4.1
T1 = initial temperature = 27 oC = 27 + 273 = 300 K
T2 = final temperature = 127 oC = 127 + 273 = 400 K
P = pressure = 1 bar = 105 pa
n = number of moles = 1
V1 = initial volume
Using the ideal gas equation
P V1 = n R T1
(105) V1 = (1) (8.314) (300)
V1 = 0.025 m3
Using the ideal gas equation
P V2 = n R T2
(105) V2 = (1) (8.314) (400)
V2 = 0.033 m3
4.2
work done is given as
W = P (V2 - V1)
W = (105) (0.033 0.025)
W = 82.5 J
4.3
internal energy change is given as
U = n Cv (T2 - T1)
U = (1) (12.5) (400- 300)
U = 1250 J
4.4 )
Enthalpy change is given as
H = n Cp (T2 - T1)
H = (1) (20.8) (400- 300)
H = 2080 J
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