# CHEM

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A) a picture shows a container that is sealed at the top by a movable piston. Inside the container is a ideal gas at 1.00atm , 20.0 *C, and 1.00L.
(The pic is of a 2L container which has a piston pushed downward to the 1L level. Under the piston, under the 1L mark, is a sample of some unknown 'ideal gas'.)

What will the pressure inside the container become if the piston is moved to the 2.20L mark while the temperature of the gas is kept constant?

P=_____atm

How do you work out this problem and the next two, as well???

B) The gas sample has now returned to its original state of 1.00atm , 20.0*C and 1.00L. What will the pressure become if the temperature of the gas is raised to 200.0 and the piston is not allowed to move?

P=_____atm?

C) The gas described in parts A and B has a mass of 1.66 grams. The sample is most likely which monoatomic gas?

• CHEM -

#1. P1V2 = P2V2

#2. P1/T1 = P2/T2. Don't forget to change T to Kevin.

#3. From 1 and 2 you should have enough information to calculate n, the number of mols. Then n = grams/atomic mass. Calculate atomic mass.

• CHEM -

for Part 1:

would the P2 just be 0.454 atm then?

I went: P1*V1=P2*V2
then: P2=P1*V1/V2?

So...1atm*1L/2.20L = P2= 0.454atm??

• CHEM -

ok...I tried my answer...I got the first part correct..it was actually 0.455atm.

• CHEM -

#2= 1.61atm:)

• CHEM -

the only prob left is: do I just add the variables? or which ones do I take into account?

PV=nRT
n=RT/PV...or should I be using:
M=mRT/PV???

• CHEM -

well...I managed to use the correct equation: M=mRt/PV...& the answer came to Ar.

Thank you very much!!

• CHEM -

OR, you could have done it in two steps.
PV = nRT
n = 0.0416 mols
Then mols = g/atomic mass and rearrange to atomic mass= g/mols = 1.66/0.0416 = 39.9 = Ar. I prefer the two step method because it makes the first equation a little easier to solve but this's just my preference.

• CHEM -

Try this formula
P1(V1)/T1=P2(V2)/T2

• CHEM -

How did Dr.Bob find n = 0.0416 mols ?

• CHEM -

Where did 0.0416 moles come from?

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