Question

The P-V diagram of path followed by one mole of perfect gas in a cylindrical container is shown in figure, the work done when the gas is taken from state A to state B is:
                                 

$\begin{array}{rl}& (A)2P_2{V}_1[1-\sqrt{{\displaystyle \frac{V_2}{V_1}}}]\\ & (B)2P_1{V}_1[1-\sqrt{{\displaystyle \frac{V_1}{V_2}}}]\\ & (C)2P_2{V}_2[1-\sqrt{{\displaystyle \frac{V_1}{V_2}}}]\\ & (D)2P_1{V}_2[1-\sqrt{{\displaystyle \frac{V_1}{V_2}}}]\end{array}$

Answer 1

Hint: We have been given the condition that, $P{V}^{{\displaystyle \frac{1}{2}}}=k$ ,where k is any arbitrary constant. Now, this is an example of a polytropic process. So we will use the formula of work done in taking a gas from initial state to a final state of a polytropic process.

Complete answer:
It has been given in the question:
The initial parameters of gas are denoted by the following terms:
Initial volume : $V_1$
Initial pressure: $P_1$
Initial temperature: $T_1$
And, the final parameters of the gas are given by the following terms:
Final volume : $V_2$
Final pressure : $P_2$
Final temperature: $T_2$
Now, once we have defined the initial and final parameters of the gas, we can proceed ahead with calculating the work done in the given process.
As it is given:
$\Rightarrow P{V}^{{\displaystyle \frac{1}{2}}}=k$
This is a Polytropic process, with the value of :
$\Rightarrow n={\displaystyle \frac{1}{2}}$
Now, the formula for work done in a polytropic process is given by the following equation:
$\Rightarrow W={\displaystyle \frac{P_1{V}_1-P_2{V}_2}{n-1}}$
Putting the value of (n) in the above equation, we get:
 $\begin{array}{rl}& \Rightarrow W={\displaystyle \frac{P_1{V}_1-P_2{V}_2}{{\displaystyle \frac{1}{2}}-1}}\\ & \Rightarrow W=2(P_2{V}_2-P_1{V}_1)\end{array}$
Taking $P_2{V}_2$common out of the bracket in the right-hand side of the equation, we get the work done as:
 $\Rightarrow W=2P_2{V}_2(1-{\displaystyle \frac{P_1{V}_1}{P_2{V}_2}})$ [Let this expression be equation number (1)]
Now, using the given polytropic function, we can write:
$\begin{array}{rl}& \Rightarrow P_1{(V_1)}^{{\displaystyle \frac{1}{2}}}=k\\ & \Rightarrow P_1{(V_1)}^{{\displaystyle \frac{1}{2}}}\times {(V)}^{{\displaystyle \frac{1}{2}}}=k{(V)}^{{\displaystyle \frac{1}{2}}}\\ & \Rightarrow P_1{V}_1=k{(V_1)}^{{\displaystyle \frac{1}{2}}}\end{array}$
Similarly, for the final state of gas we can write:
$\Rightarrow P_2{V}_2=k{(V_2)}^{{\displaystyle \frac{1}{2}}}$
Using these two equations in equation number (1), we get:
$\begin{array}{rl}& \Rightarrow W=2P_2{V}_2(1-{\displaystyle \frac{k\sqrt{V_1}}{k\sqrt{V_2}}})\\ & \therefore W=2P_2{V}_2(1-{\displaystyle \frac{\sqrt{V_1}}{\sqrt{V_2}}})\end{array}$
Hence, the work done in taking gas from $1\to 2$ under the given polytropic process is $2P_2{V}_2(1-\sqrt{{\displaystyle \frac{V_1}{V_2}}})$ .


So, the correct answer is “Option C”.

Note: In a polytropic process, (n) can take any value. If we keep on varying (n) we will see that for different values of (n), we get different processes. For example: for, $n=0$ , the process will become isobaric. For, $n=1$, the process will become isothermal, etc. Thus all the basic processes are a subset of the polytropic process.