Question

One mole of ideal gas whose adiabatic exponent $\gamma =\dfrac{4}{3}$ undergoes a process $P=200+\dfrac{1}{V}$ then change in internal energy of gas when volume changes from $2{{m}^{3}}$ to $4{{m}^{3}}$ is?
a) 400 J
b) 800 J
c) 1200 J
d) 600 J

Answer 1

Hint: Since the gas undergoes a process as: $P=200+\dfrac{1}{V}$, where P is the pressure and V is the volume of gas. For \[{{V}_{1}}=2{{m}^{3}};{{V}_{2}}=4{{m}^{3}}\], find ${{P}_{1}}$, ${{P}_{2}}$, ${{T}_{1}}$ and ${{T}_{2}}$. As we know that: internal energy is given as $\Delta U={{C}_{v}}\left( {{T}_{2}}-{{T}_{1}} \right)$ where ${{C}_{v}}$ is specific heat capacity at constant volume and is given as ${{C}_{v}}=\dfrac{R}{\gamma -1}$. Use these relations to find change in internal energy.

Complete step by step answer:
We have: \[{{V}_{1}}=2{{m}^{3}};{{V}_{2}}=4{{m}^{3}}\]
As we know that the gas undergoes the given process: $P=200+\dfrac{1}{V}$
So, by using this formula, we get:
\[\begin{align}
& {{P}_{1}}=200+\dfrac{1}{2} \\
& =200.5Pa
\end{align}\]
$\begin{align}
& {{P}_{2}}=200+\dfrac{1}{4} \\
& =200.25Pa
\end{align}$
Now, by ideal gas equation: $PV=nRT$ for n = 1,
We have:
$\begin{align}
& {{T}_{1}}=\dfrac{200.5\times 2}{R} \\
& =\dfrac{401}{R}
\end{align}$
$\begin{align}
& {{T}_{2}}=\dfrac{200.25\times 4}{R} \\
& =\dfrac{801}{R}
\end{align}$
Now, we have to find change in internal energy, i.e. $\Delta U={{C}_{v}}\left( {{T}_{2}}-{{T}_{1}} \right)$
For ${{C}_{v}}=\dfrac{R}{\gamma -1}$, we have:
$\begin{align}
& \Delta U=\dfrac{R}{\dfrac{4}{3}-1}\left( \dfrac{801}{R}-\dfrac{401}{R} \right) \\
& =3\times 400 \\
& =1200J
\end{align}$.

So, the correct answer is “Option C”.

Additional Information:
Internal energy is defined as the energy associated with the random, disordered motion of molecules. It is separated in scale from the macroscopic ordered energy associated with moving objects; it refers to the invisible microscopic energy on the atomic and molecular scale.

Note:
We do not use the value of universal gas constant (R) in the solution because the value of R changes according to the units of other parameters. So, we have pressure in atm, volume in litres, temperature in kelvin. So, the value of R is 8.314 J/mol K.