# Gas Laws - What is the final pressure in the system?

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1. Aug 31, 2015

1. The problem statement, all variables and given/known data
Two glass vessels M and N are connected by a closed valve.
M contains helium at 20 °C at a pressure of 1 × 10^5 Pa. N has been evacuated, and has three times the volume of M. In an experiment, the valve is opened and the temperature of the whole apparatus is raised to 100 °C.
What is the final pressure in the system?
A 3.18 × 10^4 Pa
B 4.24 × 10^4 Pa
C 1.25 × 10^5 Pa
D 5.09 × 10^5 Pa

2. Relevant equations
PV=nRT
Where R=8.31J/Kmol

3. The attempt at a solution
For M
T=20 °C = 298 K
P= 10^5 Pa
V= x
For N
V =3x
T =100°C = 373K
Now do I find the change in the T and V and use PV=nRT to calculate the change in the P and then find the final P? But I do not know the mass of helium, so I can not find the number of moles.

2. Aug 31, 2015

### SteamKing

Staff Emeritus
You'll have to work out the mass of the helium given the initial conditions and the fact that vessel M has a fixed volume, VM. The mass will be in terms of VM. You are given the volume of vessel N, VN, in terms of VM.

3. Aug 31, 2015

### Staff: Mentor

Try to express the number of moles of helium in terms of V (initial volume) and see, if it doesn't cancel out in the final formula.

4. Aug 31, 2015

What is the relation between mass and volume? Mass remains constant, irrespective of the gas, but how do I find the mass in terms of Vm?

5. Aug 31, 2015

But I don't know the mass. What is the relation between mass and volume?

6. Aug 31, 2015

### SteamKing

Staff Emeritus
That's what the gas law is telling you: PV = n RT, or n / V = P / RT

7. Aug 31, 2015

### Staff: Mentor

You don't need mass, number of moles is enough. But actually you don't need it either, as in the end it should cancel out, and you should be able to express the final result using given information.

Once you know number of moles and gas identity, calculating mass is trivial. What is molar mass of helium?

8. Sep 1, 2015

Okay so,
n= mass/4
then using n/V=P/RT
(m/4)/V=10^5/8.31x298
(m/4)/V=40.3815
Where V is the the volume of N
and when the gas expands into M, its volume is 4V of n
So,
(m/4)/4V=P/8.31x373
40.3815/4=P/8.31x373
P= 31800 (approx.) which is 3.18 x 10^4 Pa, option (A)