Pressure on a piston in a cylinder

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Homework Help Overview

The problem involves a vertical cylinder with a piston containing an ideal gas. Participants are tasked with calculating the pressure inside the cylinder and determining the height of the piston in equilibrium under its own weight, given specific parameters such as cross-sectional area, mass of the piston, and temperature of the gas.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of pressure using the formula F/Area and the addition of atmospheric pressure. There are attempts to calculate the volume using the ideal gas law, with some questioning the accuracy of their results. Others express confusion regarding the height calculation and its acceptance in an external system.

Discussion Status

The discussion is ongoing, with some participants confirming the calculations while others express uncertainty about the correctness of their answers in relation to an external system. There is no explicit consensus on the final outcome, as discrepancies in results have been noted.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may impose specific formatting or significant figure requirements for their answers. There is mention of an external system (UTexas) that may have specific input expectations.

Jimkatz809
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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.
 
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.025 looks good to me.
 
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
 
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.
 
Jimkatz809 said:
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
 

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