JEE Chemistry 2024 Question with Solution

Given: three moles of an ideal gas at $300 \, \text{K}$ undergo isothermal expansion from $30 \, \text{dm}^3$ to $45 \, \text{dm}^3$ against constant opposing pressure $80 \, \text{kPa}$.

Find: heat transferred $Q$.

For an isothermal process of an ideal gas, $\Delta U = 0$. Using the first law of thermodynamics,

$$\Delta U = Q + W$$

so,

$$0 = Q + W$$

Therefore,

$$Q = -W$$

The work done is calculated using external pressure:

$$W = -P_{\text{ext}} \Delta V$$

Here,

$$\Delta V = V_f - V_i = 45 - 30 = 15 \, \text{dm}^3 = 0.015 \, \text{m}^3$$

and

$$P_{\text{ext}} = 80 \, \text{kPa} = 80{,}000 \, \text{Pa}$$

Substituting,

$$W = -80{,}000 \times 0.015 = -1200 \, \text{J}$$

Hence,

$$Q = -(-1200) = 1200 \, \text{J}$$

Therefore, the amount of heat transferred is $1200 \, \text{J}$. The solution concludes this value, so the numerical answer is 1200.

Given: $P_{\text{ext}} = 80 \, \text{kPa}$, $V_i = 30 \, \text{dm}^3$, $V_f = 45 \, \text{dm}^3$, and the process is isothermal.

Find: $Q$.

First convert the volume change into SI units:

$$\Delta V = 45 - 30 = 15 \, \text{dm}^3$$

Since

$$1 \, \text{dm}^3 = 10^{-3} \, \text{m}^3$$

we get

$$\Delta V = 15 \times 10^{-3} = 0.015 \, \text{m}^3$$

Now calculate work:

$$W = -P_{\text{ext}} \Delta V = -80 \times 10^3 \times 0.015 = -1200 \, \text{J}$$

For an ideal gas in an isothermal process,

$$\Delta U = 0$$

Hence from

$$\Delta U = Q + W$$

we obtain

$$Q = -W = 1200 \, \text{J}$$

Therefore, the heat transferred to the gas is $1200 \, \text{J}$.