I don't believe I'm getting the correct answers for this problem. Could someone look over my work and verify whether it is correct or not?
One mole of an ideal gas (Cv = 3/2R) at 300. K expands adiabatically against a constant external pressure of 2.00 atm from a volume of 1.00 L to 5.00 L. Find delta-S gas, delta-S surroundings, delta-S universe for the process.
my work:
P1 = nRT/V
P1= (1.00 mol * .08206 * 300. K)/1.00 L= 24.6 atm
P2= P1(V1/V2)^Cp/Cv
P2= 24.6 atm(1.00 L/5.00 L)^5/3 = 1.68 atm
T2= PV/nR
T2= (1.68 atm * 5.00 L)/(1.00 mol * .08206)= 102.5 K
delta-S g= nRlnT2/T1 + nRlnV2/V1
delta-S g= 1.00 mol*(3/2*8.314 J/mol K)*ln(102.5K/300.K) + 1.00 mol*(8.314 J/mol K)*ln(5.00L/1.00L) = -.0120 J/K
delta-S surr= -q/T
but q=0 so delta-S surr= 0
delta-S univ= delta-S surr + delta-S gas
delta-S univ= -.0120 J/K
I noticed that I never used the external pressure given in the problem. Something is telling me that it was given for a reason. Where should I have used it? Also, is the equation for delta-S surr correct? I'm not sure if it's limited to certain situations.
Your calculations for delta-S gas and delta-S surroundings seem correct. However, you are correct in questioning the use of the external pressure given in the problem. The ideal gas expansion is adiabatic, meaning no heat exchange with the surroundings. Therefore, the external pressure is not directly relevant for calculating delta-S gas or delta-S surroundings in this case.
The equation you used for delta-S surroundings is generally correct: delta-S surr = -q/T. But as mentioned before, since no heat is exchanged with the surroundings in an adiabatic process, q = 0 and thus delta-S surr is also zero.
In this particular problem, the external pressure is only provided to calculate delta-S universe. The equation for delta-S universe is given by delta-S univ = delta-S surr + delta-S gas. Since delta-S surroundings is zero, your calculated value for delta-S gas (-0.0120 J/K) is also the value of delta-S universe in this case.
So, to summarize: Your calculations for delta-S gas and delta-S surroundings are correct, and the equation you used for delta-S surroundings is generally valid. However, in this specific adiabatic expansion problem, delta-S surr is zero because no heat is exchanged with the surroundings. You can find delta-S universe by simply adding delta-S gas to the value of zero for delta-S surr.