Question

A thermally insulated rigid container of the one-litre volume contains a diatomic ideal gas at room temperature. A small paddle installed inside the container is rotated from the outside such that the pressure rises by ${10}^5$ Pa. The change in internal energy is to:
A) 0 J
B) 67 J
C) 150 J
D) 250 J

Answer 1

Hint: The heat capacity at constant volume ($C_V$) for diatomic gas is ${\displaystyle \frac{5}{2}}R$
For ideal gas, PV = nRT
Where P is pressure, V is volume is temperature is number of moles and R is a constant Rydberg constant.

Complete Solution :
So in the question it is given that, there is a thermally insulated container with volume of 1L which contains diatomic gas and there is a change in pressure due to the action of a small paddle installed inside the container. We have to find the internal energy, for that,
We know the equation relating internal energy, molal heat capacity of gas at constant volume $C_V$, temperature and number of moles of gas as-
$$\mathrm{\Delta }U=n{C}_V\mathrm{\Delta }T$$

- Here volume remains the same. There is only change in pressure parameter. As pressure changes which results in the change in temperature also.
The heat capacity at constant volume ($C_V$) for diatomic gas is ${\displaystyle \frac{5}{2}}$R
So the equation for internal energy becomes,
-$$\mathrm{\Delta }U={\displaystyle \frac{5}{2}}nR\mathrm{\Delta }T$$
By ideal gas equation, $PV=nRT$

But in this case only volume changes so the equation is written as,
-$\mathrm{\Delta }PV=nR\mathrm{\Delta }T$
Comparing the equation of internal energy and ideal gas equation we can rearrange and write the equation as,
$\mathrm{\Delta }U={\displaystyle \frac{5}{2}}\mathrm{\Delta }PV$
Now substitute the values,
Pressure (P) = ${10}^{5}Pa$
Volume (V) = 1L = ${\displaystyle \frac{1}{1000}}{m}^3={10}^{-3}{m}^3$

Substituting the values we get,
$$\mathrm{\Delta }U={\displaystyle \frac{5}{2}}\times {10}^5\times {\displaystyle \frac{1}{{10}^{-3}}}$$
$$\mathrm{\Delta }U={\displaystyle \frac{5}{2}}\times 100=250J$$
So, the correct answer is “Option D”.

Note: If in the place of diatomic gas, monoatomic was given then the value of $C_V$ is ${\displaystyle \frac{3R}{2}}$
Values must be substituted in the final equation, after converting all the values to standard form.