A 1 mole sample of N2O (dinitrogen monoxide) gas is compressed reversibly and isothermally from an initial temperature of 298 K and an initial pressure of 1 bar (Standard state temp and pressure) to a final pressure of 5.35 bar. Assume the gas remains a gas and it is ideal. a. Calculate w, q, ΔU, ΔH, and ΔS for this process. b. Calculate the total ΔS (system plus surroundings) for this process. c. What is the value of the entropy of N2O gas at 298 K and 5.35 bar? Data found in Table 2C.7 can be used to help answer this question.
A 1 mole sample of N2O (dinitrogen monoxide) gas is compressed reversibly and isothermally from an initial temperature of 298 K and an initial pressure of 1 bar (Standard state temp and pressure) to a final pressure of 5.35 bar. Assume the gas remains a gas and it is ideal. a. Calculate w, q, ΔU, ΔH, and ΔS for this process. b. Calculate the total ΔS (system plus surroundings) for this process. c. What is the value of the entropy of N2O gas at 298 K and 5.35 bar? Data found in Table 2C.7 can be used to help answer this question.
Chemistry
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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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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A 1 mole sample of N2O (dinitrogen monoxide) gas is compressed reversibly and isothermally from an initial temperature of 298 K and an initial pressure of 1 bar (Standard state temp and pressure) to a final pressure of 5.35 bar. Assume the gas remains a gas and it is ideal.
a. Calculate w, q, ΔU, ΔH, and ΔS for this process.
b. Calculate the total ΔS (system plus surroundings) for this process.
c. What is the value of the entropy of N2O gas at 298 K and 5.35 bar? Data found in Table 2C.7 can be used to help answer this question.
Expert Solution
Step 1
Wiso, rev = -nRT ln (P1/P2)
= -1 x 8.314 x 298 ln (0.1869)
= 4155.12 J
= 4.155 kJ.
As the process is isothermal, so its u = 0 .
Now, q = - w
= -4.155 kJ
= -4155.12 J
= = 4.155kJ.
= ==-13.94 J/K
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