JEE Physics 2026 Question with Solution

Given: $n = 10$ mol, temperature changes from $30 \, ^{\circ}\text{C}$ to $40 \, ^{\circ}\text{C}$, $C_p = 7 \, \text{cal/mol} \, ^{\circ}\text{C}$, and $R = 2 \, \text{cal/mol} \, ^{\circ}\text{C}$.

Find: Change in internal energy $\Delta U$.

For an ideal gas,

$$\Delta U = n C_v \Delta T$$

This relation is valid for any process. For constant volume, heat supplied equals the change in internal energy.

Using

$$C_p - C_v = R$$

we get

$$7 - C_v = 2$$ $$C_v = 5 \, \text{cal/mol} \, ^{\circ}\text{C}$$

Now,

$$\Delta T = 40 - 30 = 10 \, ^{\circ}\text{C}$$

So,

$$\Delta U = 10 \times 5 \times 10 = 500 \, \text{cal}$$

Therefore, the change in internal energy is $500 \, \text{cal}$.

Given: The gas is heated at constant volume, with $n = 10$ mol, $C_p = 7 \, \text{cal/mol} \, ^{\circ}\text{C}$, and $R = 2 \, \text{cal/mol} \, ^{\circ}\text{C}$.

Find: The value of $\Delta U$.

The required quantity is internal energy change, so the needed molar heat capacity is $C_v$, not $C_p$.

First use the relation

$$C_p - C_v = R$$

Substituting the values,

$$7 - C_v = 2$$ $$C_v = 5 \, \text{cal/mol} \, ^{\circ}\text{C}$$

Now calculate the temperature rise:

$$\Delta T = 40 - 30 = 10 \, ^{\circ}\text{C}$$

For an ideal gas,

$$\Delta U = n C_v \Delta T$$

Hence,

$$\Delta U = 10 \times 5 \times 10$$ $$\Delta U = 500 \, \text{cal}$$

Thus, the numerical answer is 500.