The gas performs isothermal irreversible work (W).

where, $$\Delta$$U = 0 (change in internal energy)

From, 1st law of thermodynamics,

$$\Rightarrow$$ $$\Delta$$U = $$\Delta$$Q + W

$$\Rightarrow$$ 0 = $$\Delta$$Q + W

$$\Rightarrow$$ $$\Delta$$Q = $$-$$W

Now, $$W = - {p_{ext}}({V_2} - {V_1})$$

$$ = - {p_{ext}}\left( {{{nRT} \over {{p_2}}} - {{nRT} \over {{p_1}}}} \right) = - {p_{ext}} \times nRT\left( {{1 \over {{p_2}}} - {1 \over {{p_1}}}} \right)$$

Given, p_ext = 4.3 MPa, p₁ = 2.1 MPa, p₂ = 1.3 MPa, n = 5 mol, T = 293 K and R = 8.314 J mol^$$-$$1 K^$$-$$1

$$ = - 4.3 \times 5 \times 8.314 \times 293\left( {{1 \over {1.3}} - {1 \over {2.1}}} \right)$$

= $$-$$ 15347.70 J mol^$$-$$1

= $$-$$ 15.347 kJ mol^$$-$$1 $$ \simeq $$ $$-$$ 15 kJ mol^$$-$$1

$$\Rightarrow$$ $$\Delta$$Q = 15 kJ mol^$$-$$1