# Homework Help: Air pressure in a container then released

1. Jun 30, 2009

### talaroue

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

ni=PiVi/RTi

nf=PfVf/RTf

3. 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?

2. Jun 30, 2009

### talaroue

I even tried the final pressure (Pf) the same as the intital pressure (Pi) of 131690 and it still didn't work.

3. Jun 30, 2009

any ideas?

4. Jul 1, 2009

### LowlyPion

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.