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?

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  1. 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
    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
    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
    H3 = 50*.981 = 49.05 cm
    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
    V2/V1 = T2/T1
    T1 = 293 K
    1.02 = T2/293
    T2 = 298.7 K = 25.7 C

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  2. thank you so much! it's really clear now with the way you've explained it

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  3. 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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  4. It was magical 🤪😃💁‍♀️✨

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