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^$-$1. (Rounded off to the nearest integer) [Use R = 8.314 J mol^$-$1K^$-$1]

<p>The gas performs isothermal irreversible work (W).</p> <p>where, $\Delta$U = 0 (change in internal energy)</p> <p>From, 1st law of thermodynamics,</p> <p>$\Rightarrow$ $\Delta$U = $\Delta$Q + W</p> <p>$\Rightarrow$ 0 = $\Delta$Q + W</p> <p>$\Rightarrow$ $\Delta$Q = $-$W</p> <p>Now, $W = - {p_{ext}}({V_2} - {V_1})$</p> <p>$$ = - {p_{ext}}\left( {{{nRT} \over {{p_2}}} - {{nRT} \over {{p_1}}}} \right) = - {p_{ext}} \times nRT\left( {{1 \over {{p_2}}} - {1 \over {{p_1}}}} \right)$$</p> <p>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</p> <p>$$ = - 4.3 \times 5 \times 8.314 \times 293\left( {{1 \over {1.3}} - {1 \over {2.1}}} \right)$$</p> <p>= $-$ 15347.70 J mol^$-$1</p> <p>= $-$ 15.347 kJ mol^$-$1 $\simeq$ $-$ 15 kJ mol^$-$1</p> <p>$\Rightarrow$ $\Delta$Q = 15 kJ mol^$-$1</p>