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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.

User Ziqi Liu
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Answer:

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

User Dilini
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