A1-m3 tank containing air at 10°c and 350 kpa is connected through a valve to another tank containing 3 kg of air at 35°c and 150 kpa. now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 19.5°c. determine the volume of the second tank and the final equilibrium pressure of air. the gas constant of air is r = 0.287 kpa·m3/kg·k.

V₂=1.76 m³

P=222.03 KPa

Explanation:

Given that

For tank 1

V₁=1 m³

T₁= 10°C = 283 K

P₁=350 KPa

For tank 2

m₂=3 kg

T₂=35°C = 308 K

P₂=150 KPa

We know that for air

P V = m R T

P=pressure ,V= Volume,R= gas constant ,T= temperature ,m =mass

for tank 2

P₂ V₂ = m₂ R T₂

By putting the values

150 x V₂ = 3 x 0.287 x 308

V₂=1.76 m³

Final mass = m₁+m₂

m =m₁+m₂

The final volume V= V₂+V₁

V= 1.76 + 1 m³

V= 2.76 m³

The final temperature T= 19.5°C

T= 292.5 K

m =m₁+m₂

m =4.3 + 3 = 7.3 kg

Now at final state

P V = m R T

P x 2.76 = 7.3 x 0.287 x 292.5

P=222.03 KPa

answer: 500 wph.

explanation: