# Termo physics nr1 help

## Homework Statement

In 2 identical containers we have equal mass of helium and Argon at the same temperature.The ratio of the 2 pressures (pressure 1 and pressure 2) knowing that μ helium= 4 kg/kmol and μ argon= 40 kg/kmol is:
a 0.1 b 10 c 5 d 40 e 1/5

i dont know

## The Attempt at a Solution

i dont know

What macroscopic equation gives the behaviour of ideal gases?

idk tell me
i got bad marks at physics

idk =?

pv=nrt?
i dont know man physics is not my main thing

pV=nRT

Correct.

Now see what is identical in both cases and hence try to see on what physical quantity does p depend.

p=nrt/v
?
can you please give more hints?

Temp is the same for both.
What else is the same?

R constant and T
n is the quantity
EDIT:Was joking man i know how to do it p1v=m/miu RT

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Correct.
But if you read the problem CAREFULLY you will notice that there is something else constant.

HAHAAHAH man now i dont understand
p1v=U x R
??

Which gas has the bigger volume?

Which gas has the bigger volume?

Argon i think.

is the argon in a bigger container?

is the argon in a bigger container?

no. what to do?

Each gas in an IDENTICAL container. Hence the volume ...

Ok so volume is constant.Here is what i made so far.
Find p1 and p2
P1=1/μ helium *R
P2=1/μ argon *R

and then P2/P1 = 10.

so from those answers: a) 0.1 b)10 c)5 d)40 e)1/5

the right one is b?

P argon/P helium = 0.1

P helium/P argon = 10

P helium/P argon = 10

so its 10?
i did it good?
can you help me at the other problem too? :D

which problem?

At boiling whit 1K(kelvin) of a gas whit constant pressure the volume is 3 times bigger.The initial temperature of the gas in this process has the value:
a. 6K b 0.5 K c 10K d 100K e 500K

Which equation gives the behaviour of an ideal gas?

pv/t=p2v2/t2

Cross out the P because it is the same on both sides.

yup so v/t = v2/t2 .
What i do next?

Now put in the values you know to get what you do not yet know.

So T2= t1 +1k?
V2 = 3V1?
WE know that the second volume is 3x bigger then the first volume.

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