Posted by **Jessi** on Sunday, August 23, 2009 at 10:35am.

A cylinder that has a 40.0 cm radius and is 50.0 cm deep is filled with air at 20.0°C and 1.00 atm. A 21.0 kg piston is now lowered into the cylinder, compressing the air trapped inside. Finally, a 79.0 kg man stands on the piston, further compressing the air, which remains at 20°C.

(a) How far down (Ä h) does the piston move when the man steps onto it?

(b) To what temperature should the gas be heated to raise the piston and the man back to h(initial)?

I don't know where to start from here. Do you have to a change in volume type problem? Or could you do it using the ideal gas law?

- physics -
**Damon**, Sunday, August 23, 2009 at 3:04pm
radius = .4 m so piston area = pi(.16) =.506 m^2

(a) is a change in volume problem all apparently at constant temperature = 293 K

First new volume with the piston:

P1V1/T1 = P2V2/T2

T1=T2 = 293 K

so

V2 = V1 (P1/P2)

P1 = 1 atm = 10^5 Pascals

P2 = 1 atm + (21*9.8)/.506 = 10^5+407

=100,407 Pascals

V2/V1 = 100000/100407 =.996

so

H2 = .996 (50) = 49.8 cm hardly moves

now add the man

V3 = V1(P1/P3)

P3 = 1 atm + (100*9.8)/.506 =10^5+1937

=101,937 Pascals

V3/V1 = 100000/101937 = .981

so

H3 = 50*.981 = 49.05 cm

so

when the piston was added the level went down to 49.8 and then it went down to 49.05 when the man stepped on so it went down about .8 cm due to the man.

For Part V

V2/V1 = 50/49.05 = 1.02

constant pressure so temp goes up

P1V1/T1 = P2 V2/T2

so

V2/V1 = T2/T1

T1 = 293 K

1.02 = T2/293

T2 = 298.7 K = 25.7 C

- physics -
**Jessi**, Sunday, August 23, 2009 at 3:17pm
thank you so much! it's really clear now with the way you've explained it

- physics -
**bubba**, Sunday, August 23, 2009 at 5:35pm
two airplanes leave an airport at the same time. the velocity of the first airplane is 700 m/h at a heading of 52.2degrees . the velocity of the second is 600 m/h at a heading of 97 degrees . how far apart are they after 3.1 h? answer in units of m.

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