3$.

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

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