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Kinetic theory HW help

  1. Feb 4, 2004 #1
    Kinetic theory HW help :(

    I am new here and found the site while trying to find a formula. I am having problems solving some questions. I worked out what i could but dont know if i did it right. Please Help :(


    A raindrop of mass (1mg) fall vertically at a constant speed of 10 m/s, striking a horizontal skylight at the rate of 1000 drops/s and draining off. The window is 15cm X 25cm. Assume the collisions are completely inelastic.

    a) Calculate the magnitude of the average force of the raindrops on the window.

    i used this equation. F(ave) = -2mV/(2L/V)

    -2(1x10^-6kg)(10m/s) / ( 2 (.0375m^2)/(10m/s) ) = 2.66x10^-3 (is this right?)

    b) what is the resuling pressure developed by the raindrop.

    I know that P= F(average)/Area I know the area and F(average) is from above.


    The molar mass of N2 is 28g.

    a) find the mass of 1 nitrogen molecule.

    (1mol N2/ 28g N2) X ( 6.02x10^23 molecules/ mol) = 2.15x10^25 mol/g so each molecule weighs (the inverse) = 4.65x10^-23g

    b) Find the rms speed of a nitrogen molecule at a temp. of -23C.
    V(rms) = (3RT/M)^.5

    ( (3*8.315*250.15K)/ 28g ) ^.5 = 14.9

    c) H2 is also present in same container. molar mass 2g/mol. What is the temp of the Hydrogen gas?

    This is where I am stuck ? do i use the PV= nRT

    D) what is the rms speed of the hydrogen molecule?

    I used the V(rms) = 3RT/M (the T is what i am trying to find from part C right?

    e) what new temp would cause the V(rms) to increase by 2 in part b?
    New temp = part b temp x 2. right? = 250.15K x 2 = 500.3K = 227.15C.

    Problem 3

    1 mol of HE gas @ 300K is in a cubical box of 10cm sides.

    a) what is the Vrms of the particles.

    V= 3RT/M ---> 3(8.315)(300) / 4.0026 = 43.23

    b) If there were no collisins along the way, how long would it take a particle to travel from one side to the next?
    L= Vt t(time)= L/v ----> .1meter/43.23 = 2.313x10^3 sec.

    c) what is the pressure of the container?

    PV= nRT solving to P---> P= nRT/V ( (1mol)(8.315)(300K) / 10m^3 ) = 249.45

    d) what is the average force of the particle excerted on the side of the box?

    F(ave) = MV^2 / L---------> ( (4.0026)(43.23m/s)^2 ) / (10m^3) = 7.48x10^4

    Problem 4

    1 mol of a monatomic ideal gas @ temp 300K accupies a volume of 5Litters. The gas now expands adiabatically till its volume is doubled. What is the final pressure of the gas? (NOTE : @ = gamma)

    I am using the

    PV^@ = constant.

    Pfinal X Vfinal ^@ = Pinitial X Vinitial^@

    solving for Pfinal... i get P(f) = P(i) X (V(i)/V(f))^@.

    I found @ to be equal to 1.66 by the equations @ = C(p)/ C(v) where Cv = 3/2R and Cp = Cv + R.

    so P(f) = .31498 X P(i) where P(i) = 498.9 from the PV= nRT
  2. jcsd
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