Pressure on a piston in a cylinder

[Jimkatz809]

Homework Statement

A vertical cylinder of cross-sectional area 0.25 m2 is fitted with a tight-fitting, frictionless piston of mass 22 kg. The acceleration of gravity is 9.8 m/s2 . If there are 1 mol of an ideal gas in the cylinder at 310 K, find the pressure inside the cylinder. Assume that the system is in equilibrium. Answer in units of Pa. At what height will the piston be in equilibrium under its own weight? Answer in units of m.

Homework Equations

Pressure=F/Area
PV=nRT
Height= Volume /Area

The Attempt at a Solution

Pressure =F/Area (22kg x 9.8m/s^2)/(0.25m^2) Then I added atmospheric pressure to this pressure which is 1.013x10^5 Pa. I got an answer for pressure as 1.02x10^5 Pa and I got it correct.
Then for volume I have been trying V= (1 mol x 8.3145 x 310 K)/(1.02x10^5) and I get .025 as my volume but I think this number is off and I can't figure out why... I don't want to move forward until I figure this part out.

 

[Delphi51]

.025 looks good to me.

 

[Andrew Mason]

The volume is correct, in m^3. The question asks for the height, which is .025/.25 = .1 m or 10 cm. What is the answer you are given?

AM

 

[Jimkatz809]

I the calculations as you said and got .1 m just as you said and I tried puting it into the UTexas system but its said that is wrong. I don't understand because it makes no sense that should be the answer.

 

[Andrew Mason]

I the calculations as you said and got .1 m just as you said and I tried puting it into the UTexas system but its said that is wrong. I don't understand because it makes no sense that should be the answer.

Try .10 m (two sig. figures).

AM