Work done by 0.1 mole of a gas at 27^{circ}C | vedclass

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(C) For an isobaric process,the work done is given by $W = P \Delta V$.Using the ideal gas equation $PV = nRT$,we can write $W = nR \Delta T$.Given:Nu

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) For an isobaric process,the work done is given by $W = P \Delta V$.Using the ideal gas equation $PV = nRT$,we can write $W = nR \Delta T$.Given:Number of moles $n = 0.1 \, mol$.Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \, K$.Since the pressure is constant,by Charles's Law,$V \propto T$. If the volume doubles $(V_2 = 2V_1)$,the temperature must also double $(T_2 = 2T_1 = 600 \, K)$.Change in temperature $\Delta T = T_2 - T_1 = 600 - 300 = 300 \, K$.Substituting the values into the work formula:$W = 0.1 	imes 2 	imes 300 = 60 \, cal$.

Work done by $0.1$ mole of a gas at $27^{\circ}C$ to double its volume at constant pressure is ....... $cal$ $(R = 2 \, cal \, mol^{-1} \, K^{-1})$.

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