Two moles of O2 gas at 1 atm and 10 degrees C are mixed adiabatically with 4 moles of N2 gas at 1 atm and 80 degrees C to yield a mixture whose pressure is also 1 atm. Cp of O2 and N2 may be taken as 7 cal/mol-K. What is the entropy change for the process?
Physical Chemistry: Show complete solution.
Two moles of O2 gas at 1 atm and 10 degrees C are mixed adiabatically with 4 moles of N2 gas at 1 atm and 80 degrees C to yield a mixture whose pressure is also 1 atm. Cp of O2 and N2 may be taken as 7 cal/mol-K. What is the entropy change for the process?
Given data:
Number of moles of O2 gas = 2 mole
The temperature of O2 gas = 10 0C
The pressure of O2 gas = 1 atm
The temperature of N2 gas = 80 0C
The pressure of N2 gas = 1 atm
Number of moles of N2 gas = 4 mol
The entropy change of mixing is calculated as follows:
................. (1)
Where,
nA represents the number of moles of component A.
nB represents the number of moles of component B.
yA represents the mole fraction of component A in the mixture.
yB represents the mole fraction of component B in the mixture.
R represents the universal gas constant.
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