Question

A certain balloon maintains an internal gas pressure of \[{{P}_{o}}=100kPa\] until the volume reaches ${{V}_{o}}=20{{m}^{3}}$. Beyond the volume of $20{{m}^{3}}$, the internal pressure varies as $P={{P}_{o}}+2k{{\left( V-{{V}_{o}} \right)}^{2}}$, where $P$ is in$kPa$, $V$ is ${{m}^{3}}$ in and $k$ is a constant $(k=1kPa/{{m}^{3}})$. Initially the balloon contains helium gas at ${{20}^{\circ }}C$, $100kPa$with a $15{{m}^{3}}$ volume. The balloon is then heated until the volume becomes $25{{m}^{3}}$ and the pressure is $150kPa$. Assume ideal gas behaviour of helium. The work done by the balloon for the entire process in $KJ$ is
\[A.1256\]
$B.1414$
$C.1083$
$D.1512$

Answer 1

Hint: We know that this problem is from the application of First Law of Thermodynamics and Kinetic theory of gases. So, we will apply the basic idea of the first law of thermodynamics to find the work done by the balloon for the entire process. We will consider the idea of ideal gas behaviour of helium for the entire process.
Formula used:
We will use the formula $W=\int\limits_{{{v}_{i}}}^{{{v}_{f}}}{Pdv}$ to find work done in this case.

Complete answer:
From the above given problem we have following parameters:-
${{P}_{o}}=100kPa$
$P={{P}_{o}}+2k{{\left( V-{{V}_{o}} \right)}^{2}}$, where $V$ and ${{V}_{o}}$ denote the volumes at different points.
$W=$ Work done for the given process.
Now, we will use the following formula,
$W=\int\limits_{{{v}_{i}}}^{{{v}_{f}}}{Pdv}$………………… (i)
We know that, this relation is used to find the work done$W$, which is equal to the integration of Pressure from initial value of volume ${{v}_{i}}$to the final value of volume${{v}_{f}}$.
According to the case given in this problem we will calculate the work done by the balloon with the application of equation (i).
$W=\int\limits_{15}^{20}{{{P}_{o}}}dV+\int\limits_{20}^{25}{PdV}$
$W=\int\limits_{15}^{20}{{{P}_{o}}dV+\int\limits_{20}^{25}{\left[ {{P}_{0}}+2{{\left( V-{{V}_{o}} \right)}^{2}} \right]}}dV$, as\[1kPa/{{m}^{3}}\]…………… (ii)
Now, putting values in equation (ii), we get,
$W={{10}^{5}}\left[ V \right]_{15}^{20}+{{10}^{5}}\times \left[ 25-20 \right]+2\times {{10}^{3}}\int\limits_{20}^{25}{{{\left( V-{{V}_{0}} \right)}^{2}}dV}$
Calculating further,
$W=5\times {{10}^{5}}+5\times {{10}^{5}}+2\times {{10}^{3}}\left[ \dfrac{{{\left( V-{{V}_{o}} \right)}^{3}}}{3} \right]_{20}^{25}$
$W={{10}^{6}}+2\times {{10}^{3}}\left[ \dfrac{{{\left( 25-20 \right)}^{3}}}{3} \right]$
On doing more calculations, we get,
$\Rightarrow W={{10}^{6}}+2\times {{10}^{3}}\times \dfrac{{{5}^{3}}}{3}$
$\Rightarrow W={{10}^{6}}+\dfrac{2\times {{10}^{3}}\times 125}{3}$
$\Rightarrow W={{10}^{6}}+\dfrac{250\times {{10}^{3}}}{3}$
$\Rightarrow W=10.83\times {{10}^{5}}J$
$\Rightarrow W=1083kJ$.
Therefore, we got the required answer.

So, the correct answer is “Option C”.

Note:
Work done for a thermodynamic process is a path dependent process. It is calculated using the concept of the first law of thermodynamics. We should take care of the given process as the different thermodynamic process requires different concepts. Different types of thermodynamic processes are isobaric process, isochoric process, adiabatic process and isothermal process. Therefore, identification of thermodynamic processes is an important task in solving these types of problems.