Respuesta :
Answer:
The final pressure is 3.16 torr
Solution:
As per the question:
The reduced pressure after drop in it, P' = 3 torr = [tex]3\times 0.133\ kPa[/tex]
At the end of pumping, temperature of air, [tex]T = 5^{\circ}C = 278 K[/tex]
After the rise in the air temperature, [tex]T' = 20^{\circ}C = 293 K[/tex]
Now, we know the ideal gas eqn:
PV = mRT
So
[tex]P = \frac{m}{V}RT[/tex]
[tex]P = \rho_{a}RT[/tex] Â Â Â Â Â (1)
where
P = Pressure
V = Volume
[tex]\rho_{a} = air\ density[/tex]
R = Rydberg's constant
T = Temperature
Using eqn (1):
[tex]P = \rho_{a}RT[/tex]
[tex]\rho_{a} = \frac{P}{RT}[/tex]
[tex]\rho_{a} = \frac{3 times 0.133\times 10^{3}}{0.287\times 278} = 0.005 kg/m^{3} [/tex]
Now, at constant volume the final pressure, P' is given by:
[tex]\frac{P}{T} = \frac{P'}{T'}[/tex]
[tex]P' = \frac{P}{T}\times T'[/tex]
[tex]P' = \frac{3}{278}\times 293 = 3.16 torr[/tex]
