n = 3.35 mol of Hydrogen gas is initially at T = 350 K temperature and pi = 1.52
ID: 1483479 • Letter: N
Question
n = 3.35 mol of Hydrogen gas is initially at T = 350 K temperature and pi = 1.52×105 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches pf = 9.38×105 Pa.
1. What is the volume of the gas at the end of the compression process?
2. How much work did the external force perform?
3. How much heat did the gas emit?
4. How much entropy did the gas emit?
5. What would be the temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure?
PLEASE SHOW ALL YOUR WORK
Explanation / Answer
Given
n = 3.35
T = 350 K
Pi = 1.52*10^5 pa
Pf = 9.38*10^5 pa
1) use, Pi*Vi = n*R*T
==> Vi = n*R*T/Pi
= 3.35*8.314*350/(1.52*10^5)
= 0.0641 m^3
we know, In Isothermal process, P*V = constant
Pi*Vi = Pf*vf
==> vf = Vi*(Pi/Pf)
= 0.0641*(1.52/9.38)
= 0.01039 m^3
2) workdone by external force = n*R*T*ln(Pf/Pi)
= 3.35*8.314*350*ln(9.38/1.52)
= 17740.4 J
3) Q = 17740.4 J (since dU = 0)
4) dS = Q/T
= 17740.4/350
= 50.7 J/k
5) P1 = 9.38*10^5 pa
P2 = 1.52*10^5 pa
T1 = 350 K
T2 = ?
In adiabatic process , P^(1-gamma)*T^gmma = constant
so, P2^(1-gamma)*T2^gamma = P1^(1-gamma)*T1^gamma
T2 = T1^gamma*(P1/P2)^(1-gamma)
= 350^1.41*(9.38/1.52)^(1 - 1.41)
= 1832.7 K
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