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
[tex]m=\dfrac{PV}{RT}[/tex]
[tex]m_1=\dfrac{P_1V_1}{RT_1}[/tex]
[tex]m_1=\dfrac{350\times 1}{0.287\times 283}[/tex]
[tex]m_1=4.3\ kg[/tex]
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:
a
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answer:
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