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Thermodynamics, ideal gas, probability distributions

  1. Jan 3, 2012 #1
    I need some help with e), but could someone also check to see if the rest is correct?

    1. The problem statement, all variables and given/known data
    The velocity component [itex]v_x[/itex] of gas particles in the x-direction is measured and the probability distribution for [itex]v_x[/itex] is found to be [itex]P \propto e^{-\frac{-m v_x^2}{2 k_B T}}[/itex] with m the mass of a gas particle and T the temperature.

    a) Calculate the normalized probability distribution.

    b) Calculate [itex]\langle v_x \rangle[/itex] and [itex]\langle v_x^2 \rangle[/itex]

    c) Assume the gas is ideal and consists of N identical point-particles. The gas is isothermally compressed at temperature T from volume V to volume V/2, with N unchanged. Calculate the heat Q that is transferred to the surroundings.

    d) Now from the volume V/2 the gas is adiabatically expanded to the original volume V. Calculate the end temperature T', the work done by the gas W' and the added heat Q' during this expansion.

    e) After this expansion to volume V the gas has a temperature of T = 27 degrees Celsius and a pressure of 9 atmosphere. The number of gas particles is equal to Avogadro's constant [itex]N = 6 \times 10^23[/itex]. Calculate the volume V in litres.

    3. The attempt at a solution

    a) [tex]1 = \int_{-\infty}^{\infty} e^{-\frac{-m v_x^2}{2 k_B T}} dv_x = A \sqrt{\frac{2 k_B T \pi}{m}}[/tex]
    So [itex]P = \sqrt{\frac{m}{2 k_B T \pi}}e^{-\frac{-m v_x^2}{2 k_B T}}[/itex]

    b) [tex] \langle v_x \rangle = \sqrt{\frac{m}{2 k_B T \pi}} \int_{-\infty}^{\infty} v_x e^{-\frac{-m v_x^2}{2 k_B T}} dv_x = 0 [/tex]
    [tex] \langle v_x^2 \rangle = \sqrt{\frac{m}{2 k_B T \pi}} \int_{-\infty}^{\infty} v_x^2 e^{-\frac{-m v_x^2}{2 k_B T}} dv_x = \sqrt{\frac{m}{2 k_B T \pi}} \frac{1}{2} \sqrt{\frac{m^3 \pi}{8 k_B^3 T^3}} [/tex]
    c) [tex]\Delta W = - \int_{V}^{V/2} \rho dV = - \int_{V}^{V/2} \frac{N k_B T}{V} dV = -N k_B T \ln{\frac{1}{2}}[/tex]
    This is equal (save for the minus sign) to the amount of heat that is transferred to the surroundings, because we have that [itex]dW = - dQ[/itex].

    d) In an adiabatic expansion there is no flow of heat, so [itex]Q' = 0[/itex] and [itex]dU = dW = C_V dT = -\rho dV = -\frac{N k_B T}{V} dV[/itex]. This leads to:

    [tex]\ln \frac{T'}{T} = - \frac{N k_B}{C_V} \ln \frac{V/2}{V}[/tex]
    Thus:

    [tex]T' = \frac{1}{2} e^{- \frac{N k_B}{C_V}}[/tex]
    The work W' is:

    [tex]W' = -\int_{V/2}^{V} \frac{N k_B T}{V} dV = N k_B T \ln 2[/tex]
    e) I can't just plug the numbers into the ideal gas equation since I'm missing the pressure, but I don't know how to do this one.
     
  2. jcsd
  3. Jan 7, 2012 #2
    Anyone got any idea about e)?
     
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