[SOLVED] volume of gas in a cylindrical vessel on which a piston has been placed

merbear

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

A cylindrical vessel with a tight but movable piston is placed in a vertical position so that the piston, whose area is 60 cm2, is subject to atmospheric pressure. When the gas in the vessel is heated from 20oC to 80oC, a 0.6 kg mass must be placed on top of the piston to hold it at the position it occupied at lower temperature.
What is the volume of 0.3 mol of the gas?

2. Relevant equations

Pv=nRT



3. The attempt at a solution


Since we were given the moles of gas (.3 mol) and we know R since it's a constant, I assumed that the pv=nRT would be the correct equation. I found the pressure of the cylinder with the weight on to be the pressure by the weight: P=F/A= [(.6 kg * 9.8 m/s^2)]\.6m^2. The temperature I put was 353K - that of the second cylinder.

I got a volume of 53.9 m^2 but I know that the volume must be much greater, given the size of the cylinder in the problem. I don't know how else to go about the problem so I would appreciate any help.

Thank-you!!

Shooting Star

> What is the volume of 0.3 mol of the gas?

Under what T and P? Without that, the question is meaningless. Also, it is not mentioned in the problem that the amount of gas is 0.3 moles. The length of the cylinder is also not given.

What are known are these:

P1/T1 = P2/T2, since the volume remains the same.
P2 - p1 = W/A, where W is the weight of the mass, A = area of piston.

So, P1 and P2 both may be found.

merbear

I don't understand what you mean when you say, 'under what temperature and pressure'. The question stated that you had one cylinder at 20 degrees and the other one with the weight on it at 80 degrees. It also stated that the piston is subjected to atmospheric pressure. Does the atmospheric pressure not need to be accounted for?

I see how you were saying to solve p1 and p2, but I don't see where to go from there. Would you use the pressure difference to solve for the volume of gas using pv=nRT?

Shooting Star

The question should have been formulated more precisely. In the main problem, it is not mentioned that there were 3 moles of gas. At the very last, it is asked, "What is the volume of 0.3 mol of the gas?", which does not imply that there were 3 moles inside. Anyway, let it go.

The P_atm is acting on both cases and is included in P1 and P2. In fact, if the piston is very light, P1 = P_atm. But we can do without knowing the value of P_atm.

P1V = nRT1 and P2V = nRT2 =>
(P2-P1)V= nR(T2-T1).

You can get V now.