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

The final pressure is 3.16 torr

Solution:

As per the question:

The reduced pressure after drop in it, P' = 3 torr =
`3* 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 = (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) = (P)/(RT)`

`\rho_(a) = (3 times 0.133* 10^(3))/(0.287* 278) = 0.005 kg/m^(3)`

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

`(P)/(T) = (P')/(T')`

`P' = (P)/(T)* T'`

`P' = (3)/(278)* 293 = 3.16 torr`