Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external pressure 4.3 MPa. The heat transferred in this process is ______ kJ mol$$^{-1}$$.
(Rounded-off to the nearest integer)
[Use $$R = 8.314$$ J mol$$^{-1}$$ K$$^{-1}$$]

For an isothermal process involving an ideal gas, the internal energy change is zero ($$\Delta U = 0$$), so from the first law of thermodynamics, $$q = -w$$, where $$w$$ is the work done on the gas.

For an irreversible expansion against a constant external pressure $$P_{ext}$$, the work done on the gas is $$w = -P_{ext}(V_2 - V_1)$$, and therefore $$q = P_{ext}(V_2 - V_1)$$.

Using the ideal gas law, the initial and final volumes for 5 moles at 293 K are: $$V_1 = \frac{nRT}{P_1} = \frac{5 \times 8.314 \times 293}{2.1 \times 10^6} = \frac{12180}{2.1 \times 10^6} = 5.80 \times 10^{-3}$$ m$$^3$$ and $$V_2 = \frac{nRT}{P_2} = \frac{12180}{1.3 \times 10^6} = 9.37 \times 10^{-3}$$ m$$^3$$.

So $$\Delta V = V_2 - V_1 = (9.37 - 5.80) \times 10^{-3} = 3.57 \times 10^{-3}$$ m$$^3$$.

The total heat transferred is $$q = P_{ext} \times \Delta V = 4.3 \times 10^6 \times 3.57 \times 10^{-3} = 15351$$ J $$= 15.35$$ kJ.

Since the question asks for heat transferred per mole: $$q_{\text{per mol}} = \frac{15.35}{5} = 3.07 \approx 3$$ kJ mol$$^{-1}$$.

Rounding to the nearest integer, the heat transferred is $$\boxed{3}$$ kJ mol$$^{-1}$$.