6. a) Show for constant temperature the maximum work is given by \[ dw_{\text{max}} = dU - TdS = dA \] Hint: Start with the Clausius inequality, the definition of the internal energy \( dU \), and remember \( w \) is negative when done by the system on the surroundings.b) A similar process at constant \( T \) and \( P \) leads to the expression \[ dG = dw_{\text{max,add}} \]Where \( w_{\text{max,add}} \) is all work except expansion work. For measurable changes we can write \(\Delta G = w_{\text{max,add}} \). Use this result to calculate the maximum non-expansion work from the combustion of \( \text{C}_3\text{H}_8 \) at STP given \(\Delta_f G^\circ(\text{H}_2\text{O}) = -237 \text{ kJ mol}^{-1}\), \(\Delta_f G^\circ(\text{CO}_2) = -394 \text{ kJ mol}^{-1}\), \(\Delta_f G^\circ(\text{C}_3\text{H}_8) = -23 \text{ kJ mol}^{-1}\).

a) Show for constant temperature the maximum work is given by dwmax = dU – TdS = dA Hint: start with the Clausius inequality, the definition of the internal energy dU, and remember w is negative when done by the system on the surroundings. b) A similar process at constant T and P leads to the expression dG = dwmax,add Where wmax,add is all work except expansion work. For measurable changes we can Wmax,add. Use this result to calculate the maximum non-expansion work write AG = from the combustion of C3H8 at STP given A,G°(H20) = -237 kJ mol-1,A,G°(CO2) = –394 kJ mol-1,A;G°(C3H3) = –23 kJ mol-1.

a) Show for constant temperature the maximum work is given by dwmax = dU – TdS = dA Hint: start with the Clausius inequality, the definition of the internal energy dU, and remember w is negative when done by the system on the surroundings. b) A similar process at constant T and P leads to the expression dG = dwmax,add Where wmax,add is all work except expansion work. For measurable changes we can Wmax,add. Use this result to calculate the maximum non-expansion work write AG = from the combustion of C3H8 at STP given A,G°(H20) = -237 kJ mol-1,A,G°(CO2) = –394 kJ mol-1,A;G°(C3H3) = –23 kJ mol-1.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

6.

a) Show for constant temperature the maximum work is given by
\[ dw_{\text{max}} = dU - TdS = dA \]
Hint: Start with the Clausius inequality, the definition of the internal energy \( dU \), and remember \( w \) is negative when done by the system on the surroundings.

b) A similar process at constant \( T \) and \( P \) leads to the expression
\[ dG = dw_{\text{max,add}} \]

Where \( w_{\text{max,add}} \) is all work except expansion work. For measurable changes we can write \(\Delta G = w_{\text{max,add}} \). Use this result to calculate the maximum non-expansion work from the combustion of \( \text{C}_3\text{H}_8 \) at STP given \(\Delta_f G^\circ(\text{H}_2\text{O}) = -237 \text{ kJ mol}^{-1}\), \(\Delta_f G^\circ(\text{CO}_2) = -394 \text{ kJ mol}^{-1}\), \(\Delta_f G^\circ(\text{C}_3\text{H}_8) = -23 \text{ kJ mol}^{-1}\).

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