Question

. An ideal gas initially contained at 1 bar and 25ºC, is compressed to a final state of 5 bar and 25ºC. Although the initial and final temperatures are the same, the process is NOT isothermal. Rather, the change of state occurs in two steps. First, the gas is cooled at constant pressure (i.e., 1 bar). Second, the gas is heated at constant volume. Please calculate Q, W, DU, and DH for the two steps. Note: Cp*=29.1 J/mol/K, and 1 mole of an ideal gas occupies 0.0227 m³ at 0ºC and 1 bar. PV/T is a constant of ideal gases.

Answer 1

Calculate the Volume at 25°C (298.15K)

V1 = T1 x V2/T2

= 298.15 x 0.0227/273.15

= 0.02479 m3

Part a

Final volume of the gas

V2 = V1 x P1/P2

= 0.02479 x (1/5)

= 0.004958 m3

Temperature after cooling

T' = T1 x V2/V1

= 298.15 x 0.004958/0.02479

= 59.63 K

Cooling at constant pressure

Q = H = Cp ( T' - T1)

= 29.1 J/mol·K x (59.63 - 298.15) K

= - 6941 J

Q = ∆H − ∆(PV) = ∆H − P∆V

= −6941 − (1 × 10^5) x (0.004958 − 0.02479)

= −4958J

U = Q = Cv (T1 - T')

= (Cp - R) x (298.15-59.63) = 4958 J

Total energy Q = -6941 + 4958 = -1983J

ΔU = -4958 + 4958 = 0 J

From the first law of thermodynamics

ΔU = Q + W

0 = -1983 + W
W = 1983 J

For the overall process

Initial Temperature T1 = final temperature T2

P1 V1 = P2 V2

H = U + (PV)

H = U = 0

Part b

Temperature of air at final stage

T' = T1 x P2/P1

= 298.15 x (5/1)

= 1490.75 K

At constant volume

U = Q = Cv (T' - T1)

= (Cp - R) x (1490.75 - 298.15) = 24788 J

Q = H = Cp ( T1 - T')

= 29.1 ( 298.15 - 1490.75)

= - 34703 J

U = H - (PV) = H - PV

= - 34703 - (5*10^5) x (0.004958 - 0.02479)

= - 24788 J

Q = 24788 - 34703 = - 9915 J

U = 24788 - 24788 = 0 J

W = U - Q = 0 - (-9915) = 9915 J

H = U = 0