If three moles of an ideal gas at $300 \mathrm{~K}$ expand isothermally from $30 \mathrm{~dm}^3$ to $45 \mathrm{~dm}^3$ against a constant opposing pressure of $80 \mathrm{~kPa}$, then the amount of heat transferred is _______ J.

<p>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:</p> <p>$\Delta U = 0$ because temperature is constant.</p> <p>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, $Q = -W$.</p> <p>The work done by the gas during the isothermal expansion against a constant external pressure ($P_{ext}$) is given by:</p> <p>$W = P_{ext} \times \Delta V$</p> <p>where:</p> <ul> <li>$P_{ext}$ = constant opposing pressure = $80$ kPa = $80 \times 10^3$ Pa (since $1$ kPa = $10^3$ Pa)</li> <li>$\Delta V$ = change in volume = $45$ dm³ - $30$ dm³ = $15$ dm³ = $15 \times 10^{-3}$ m³ (since $1$ dm³ = $10^{-3}$ m³)</li> </ul> <p>$W = -80 \times 10^3 \times 15 \times 10^{-3} = -1200$ J</p> <p>Therefore, the amount of heat transferred ($Q$) is $1200$ J, taking into consideration the sign convention that work done by the system is negative.</p>