Thermodynamics Question: Please show steps and work for the following questions.

Question

Thermodynamics Question: Please show steps and work for the following questions.

Consider the following reaction at constant pressure at 600 K: CO (g) + 1/4 02 (g) CO2 (g) a) Calculate deltaH degree R and deltaG degree R at 298.15 K for this reaction. Hint: Use the thermodynamic table. b) In order to calculate deltaG degree R at different temperature, derive the expression delta GR in terms of temperature and delta H degree R assuming the enthalpy is constant in the temperature interval. Hint: Use Gibbs-Helmholtz equation. c) Calculate delta GR at 600 K using the equation from (b).

Explanation / Answer

First, PV^k = constant for this process, and k is Cp/Cv. Since He is ideal, Cv = 1.5R and Cp = 2.5R, so k = 5/3. Therefore, (P_2)/(P_1) = (V_2)/(v_1)^(-k). Since (P_2)/(P_1) = 7, it follows that (V_2)/(V_1) = 7^(-5/3).

Since He is ideal, (T_2)/(T_1) = (P_2)/(P_1)*(V_2_/(V_1) = 7*7^(-5/3) = 7^(-2/3). Since you know T_1 and P_1, you now know P_2 and T_2 as well. The change in internal energy is Cv*(T_2-T_1), so this is now known, too.

To find the work, you must integrate PdV from the initial state to the final state. Since PV^k is constant, it must be true that PV^k = (P_1)(V_1)^k. But V_1 = mR(T_1)/(P_1), so you get PV^k = (mrT_1)^k*(P_1)^(k-1). You know all these numbers, so you can calculate this constant. Let's call it A. Then P=A*V^(-k) , so the work integral turns into integral(A*V^(-k)dV) between the initial and final volumes. You already have the initial V, and the final is found the same way, by V_2 = mR(T_2)/(P_2), so you have the limits of integration too. The integral is easy, and once you have the work, you can get the heat from delta U = Q-W.

Hope this helps!

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