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.