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Calculating work done on a gas

  1. Oct 13, 2016 #1
    1. The problem statement, all variables and given/known data
    n moles of an ideal gas are placed in a frictionless piston with weight ##w_p## and cross-sectional area ##A##. The quantity ##\gamma = \frac{c_p}{c_v}## is a constant, the gas is originally at equilibrium values##(P_i, V_i, \theta_i)## and the external pressure is taken to be zero. Internal Energy given by ##U = nc_v\theta## Find the irreversible work done if grain of weight ##w_g## is added the following ways:

    a) All of the grain is added at once to the top of the piston, the piston oscillates and eventually reaches a new equilibrium at ##\theta_i##, show that ##W_{on}^{irrev}=nR\theta_i(\frac{w_g}{w_p})##

    b) If no heat is allowed in or out of the gas (adiabatic process) and and all of the grain is added at once to the top of the piston show that ##W_{on}^{irrev}=\frac{nR\theta_i}{\gamma}(\frac{w_g}{w_p})##

    2. Relevant equations

    ##\delta Q = \Delta U + W##

    ##W_{on} = - \int_{V_i}^{V_f} PdV##

    ##PV=nR\theta##

    Adiabatic Process:

    ##PV^{\gamma} = constant##

    3. The attempt at a solution

    For part a) we have ##W_{on} = -\int_{V_i}^{V_f}PdV=-\int_{V_i}^{V_f}P_{ext}dV## and ##P_{ext} = \mbox{constant} = \frac{w_g+w_p}{A}## so

    ##W_{on}=-\frac{w_g+w_p}{A}\int_{V_i}^{V_f}dV= -\frac{w_g+w_p}{A}(V_f-V_i) \Longleftarrow (1)##

    using the ideal gas law ##V_f = \frac{nR\theta_i}{P_f}=\frac{nR\theta_i A}{w_g+w_p}## and ##V_i =\frac{nR\theta_i A}{w_p}## so ##V_f - V_i = nR\theta_i A(\frac{w_p}{w_p(w_g+w_p)}-\frac{w_g+w_p}{w_p(w_g+w_p)}) = \frac{nR\theta_i A w_g}{(w_g+w_p)w_p} \Longleftarrow (2)##

    Subbing (2) into (1) ##W_{on} = -\frac{w_g+w_p}{A}\frac{nR\theta_i A w_g}{w_p(w_g+w_p)}=nR\theta_i ( \frac{w_g}{w_p} )##

    For part b) It seems to me like the exact same reasoning applies that ##W_{on} = -\int_{V_i}^{V_f} P_{ext}dV## but this will just produce the exact same result as I got in a), i.e. no ##\gamma## will appear in the denominator.

    I'm pretty sure I'm missing a key concept here about it being an adiabatic process but I just can't see how it being adiabatic renders this thought process for calculating the work done incorrect.
     
    Last edited: Oct 13, 2016
  2. jcsd
  3. Oct 13, 2016 #2
    In the 2nd scenario, the internal energy changes because the temperature changes. In this scenario, Q is equal to zero.
     
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