Question

Air that initially occupies $0.140{{m}^{3}}$ at a gauge pressure of $103.0kPa$ is expanded isothermally to a pressure of $101.3kPa$ and then cooled at a constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)

Answer 1

Hint: Here we have to use the ideal gas equation and by using this formula we can find the temperature T and also we use the formula of work for the isothermal process. Isothermal is a process in which temperature of a system remains constant and the SI unit of pressure is $joule$.

Complete solution:
Let ${{V}_{i}}$ be the initial volume and ${{V}_{f}}$ be the final volume, and let us assume that the gas expands from ${{V}_{i}}$ to ${{V}_{f}}$ when the isothermal process is being done in the system. Thus, the work done by it is,
$W=\int\limits_{Vi}^{{{V}_{f}}}{pdV}$ eq. (1)

And now from the ideal gas equation we put the pressure equation by replacing P with$\dfrac{nRT}{V}$.
$W=nRT\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{\dfrac{dV}{V}}$
$W=nRT\ln \dfrac{{{V}_{f}}}{{{V}_{i}}}$ eq. (2)

So now from the equation, final volume is written as,
${{V}_{f}}=\dfrac{nRT}{{{P}_{f}}}$

And the initial volume is written as,
${{V}_{i}}=\dfrac{nRT}{{{P}_{i}}}$

And hence the ratio between both the volume will be,
$\dfrac{{{V}_{f}}}{{{V}_{i}}}=\dfrac{{{P}_{i}}}{{{P}_{f}}}$.

Now replace $nRT$ by ${{P}_{{{i}_{{}}}}}{{V}_{i}}$ hence the above equation (2) becomes,
$W={{P}_{i}}{{V}_{i}}\ln \dfrac{{{P}_{i}}}{{{P}_{f}}}$
The given initial gauge pressure is $1.03\times {{10}^{5}}Pa$.

Then the total gauge pressure is defined as,
${{P}_{t}}=1.03\times {{10}^{5}}+1.013\times {{10}^{5}}Pa$
$\Rightarrow {{P}_{t}}=2.04\times {{10}^{5}}Pa$

The final pressure is atmospheric pressure,
${{P}_{t}}=1.013\times {{10}^{5}}Pa$
Thus,
$W=(2.04\times {{10}^{5}}Pa)(0.14{{m}^{3}})\ln \dfrac{(2.04\times {{10}^{5}}Pa)}{(1.013\times {{10}^{5}}Pa)}$
$\Rightarrow W=2.00\times {{10}^{4}}J$

As it is mentioned above that during constant pressure when it reaches to its initial volume, so the work done by the gas is,
$W={{P}_{f}}({{V}_{i}}-{{V}_{f}})$

The gas starts in a state with pressure ${{P}_{f}}$, so throughout the process this is the pressure. We have to also notice that the volume decreases from ${{V}_{f}}$ to ${{V}_{i}}$. Now,
${{V}_{f}}=\dfrac{{{P}_{i}}{{V}_{i}}}{{{P}_{f}}}$

Thus the work done is,
$\begin{align}
  & W={{P}_{f}}({{V}_{i}}-\dfrac{{{P}_{i}}{{V}_{i}}}{{{P}_{f}}}) \\
 & \Rightarrow W=({{P}_{f}}-{{P}_{t}}){{V}_{i}} \\
 & \Rightarrow W=(1.013\times {{10}^{5}}Pa-2.04\times {{10}^{5}}Pa)(0.14{{m}^{3}}) \\
 & \therefore W=-1.44\times {{10}^{4}}J..............eq(3) \\
\end{align}$

The total work done by the gas over the entire process is,
$W=2.00\times {{10}^{4}}J-1.44\times {{10}^{4}}J$

$\therefore W=5.60\times {{10}^{3}}J$

Note:
Now one question arises why the total pressure is positive and why the work done we find in equation (3) is negative, as in pressure it is positive due to this is an expansion and total pressure is negative because there is compression.