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Homework Help: C_V or C_p for this question

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data

    A quantity of 35.0 of water at 25.0 C (called A) is mixed with 160.0 g of water at 86.0 C (called B). a) calcuate the final temperature of the system, assuming that the mixing is carried out adiabatically and b) calculate the entropy change of A, b, and the entire system

    2. Relevant equations
    delta S = n x C_V or C_p ln (Tf/Ti)

    3. The attempt at a solution

    The first part is very easy. Basically, before anything happens, the two gases need to be in thermal equilibrium.

    I just did q_A = q_b and figured the temperature at thermal equilbirum, which is 75 C or 348 K.

    part b was kind of confusing. I had Gibb's paradox in mind.

    I guess I was thinking along the lines of..

    if you take 1 L of water and a 2L bucket of water and mix them together, you are going to get three L of water, which is same as before. Thus, no volume change occurs and essentially the only entropy change you are going to get is due to heating.

    Which is delta S = n x C_V or C_p ln (Tf/Ti)

    I was debating of using C_v or C_p for this question. C_p looked really tempting but I didn't know what pressure it was occuring.. (it didn't say the mixing took place at atmospheric pressure ) so I supposed that it wouldn't be sort of right to using C_p.. (I guess what I missed here is that the pressure would equlibriate between A and B when you mix them and you would get a constant pressure.

    Finally I settled for C_v, because I recalled my previous argument that the volume doesn't change so the C_v applies.

    However, what I missed here is that the two waters are not at the same temperature. Thus, when the two waters come into thermal equilibrium, the volume of the colder water is going to expand a bit while the volume of the hotter water is going to shrink a bit. Thus C_v doesn't really apply here.

    The correct approach would have been using C_p.

    but then I ask, if volume do change.. what about adibatic expansion?

    In adibatic expansion..

    [tex]T2/T1 = (V1/V2) ^ (\gamma -1) [/tex]

    If there is a volume change say in A.. wouldn't that mean that the temperature of A is also going to change (drop a bit) .. not only that.. entropy is also given as delta S = nR ln (Vf/Vi)... so wouldn't this contribute to the entropy change of the A too?

    but then I was like... if the temperature of A is going to drop a bit.. then temperature of B is going to rise a bit because its volume is going to drop a little since it is in contact with the colder water.. so the two temperature changes would even out and you still would be back at 75 C.

    now to address delta S = nR ln (Vf/Vi) for B.. that would be like the opposite of the delta S = nR ln (Vf/Vi) for A so they would really cancel out?

    So I guess if instead of using C_p to figure out the entropy change for A..
    I would use C_v but I would just have to take the delta S = nR ln (Vf/Vi) into consideration...

    so.. that means

    nC_pln (Tf/Ti) = n C_vln (Tf/Ti) + nR ln (Vf/Vi)

    now.. when you figure out delta S for A and delta S for B, you see that they are just the opposite of each other.. I had Gibb's paradox in mind.. in which I was thinking that since they are all the same material, entropy should be 0 but that wouldn't really be true in this cause due to the temperature difference between the two waters..?

    So I guess my question really is.. is my way of thinking correct?

    Thanks for the patience!
  2. jcsd
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