QJEEM24Che3cd: JEE Main 2024 Chemistry - Class 11 JEEM24Che3cd Solution: If three moles of an ideal gas at expand isothermally from to…

The process involves an ideal gas expanding isothermally, meaning temperature ( $T$ ) remains constant. In an isothermal process for an ideal gas, the change in internal energy ( $\Delta U$ ) is zero:

$\Delta U = 0$ because temperature is constant.

According to the first law of thermodynamics, $\Delta U = Q + W$ , where $Q$ is the heat transferred to the system and $W$ is the work done by the system. Since $\Delta U = 0$ in an isothermal process, $Q = -W$ .

The work done by the gas during the isothermal expansion against a constant external pressure ( $P_{ext}$ ) is given by:

$W = P_{ext} \times \Delta V$

where:

• $P_{ext}$ = constant opposing pressure = $80$ kPa = $80 \times 10^3$ Pa (since $1$ kPa = $10^3$ Pa)
• $\Delta V$ = change in volume = $45$ dm³ - $30$ dm³ = $15$ dm³ = $15 \times 10^{-3}$ m³ (since $1$ dm³ = $10^{-3}$ m³)

$W = -80 \times 10^3 \times 15 \times 10^{-3} = -1200$ J

Therefore, the amount of heat transferred ( $Q$ ) is $1200$ J, taking into consideration the sign convention that work done by the system is negative.