frac{1}{24.63} mole of an ideal mono-atomic g | vedclass

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(B) Given: $n = \frac{1}{24.63} \ mol$,$PV^3 = C$,$V_1 = 1 \ L$,$T_1 = 300 \ K$,$V_2 = 2 \ L$.Since $PV = nRT$,$P = \frac{nRT}{V}$. Substituting into

Vedclass · Accepted Answer

$75.9$

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(B) Given: $n = \frac{1}{24.63} \ mol$,$PV^3 = C$,$V_1 = 1 \ L$,$T_1 = 300 \ K$,$V_2 = 2 \ L$.Since $PV = nRT$,$P = \frac{nRT}{V}$. Substituting into $PV^3 = C$,we get $\frac{nRT}{V} \cdot V^3 = C$,so $TV^2 = K'$ (constant).$T_1 V_1^2 = T_2 V_2^2$ $\Rightarrow 300 	imes (1)^2 = T_2 	imes (2)^2$ $\Rightarrow T_2 = \frac{300}{4} = 75 \ K$.Work done $w = -\int_{V_1}^{V_2} P \ dV = -\int_{1}^{2} \frac{C}{V^3} \ dV$. Since $P_1 V_1^3 = C$,$P_1 = \frac{nRT_1}{V_1} = \frac{(1/24.63) 	imes 0.0821 	imes 300}{1} = 1 \ atm$. Thus $C = 1 	imes 1^3 = 1 \ L^2 \cdot atm$.$w = -\int_{1}^{2} V^{-3} \ dV = -[\frac{V^{-2}}{-2}]_1^2 = \frac{1}{2} [\frac{1}{4} - 1] = -0.375 \ L \cdot atm$.Change in internal energy $\Delta U = nC_v \Delta T = \frac{1}{24.63} 	imes \frac{3}{2} 	imes 0.0821 	imes (75 - 300) = -1.125 \ L \cdot atm$.Using $\Delta U = q + w$,we get $-1.125 = q - 0.375 \Rightarrow q = -0.75 \ L \cdot atm$.Since $1 \ L \cdot atm = 101.3 \ J$,$q = -0.75 	imes 101.3 = -75.975 \ J$.The heat lost is $75.9 \ J$.

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