One mole of N 2 (g), C V =5/2 R, at 300 K and 10 bar expands adiabatically again

Question

One mole of N2(g), CV =5/2 R, at 300 K and 10 bar expands adiabatically against a constant opposing pressure of 1 bar until its pressure equals the opposing pressure. Calculate the temperature of the N2 after this process.

To calculate the entropy change of the universe for this process we must replace this spontaneous process with a quasi-static process that goes between the same initial and final conditions of the N2.

For a quasi-static, adiabatic expansion p1V1 = p2V2.

Show that this equation may be written as p1(1-)/T1 = p2(1-)/T2 .

Calculate the T2 if the actual process had been quasi-static. What additional quasi-static process must be done to get to the actual final conditions of the spontaneous process? Having replaced the spontaneous process with a two-step, quasi-static process calculate the entropy change of the universe in J/K for the spontaneous process.

Explanation / Answer

p1V1 = p2V2.

ALSO

p V=nRT

v=nRT/ p  

put in above equation we get

p1(1-)/T1 = p2(1-)/T2 .

using above relation

300*10-2/7 =1*T

T=155.38 K

the next process is reverse of above that is adiabatic compression from initial pressure of 1 bar to 10 bar

since in case of adiabatic process energy transfer is zero so changein entropy is ZERO.

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