Question

Two moles of an ideal gas at temperature, ${{T}_{0}}=300K$ was cooled down isochorically so that the pressure was reduced to half. Then, in an isobaric process, the gas expanded till its temperature got back to the initial value. Find work done in the isobaric process.

Answer 1

Hint: Since the first process is an isochoric process, that is in this process there will be no change in the volume of the gas, therefore the net work done in this step will be equal to Zero. So, the total heat released will bring about a change in its internal energy only. Also, in the second process we can make use of properties under isobaric conditions that is the pressure on the gas remains constant with time.

Complete answer:
Let the initial thermodynamic parameters of the gas be $({{P}_{0}},{{V}_{0}},{{T}_{0}})$ ,then using Ideal gas equation we can write:
$\Rightarrow {{P}_{0}}{{V}_{0}}=nR{{T}_{0}}$
It is given in the problem:
$\begin{align}
  & \Rightarrow n=2 \\
 & \Rightarrow {{T}_{0}}=300K \\
\end{align}$
Then, ${{V}_{0}}$ can be written as:
$\Rightarrow {{V}_{0}}=\dfrac{2R{{T}_{0}}}{{{P}_{0}}}$ [Let this be equation number (1)]
Since, first process is isochoric, then:
$\Rightarrow {{W}_{isochoric}}=0$
Also, at constant volume:
Pressure is directly proportional to temperature.
Therefore, at the end of first step the thermodynamic parameters will be:
Pressure, ${{P}_{1}}=\dfrac{{{P}_{0}}}{2}$
Volume, ${{V}_{1}}={{V}_{0}}$
Temperature, ${{T}_{1}}=\dfrac{{{T}_{o}}}{2}$
Now, in the second step since the process is isobaric, then:
Volume is directly proportional to temperature.
Therefore, at the end of second step:
Pressure, ${{P}_{2}}={{P}_{1}}=\dfrac{{{P}_{0}}}{2}$
Volume, ${{V}_{2}}=2{{V}_{0}}=\dfrac{4R{{T}_{0}}}{{{P}_{0}}}$ [using the value of ${{V}_{0}}$ from equation number (1)]
Temperature, ${{T}_{2}}=2{{T}_{1}}={{T}_{0}}$
The work done will be given by:
$\Rightarrow {{W}_{isobaric}}={{P}_{ext}}(\vartriangle V)$
Here, ${{P}_{ext}}=\dfrac{{{P}_{0}}}{2}$
And,
$\begin{align}
  & \Rightarrow \vartriangle V={{V}_{2}}-{{V}_{1}} \\
 & \Rightarrow \vartriangle V=\left( \dfrac{4R{{T}_{0}}}{{{P}_{0}}}-\dfrac{2R{{T}_{0}}}{{{P}_{0}}} \right) \\
\end{align}$
Let the work done in isobaric process be denoted as ${{W}_{ib}}$
Therefore, work done in the isobaric process will be equal to:
$\begin{align}
  & \Rightarrow {{W}_{ib}}=\left( \dfrac{{{P}_{0}}}{2} \right)\left( \dfrac{4R{{T}_{0}}}{{{P}_{0}}}-\dfrac{2R{{T}_{0}}}{{{P}_{0}}} \right) \\
 & \Rightarrow {{W}_{ib}}=\dfrac{{{P}_{0}}}{2}\times \dfrac{2R{{T}_{0}}}{{{P}_{0}}} \\
 & \Rightarrow {{W}_{ib}}=R{{T}_{0}} \\
 & \Rightarrow {{W}_{ib}}=8.3\times 300J \\
 & \Rightarrow {{W}_{ib}}=2490J \\
\end{align}$
Hence, the work done in the Isobaric process is $2490J$ .

Note:
We should always remember what these thermodynamic terms stand for and their proper use in a question. We should not confuse one with the other. Also in these types of lengthy problems, we should always check our calculation at every step and make sure we are using the correct formula.