Answer:
The final pressure is 3.16 torr
Solution:
As per the question:
The reduced pressure after drop in it, P' = 3 torr =
At the end of pumping, temperature of air,
After the rise in the air temperature,
Now, we know the ideal gas eqn:
PV = mRT
So
(1)
where
P = Pressure
V = Volume
R = Rydberg's constant
T = Temperature
Using eqn (1):
Now, at constant volume the final pressure, P' is given by:
Answer:
the world will not be the same it will change dramticily and
things will not be the same
Explanation:
Answer:
(a) The area of the duct exit is 1.67 m²
(b) The air pressure at the exit is 1.195 x 10⁵ N/m²
Explanation:
Given;
inlet area, A₁ = 5 m²
inlet velocity, V₁ = 10 m/s
exit velocity, V₂ = 30 m/s
(a) The area of the duct exit is determined by applying continuity equation;
A₁V₁ = A₂V₂
Where;
A₂ is the area of the duct exit
A₂ = (A₁V₁) / (V₂)
A₂ = (5 x 10) / (30)
A₂ = 1.67 m²
(b) Apply Bernoulli’s equation to determine the pressure at the exit;
Density of air at 300k = 1.177 kg/m³
Therefore, the air pressure at the exit is 1.195 x 10⁵ N/m²