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Gas mixture final temperature

  1. May 2, 2012 #1
    1. The problem statement, all variables and given/known data

    I need some help with the following problem:

    Consider a thermally insulated enclosure containing an ideal gas initially separated by a thermally insulated partition into two separate regions. One region is at pressure p1=100 Pa, temprature T1=200 K, and volume V1 = 10 m3. The other is at p2=50 Pa, T2=300 K, and V2=20 m3.

    If the partition is removed and the gas from each region is allowed to mix freely, what is the final temprature of the mixture?

    2. Relevant equations

    Ideal gas law:

    PV=NKT

    p=ρRT

    3. The attempt at a solution

    I've first found the final pressure as follows

    [itex]U_i \propto P_1 V_1 + P_2 V_2[/itex]
    [itex]U_f \propto P_f (V_1+V_2)[/itex]

    [itex]P_1 V_1 + P_2 V_2 = P_f (V_1 + V_2)[/itex]

    [itex]P_f = \frac{V_1}{V_1 + V_2} P_1 + \frac{V_2}{V_1 + V_2} P_2 = 66.66 \ Pa[/itex]

    Now to find the temprature, I wrote

    [itex](N_1 + N_2) k T_f = (N_1T_1 + N_2T_2) k = P_f(V_1+V_2) = 66.66[/itex]

    But how can I solve for "Tf"? The problem is that I do not know the number of moles of gas in each partition. Is there a way to calculate the number of moles, or should I use a different method to find Tf? Any helps is appreciated.
     
    Last edited: May 2, 2012
  2. jcsd
  3. May 2, 2012 #2
    Hint:

    There is no work being done so initial internal energy equals final internal energy. While you cannot determine a numerical value for the mass in each section, you can determine the ratio of the masses.
     
  4. May 3, 2012 #3
    Thank you for the hint, but I think we still need the numerical values for the masses in each section in order to find Tf because

    Qi=Qf

    m1c(T1) + m2c(T2) = (m1 + m2)cTfinal

    [itex]T_{final} = \frac{m_1(T_1) + m_2(T_2)}{m_1 + m_2}[/itex]

    This is how I would factor mass into my equations, but I can't solve this without knowing the two masses. So, how can we find the ratio of the masses that you mentioned (given volume and pressure)? And how does that help in solving this equation? :confused:
     
    Last edited: May 3, 2012
  5. May 4, 2012 #4
    "m1c(T1) + m2c(T2) = (m1 + m2)cTfinal"

    Use the ideal gas law to relate m1 to m2. One is the other multiplied by a factor. Then each term has the same mass (either m1 or m2) in it and can be removed from the equation.
     
  6. May 4, 2012 #5
    I tried this but I'm still getting stuck. Unless I misunderstood you, this what I did using the ideal gas law PV=ρRTV with ρ=mN/V, so I can write

    [itex]PV=\frac{mN}{V} RTV[/itex]

    [itex]P_1V_1 = P_2 V_2[/itex]

    [itex]m_1 N_1 T_1 = m_2 N_2 T_2[/itex]

    so I can write each mass as the other one times a factor:

    [itex]\left\{\begin{matrix}m_1 = m_2 \frac{N_2T_2}{N_1T_1}\\ m_2 = m_1 \frac{N_1 T_1}{N_2T_2} \end{matrix}\right.[/itex]

    And so I substituted those in the equation:

    [itex]T_f = \frac{m_1T_1+m_2T_2}{m_1+m_2}[/itex]

    and I get:

    [itex]T_f = \frac{m_2 \frac{N_2T_2}{N_1} + m_1 \frac{N_1 T_1}{N_2}}{m_1+m_2}[/itex]

    I tried to simplify this but I can't get rid of N1, 2 and m1, 2. So what can I do?
     
  7. May 5, 2012 #6
    Hint:

    Let's write the conservation of energy equation:

    M1cvT1 + M2cvT1 = M1cvTf + M2cvTf, specific heat a common factor so bye, bye.

    rho1 = P1/(R*T1) where rho is density
    rho2 = P2/(R*T2)

    M1 = rho1 * V1
    M2 = rho2 * V2
     
  8. May 5, 2012 #7
    Thank you very much, I get it now! I substituted the masses into Tf=(m1T1+m2T2)/(m1+m2) and the answer came out 240 K, hopefully that is correct. Thanks for the help!
     
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