20.0 dm^{3} of an ideal gas 'X' at 600 K an | vedclass

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Question

(A) For an isothermal process involving an ideal gas, the internal energy change $\Delta U$ and enthalpy change $\Delta H$ are zero because they depen

Vedclass · Accepted Answer

$w=-9.1 \ kJ, \Delta U=0, \Delta H=0, q=9.1 \ kJ$

Vedclass · Answer

(A) For an isothermal process involving an ideal gas, the internal energy change $\Delta U$ and enthalpy change $\Delta H$ are zero because they depend only on temperature, which remains constant.For an isothermal reversible expansion, the work done $w$ is given by $w = -nRT \ln(V_2/V_1) = -P_1V_1 \ln(P_1/P_2)$.Given $P_1 = 0.5 \ MPa = 0.5 	imes 10^6 \ Pa$, $V_1 = 20.0 \ dm^3 = 20.0 	imes 10^{-3} \ m^3$, and $P_2 = 0.2 \ MPa$.$w = -(0.5 	imes 10^6 \ Pa) 	imes (20.0 	imes 10^{-3} \ m^3) 	imes \ln(0.5/0.2)$.$w = -10000 \ J 	imes \ln(2.5) = -10000 	imes 2.303 	imes \log(2.5)$.Since $\log(2.5) = \log(5/2) = \log(5) - \log(2) = 0.6989 - 0.3010 = 0.3979$.$w = -10000 	imes 2.303 	imes 0.3979 \approx -9166 \ J \approx -9.1 \ kJ$.Since $\Delta U = q + w = 0$, then $q = -w = 9.1 \ kJ$.

$20.0 \ dm^{3}$ of an ideal gas '$X$' at $600 \ K$ and $0.5 \ MPa$ undergoes isothermal reversible expansion until the pressure of the gas is $0.2 \ MPa$. Which of the following options is correct? (Given: $\log\,2=0.3010$ and $\log\,5=0.6989$)

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