One mole of an ideal monoatomic gas initially at 1200 K and 64 atm is expanded to a final state at 3 — Thermodynamics and Thermochemistry Chemistry Question
Question
One mole of an ideal monoatomic gas initially at 1200 K and 64 atm is expanded to a final state at 300 K and 1 atm. To achieve the above change, a reversible path is constructed that involve an adiabatic expansion in the beginning followed by an isothermal expansion to the final state. The magnitude of net work done by the gas (in cal) is
💡 Solution & Explanation
Step 1 (Adiabatic to 300K): $T_2 = 300\text{ K}$. $W_{\text{adia}} = n C_v (T_2 - T_1) = 1 \times (1.5 \times 2) \times (300 - 1200) = 3 \times (-900) = -2700\text{ cal}$. $P_2$ is found by $T_1 P_1^{\frac{1-\gamma}{\gamma}} = T_2 P_2^{\frac{1-\gamma}{\gamma}} \implies 1200 (64)^{-2/5} = 300 P_2^{-2/5} \implies P_2 = 2\text{ atm}$. Step 2 (Isothermal to 1 atm): $W_{\text{iso}} = -n R T_2 \ln(P_2/P_3) = -1 \times 2 \times 300 \ln(2) = -600 \times 0.7 = -420\text{ cal}$. Total work = $-2700 - 420 = -3120\text{ cal}$. Magnitude is 3120.