Air pressure in a container then released

AI Thread Summary
In the discussion about air pressure in a sealed container, the initial conditions are set at 22°C and 1.3 atm in a 2.05 L container. The user calculates the initial moles of air (ni) using the ideal gas law, obtaining 0.110 moles. After heating the container to boiling water temperature, they attempt to find the final moles (nf) assuming a pressure of 1 atm and temperature of 100°C, resulting in 0.066 moles. The user is confused about why subtracting ni from nf does not yield the expected result, prompting a consideration of equalized pressure after air escapes. The final suggestion is to use the relationship of moles and temperature to determine how much air escapes when pressure equalizes.
talaroue
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Homework Statement



The air temperature and pressure in a laboratory are 22°C and 1.3 atm. A 2.05 L container is open to the air. The container is then sealed and placed in a bath of boiling water. After reaching thermal equilibrium, the container is open. How many moles of air escape?


Homework Equations



ni=PiVi/RTi

nf=PfVf/RTf



The Attempt at a Solution



I easily found ni by plugging in for the top equation (131690 Pa*.00205 m^3) / (8.31 * 295K) and got .110 mols

then I found the nf by doing the same thing except assuming that Pf is 1 atm (101300 Pa) and that the Tf is 100 deg C(273 K), with constant volume. I got .066 Then I subtracted the ni-nf to get how many mols are left. Why wouldn't this work?
 
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I even tried the final pressure (Pf) the same as the intital pressure (Pi) of 131690 and it still didn't work.
 
any ideas?
 
talaroue said:
I even tried the final pressure (Pf) the same as the intital pressure (Pi) of 131690 and it still didn't work.

The final pressure in the vessel will be equal after letting the air escape.

Consider then how many moles will be in the container when the pressure equalizes.

With P and V the same then don't you have ...

nf * Tf = ni * Ti

nf = .110 moles * (295/373)

The difference is how much escapes.
 
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