Chemical Thermodynamics Part 3

Concept:For an ideal gas undergoing isothermal reversible expansion, the internal energy ($U$) and enthalpy ($H$) depend only on temperature.Since temperature is constant, $\Delta U = 0$ and $\Delta H = 0$.Explanation:Step 1: Convert the given data to SI units.Volume $V_1 = 20.0\ \text{dm}^3 = 20.0 \times 10^{-3}\ \text{m}^3 = 0.020\ \text{m}^3$Initial pressure $P_1 = 0.5\ \text{MPa} = 0.5 \times 10^6\ \text{Pa} = 5.0 \times 10^5\ \text{Pa}$Final pressure $P_2 = 0.2\ \text{MPa}$Temperature $T = 600\ \text{K}$Step 2: Calculate the number of moles ($n$) using the ideal gas equation $PV = nRT$.$$n = \frac{P_1 V_1}{RT} = \frac{(5.0 \times 10^5)(0.020)}{(8.314)(600)} = \frac{10000}{4988.4} \approx 2.0\ \text{mol}$$Step 3: For a reversible isothermal expansion, work done is given by:$$w = -nRT \ln\left(\frac{V_2}{V_1}\right) = -nRT \ln\left(\frac{P_1}{P_2}\right)$$Convert to base-10 logarithm:$$w = -2.303\ nRT\ \log\left(\frac{P_1}{P_2}\right)$$Calculate the pressure ratio:$$\frac{P_1}{P_2} = \frac{0.5}{0.2} = 2.5$$Using the given logarithms:$$\log(2.5) = \log\left(\frac{5}{2}\right) = \log 5 - \log 2 = 0.6989 - 0.3010 = 0.3979$$Now compute $nRT$:$$nRT = 2.0 \times 8.314 \times 600 = 9976.8\ \text{J}$$Thus, work done:$$w = -2.303 \times 9976.8 \times 0.3979 \approx -9130\ \text{J} = -9.1\ \text{kJ}$$Step 4: For an isothermal process involving an ideal gas, $\Delta U = 0$ and $\Delta H = 0$.Step 5: Apply the first law of thermodynamics: $\Delta U = q + w$.Since $\Delta U = 0$, we get $0 = q + (-9.1)$, so $q = +9.1\ \text{kJ}$.Answer:Option A:$w = -9.1\ \text{kJ}$, $\Delta U = 0$, $\Delta H = 0$, $q = +9.1\ \text{kJ}$