Question

Air is pumped from a vacuum chamber until the pressure drops to 3 torr. If the air temperature at the end of the pumping process is 5°C, calculate the air density. Eventually, the air tem- perature in the vacuum chamber rises to 20°C because of heat transfer with the surroundings. Assuming the volume is constant, find the final pressure, in torr.

Answer 1

Answer:

The final pressure is 3.16 torr

Solution:

As per the question:

The reduced pressure after drop in it, P' = 3 torr = $3\times 0.133\ kPa$

At the end of pumping, temperature of air, $T = 5^{\circ}C = 278 K$

After the rise in the air temperature, $T' = 20^{\circ}C = 293 K$

Now, we know the ideal gas eqn:

PV = mRT

So

$P = \frac{m}{V}RT$

$P = \rho_{a}RT$          (1)

where

P = Pressure

V = Volume

$\rho_{a} = air\ density$

R = Rydberg's constant

T = Temperature

Using eqn (1):

$P = \rho_{a}RT$

$\rho_{a} = \frac{P}{RT}$

$\rho_{a} = \frac{3 times 0.133\times 10^{3}}{0.287\times 278} = 0.005 kg/m^{3}$

Now, at constant volume the final pressure, P' is given by:

$\frac{P}{T} = \frac{P'}{T'}$

$P' = \frac{P}{T}\times T'$

$P' = \frac{3}{278}\times 293 = 3.16 torr$