Compressibility and heights

  • #1
thenewbosco
187
0
Here is my problem:

A cylinder with 0.4m radius and 0.5m depth is filled with air at 20C and 1atm.
A 20.0 kg piston is lowered into the cylinder, compressing the air inside. Finally a 75 kg man stands on the piston further compressing the air which remains at 20C.

how far down does the piston move when the man steps on it and to what temp. must the gas be heated to return the man back to the height when the 20kg piston was placed on.

So far i have calculated the number of moles of gas in the container, which will be constant throughout. and i can calculate the force applied when the man and piston are compressing it as mgh. i suppose the h will give me the distance it compresses but i have nothing to equate mgh to...
can someone help me out here
 

Answers and Replies

  • #2
thenewbosco
187
0
...anyone?
 
  • #3
thenewbosco
187
0
no one can help me with this?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
43,021
971
PV= NRT.

I'm not sure why you would calculate N, the number of moles. Since it remains constant, it will cancel out anyway. You know P, V, and T initially. You add 95kg mass so you can calculate the additional force (weight) and divide by the cross section area of the cylinder to find the additional pressure. I am assuming any heat generated conducts away so the temperature remains the same.

Calling the initial P and V P1 and V1 and the later P and V P2 and V2, you have
[tex]\frac{P_1V_1}{P_2V_2}= \frac{NRT}{NRT}= 1[/tex]
That is, V2= P1V1/P2.

Once you know the volume, of course, you divide by the cross section area of the cylinder to find the height, subtract that from the original height to find how far the piston moves.

For the second part, once again PV= NRT. Now you are maintaining the same pressure while changing the temperature. Writing T1 and T2 for the temperatures before and after (you know that T1= 20 C), we have
[tex]\frac{PV_1}{PV_2}= \frac{NRT_1}{NRT_2}[/tex]
so that T2= T1(V2/V1).
 

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