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Archived Thermodynamics: Calculate the volume and temperature

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data
    In an isolated chamber closed by a moving piston is an capsule with 0.5 l of air at 3 bar and 100 °C. Chamber is filled with 2 l of air at 1 bar and 20 °C. The same pressure and temperature surround the piston from the other side. Capsule breaks. What is the volume and temperature now? ##\kappa =1.4##


    2. Relevant equations

    ##\Delta U= \Delta W + \Delta Q = \Delta W + 0## because of the isolation

    ##p_cV_c=\frac{m_c}{M}RT_c## - c for capsule

    ##p_0V_0=\frac{m_0}{M}RT_0## - 0 for gas before the capsule breaks

    ##p_0V=\frac{m_0+m_c}{M}RT## - no index for gas after the capsule breaks

    3. The attempt at a solution

    I would be really happy if somebody could check what am I doing wrong because I just can't get the same result as written in the book...

    ##\Delta U=m_cc_v(T-T_c)+m_0c_v(T-T_0)=(m_c+m_0)c_vT-m_cc_vT_c-m_0c_vT_0##

    ##\Delta W=-p_0(V-(V_c+V_0))##

    ##(m_c+m_0)c_vT-m_cc_vT_c-m_0c_vT_0=-p_0(V-(V_c+V_0))##

    Now for all the indexes: ##m_iT_i=\frac{p_iV_iM}{R}## . Than:

    ##\frac{M}{R}c_v(p_0V-p_1V_1-p_0V_0)=-p_0V+p_0(V_c+V_0)##

    for ## cv=\frac{R}{M(\kappa -1)}##

    ##\frac{1}{\kappa -1}(p_0V-p_1V_1-p_0V_0)+p_0V=p_0(V_c+V_0)##

    ##p_0V(\frac{1}{\kappa -1}+1)-\frac{1}{\kappa -1}(p_1V_1+p_0V_0)=p_0(V_c+V_0)##

    ##p_0V(\frac{\kappa }{\kappa -1}=p_0(V_c+V_0)+\frac{1}{\kappa -1}(p_1V_1+p_0V_0)##

    ##V=\frac{(\kappa -1)p_0(V_c+V_0)+(p_1V_1+p_0V_0)}{p_0\kappa }##



    For data above, this gives me 3.2 l instead of 2.71 l.

    http://web2.0calc.com/?q=(0.4*10**5*(0.0005+0.002)+3*10**5*0.0005+10**5*0.002)/(10**5*1.4


    WHAT ON EARTH is wrong here? :/
     
  2. jcsd
  3. Sep 11, 2016 #2
    I get 3.21 liters also. It looks like they forgot to include the 0.5 liters for the capsule. So I confirm skrat's answer. I get 23 C for the final temperature.
     
    Last edited: Sep 12, 2016
  4. Sep 13, 2016 #3
    This is a completely incorrect interpretation of what happens. The gas originally in the capsule has a temperature of 100 C, and its temperature changes to 23 C. So pV is not constant. Also, the capsule gas occupies the entire 3.2 liters in the final state, not 0.7 L. And finally, it's final partial pressure is not 1 Bar. It is only about 1/3 bar. The number of moles of gas originally in the capsule is about 0.043. Try pV=(3.2)(0.33)=nRT=(0.043)(0.082)(296) and see what you get.
     
  5. Sep 13, 2016 #4
    The gases are intermingled in the final state, and molecules of both gases are present throughout the entire chamber.
    Who said that the final pressure is not 1 bar? I said that the final partial pressure of the gas that was originally in the capsule is 1/3 bar in the final state. Are you familiar with the concept of partial pressure?
    Who says? You?
    Yes, I do, because it really is completely incorrect.
    No. I stand by what I said. Now, if you continue to pursue this incorrect avenue of discussion, I will have to issue you a misinformation warning.
     
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