JEE Chemistry 2025 Question with Solution

Given: $q = 500 \, \text{J}$, $n = 0.5 \, \text{mol}$, $T_1 = 298 \, \text{K}$, pressure is $1.00 \, \text{atm}$, and Argon is a monoatomic ideal gas.

Find: final temperature and change in internal energy.

From the solution working, for a monoatomic gas:

and at constant pressure:

Substitute the given values:

Hence the final temperature is:

Now calculate the change in internal energy using:

Therefore, the final temperature is $348 \, \text{K}$ and the change in internal energy is $300 \, \text{J}$. The correct option is A.

The solution contains a discrepancy because one section states C, but the worked constant-pressure calculation clearly gives $348 \, \text{K}$ and $300 \, \text{J}$, which matches option A.

Given: $q = 500 \, \text{J}$ at constant pressure, with Argon as a monoatomic ideal gas.

Find: $T_2$ and $\Delta U$.

The first law is:

For a constant-pressure heating process of an ideal gas, the heat absorbed is:

For monoatomic gas:

So,

and therefore,

Now use internal energy relation:

Thus, the correct result from the extracted working is $348 \, \text{K}$ and $300 \, \text{J}$.