QJEEM25Cheecf: JEE Main 2025 Chemistry - Class 11 JEEM25Cheecf Solution: 500 J of energy is transferred as heat to 0.5 mol of Argon gas…

$C_P = \frac{5}{2} R \qquad \text{and} \qquad C_V = \frac{3}{2} R$

For a constant pressure process, the heat added is related to the temperature change by

$Q = n C_P \Delta T.$

Given that

$Q = 500 \, \text{J},$

$n = 0.5 \, \text{mol},$

$R = 8.3 \, \text{J K}^{-1} \text{mol}^{-1},$

and the initial temperature $T_i = 298 \, \text{K},$

we first calculate $C_P$ :

$C_P = \frac{5}{2} \times 8.3 = 20.75 \, \text{J K}^{-1} \text{mol}^{-1}.$

Next, the temperature change is

$\Delta T = \frac{Q}{n C_P} = \frac{500 \, \text{J}}{0.5 \, \text{mol} \times 20.75 \, \text{J K}^{-1} \text{mol}^{-1}} \approx 48.2 \, \text{K}.$

So, the final temperature is

$T_f = T_i + \Delta T \approx 298 \, \text{K} + 48.2 \, \text{K} \approx 346.2 \, \text{K}.$

Rounded to the nearest units provided in the options, this is approximately $348 \, \text{K}.$

The change in internal energy is determined by

$\Delta U = n C_V \Delta T,$

where

$C_V = \frac{3}{2} R = \frac{3}{2} \times 8.3 = 12.45 \, \text{J K}^{-1} \text{mol}^{-1}.$

Then,

$\Delta U = 0.5 \, \text{mol} \times 12.45 \, \text{J K}^{-1} \text{mol}^{-1} \times 48.2 \, \text{K} \approx 300 \, \text{J}.$

Thus, the final answers are:

Final temperature: approximately $348 \, \text{K}$

Change in internal energy: approximately $300 \, \text{J}$

These results correspond to Option B.