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(C) Given the condition $V \propto T^{2/3}$.Using the ideal gas equation $PV = nRT$,we have $P = \frac{nRT}{V}$.Since $V \propto T^{2/3}$,we can write

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$166.2$

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(C) Given the condition $V \propto T^{2/3}$.Using the ideal gas equation $PV = nRT$,we have $P = \frac{nRT}{V}$.Since $V \propto T^{2/3}$,we can write $V = cT^{2/3}$,which implies $T \propto V^{3/2}$.Substituting this into the ideal gas equation: $P \propto \frac{T}{V} \propto \frac{V^{3/2}}{V} = V^{1/2}$.Thus,$P = kV^{1/2}$ for some constant $k$.The work done $W$ is given by $W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} kV^{1/2} \, dV$.$W = k \left[ \frac{V^{3/2}}{3/2} ight]_{V_1}^{V_2} = \frac{2}{3} [kV_2^{3/2} - kV_1^{3/2}] = \frac{2}{3} [P_2V_2 - P_1V_1]$.Using $PV = nRT$,we get $W = \frac{2}{3} nR(T_2 - T_1) = \frac{2}{3} nR \Delta T$.Given $n = 1 \ mol$,$\Delta T = 30 \ K$,and $R = 8.314 \ J/mol \cdot K$ (approximated as $8.3$ in calculation).$W = \frac{2}{3} 	imes 1 	imes 8.314 	imes 30 = 20 	imes 8.314 = 166.28 \ J$.Rounding to the nearest option,$W = 166.2 \ J$.

Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} C$,if it is expanding under the condition $V \propto T^{2/3}$ is,$(R = 8.314 \ J/mol \cdot K)$ (in $J$)

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