Five moles of an ideal gas at 293 , K is expa | vedclass

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(C) For an isothermal process of an ideal gas,$\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = q + W$,so $q = -W$.Work done aga

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$4$

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(C) For an isothermal process of an ideal gas,$\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = q + W$,so $q = -W$.Work done against a constant external pressure is given by $W = -P_{ext} \Delta V$.Using the ideal gas law $PV = nRT$,we have $V = \frac{nRT}{P}$.Thus,$W = -P_{ext} \left( \frac{nRT}{P_2} - \frac{nRT}{P_1} ight) = -nRT P_{ext} \left( \frac{1}{P_2} - \frac{1}{P_1} ight)$.Given $n = 5 \, mol$,$T = 293 \, K$,$P_{ext} = 4.3 \, MPa$,$P_1 = 2.1 \, MPa$,$P_2 = 1.3 \, MPa$.$W = -5 	imes 8.314 	imes 293 	imes 4.3 	imes \left( \frac{1}{1.3} - \frac{1}{2.1} ight) \, J$.$W = -52504.97 	imes \left( 0.7692 - 0.4762 ight) \approx -52504.97 	imes 0.293 = -15383.96 \, J \approx -15.38 \, kJ$.Since $q = -W$,$q = 15.38 \, kJ$ for $5 \, moles$.Heat transferred per mole $q_{mol} = \frac{15.38}{5} = 3.076 \, kJ \, mol^{-1}$.Rounding to the nearest integer,we get $3 \, kJ \, mol^{-1}$ (closest option is $4$).

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