QJEEM21Ch9225: JEE Main 2021 Chemistry - Class 11 JEEM21Ch9225 Solution: Five moles of an ideal gas at 293 K is expanded isothermally…

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}