Question

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50 bar to p2 = 0.25 bar. Assume that N2(g) under such conditions can be considered an ideal gas with CV,m = 3 R.

A) What is the final temperature T2 of the gas?B) What is ∆U for the gas in the process?C) What are q and w exchanged by the system in the process?D) What is ∆H for the gas in the process?E) What is ∆S for the gas in the process?F) What is ∆Suni for the universe in the process?

Answer 1

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C
`_(V)` You can work out C
`_(V)`

Part C

Part D

Part E

Part F

Step-by-step explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC
`_(V)` ∆T

Polyatomic gas: C
`_(V)` = 3R

∆U= nC
`_(V)` ∆T

∆U= 28g x C
`_(V)` x (350K - 58.3K)

∆U = 28C
`_(V)` x 291.7

∆U = 10967.6 x C
`_(V)`