JEE Chemistry 2026 Question with Solution

Given: $V_1 = 20.0\,\text{dm}^3$, $T = 600\,\text{K}$, $P_1 = 0.5\,\text{MPa}$, $P_2 = 0.2\,\text{MPa}$. The process is isothermal reversible expansion of an ideal gas.

Find: The correct combination of $w,\ \Delta U,\ \Delta H,$ and $q$.

For an ideal gas in an isothermal process,

$$\Delta U = 0, \qquad \Delta H = 0$$

For reversible isothermal expansion,

$$w = -nRT \ln\left(\frac{V_2}{V_1}\right) = -nRT \ln\left(\frac{P_1}{P_2}\right)$$

and

$$q = -w$$

First,

$$\frac{P_1}{P_2} = \frac{0.5}{0.2} = 2.5$$

Using the given logarithms,

$$\ln(2.5) = 2.303(\log 5 - \log 2)$$ $$= 2.303(0.6989 - 0.3010)$$ $$= 0.916$$

Now calculate the number of moles:

$$n = \frac{PV}{RT}$$ $$= \frac{0.5 \times 10^6 \times 20 \times 10^{-3}}{8.314 \times 600}$$ $$\approx 2.0$$

Then,

$$w = -nRT\ln\left(\frac{P_1}{P_2}\right)$$ $$= -2 \times 8.314 \times 600 \times 0.916$$ $$\approx -9.1\,\text{kJ}$$

Therefore,

$$\Delta U = 0, \qquad \Delta H = 0, \qquad q = +9.1\,\text{kJ}$$

So, the correct option is C.

Given: The gas is ideal and the process is isothermal.

Find: How heat and work are related in this process.

For an ideal gas, internal energy and enthalpy depend only on temperature. Since temperature remains constant,

$$\Delta U = 0, \qquad \Delta H = 0$$

From the first law of thermodynamics,

$$\Delta U = q + w$$

So,

$$0 = q + w$$ $$q = -w$$

Hence, during reversible expansion, work done by the system is negative in chemistry sign convention, and an equal amount of heat is absorbed.

Using the calculated reversible isothermal work,

$$w \approx -9.1\,\text{kJ}$$

therefore,

$$q \approx +9.1\,\text{kJ}$$

Thus the final set is $w=-9.1\,\text{kJ},\ \Delta U=0,\ \Delta H=0,\ q=9.1\,\text{kJ}$, so the correct option is C.