A rigid adiabatic container in a room with a barometric pressure of 100 kPa is d

Question

A rigid adiabatic container in a room with a barometric pressure of 100 kPa is divided into two unequal volumes as shown. V1 = 0.108 m3 and contains air (R = 0.2870 kJ/kg-K, k = 1.40) at a gauge pressure of 32 kPa and temperature of -3°C. V2 = 0.264 m3 and contains helium (R = 2.0769 kJ/kg-K, k = 1.667) at a vacuum pressure of 40 kPa and a temperature of 23°C. The membrane separating the two volumes bursts and the two gases mix together. When equilibrium is restored,

1- what is the gauge pressure in the container?

2- What is the final temperature? In C

Explanation / Answer

For air:

P1 = 100+32 = 132 kPa

V1 = 0.108 m^3

T1 = -3 deg C = -3+273 K = 270 K

R = 0.287 kJ/kg-K

k = 1.4

For Helium :

P2 = 100 -40 = 60 kPa

T2 = 23 deg C = 23+273 K = 296 K

V2 = 0.264 m^3

R = 2.0769 J/kg-K

k = 1.667

P1*V1 = m1*R*T1

132*0.108 = m1*0.287*270

m1 = 0.184 kg

P2*V2 = m2*R*T2

60*0.264 = m2*2.0769*296

m2 = 0.0257 kg

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