Question

Work done by air when it expands from 50L to 150L at a constant pressure of 2 atmosphere?
A. $2 \times {10^4}J$
B. $2 \times {10^2}J$
C. $2 \times {10^{11}}J$
D. $2 \times {10^{15}}J$

Answer 1

Hint: In the question we have given that a gas is expanding from 50L to 150L at constant pressure. Here, the expansion of gas is done at constant pressure, thus the external work done can be easily calculated using the work done formula $W = P\Delta V$ (Pressure-volume work). Before starting with the calculations we must convert 50L and 150L in terms of the $m^3$ as the options provided here are in terms of Joules.

Formula used:
Workdone by a gas is given by:
$W=P \Delta V=P \left({V_2}-{V_1}\right)$ Where,
${V_1}$-The initial volume
${V_2}$-The final volume

Complete answer:
Since the gas expands at a constant pressure of $P = 2atm \approx 2 \times {10^5}N/{m^2}$
During the expansion of gas, the volume changes as:
${V_1} = 50L = 50 \times {10^{ - 3}}{m^3}$ and
${V_2} = 150L = 150 \times {10^{ - 3}}{m^3}$ (given)
$\left( {\therefore 1L = {{10}^{ - 3}}{m^3}} \right)$

Therefore, change in volume can be given as:
$\Delta V = \;\;{V_2} - {V_1} = \left( {150 - 50} \right){10^{ - 3}} = 100 \times {10^{ - 3}}{m^3}$

Now, we know that Pressure-Volume Work in thermodynamics is defined as:
$Workdone = W = P\Delta V$

Substituting the value of $P$ and $\Delta V$ in the above expression, we get
$ \Rightarrow W = 2 \times {10^5}\left( {100 \times {{10}^{ - 3}}} \right)$
$ \Rightarrow W = 2 \times {10^8} \times {10^{ - 3}}$

Approximating the values, we get
$ \Rightarrow W = 2 \times {10^4}J = 20KJ$
$\left( {\therefore 1KJ = {{10}^3}J} \right)$
Thus, the work done by air when it expands at constant pressure is about $2 \times {10^4}J$.

Hence, the correct solution is (A) $2 \times {10^4}J$.

Note: We know that the air is made up of constantly moving molecules. Therefore, when the air expands and becomes less compressed, the molecules need more space to move around. As a result, work is done by the molecules of air in expanding the space around each molecule which is calculated by using the above stated relation.