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Consider one mole of helium at temperature 140K, that suffers a free expansion of Joule between 1L and 2L. Admit that the specific heat at constant value is Cv=(3/2)R. Using the general expression of the variation of internal energy $$dU = n C_V dT + [T(\frac{dp}{dt})-p]dV$$ where $$p=\frac{nRT}{V-nb}-\frac{n^2a}{V^2}$$ $$a=3.44 \times 10^{-3} Pa m^6 mol^{-2}$$ $$b= 23.40 \times 10^{-6} m^3 mol^{-1}$$

My attempt: I first calculated $$\frac{dp}{dt}= \frac{nR}{V-nb}$$

Substituting in the equation of energy $$dU = n C_V dT + [\frac{nRT}{V-nb}-p]dV$$

Integrating $$U = n C_V T + [nRT\ln{V-nb}-pV]$$

Now in a free expansion of Joule the variation of U is zero. Then: $$n C_V T_1 + [nRT_1\ln{V_1-nb}-p_1V_1]= n C_V T_2 + [nRT_2\ln{V_2-nb}-p_2V_2]$$

Solving for T2, we obtain

$$T_2=\frac{n3/2RT_1+nRT_1\ln(V_1-nb)-\frac{nRT_1V_1}{V_1-nb}+\frac{n^2a}{V_1}-\frac{n^2a}{V_2}}{n3/2R+nR\ln(V_2-nb)-\frac{nRV_2}{V_2-nb}}$$

Substituting all the values I obtain 157.5 K. However the answer should 139.9. Am I missing something?

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    $\begingroup$ Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. $\endgroup$
    – John Rennie
    Commented Jun 12, 2018 at 15:30
  • $\begingroup$ Yes I know you don't answer homework questions, I'm asking if I'm mistaking anything conceptually. If you tell me what's wrong I'll eventually figure out the answer. $\endgroup$
    – WatsonHolmes
    Commented Jun 12, 2018 at 15:38
  • $\begingroup$ @WatsonHolmes The usual expectation is 1) OP shows effort to work through the problem and 2) The question is about the concept. Here, you've shown us your working, but there's no indication that you're asking about the clarification of a certain topic. It resembles a question where people are asked to check your work, particularly because you've included both your answer and the expected final answer... $\endgroup$
    – user191954
    Commented Jun 12, 2018 at 15:50
  • $\begingroup$ The clarification I need is if I'm correctly relating expressions or what am I missing that's giving me an incorrect value. $\endgroup$
    – WatsonHolmes
    Commented Jun 12, 2018 at 15:52

1 Answer 1

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Your error is in integrating pdV to get pV (p is changing because T is changing). You should substitute the equation of state (actually the Van der Waals equation) into the term in brackets in your second equation, which then yields $n^2a/V^2$. This leads to an exact differential which you can then properly integrate with respect to V.

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  • $\begingroup$ Oh right! I didn't realize it because at first I was wrongly substituting pV with nRT (wrong, because this is not an ideal gas) and then when I realized that mistake I didn't come back do integrate again. I'll try again. Thank you so much. $\endgroup$
    – WatsonHolmes
    Commented Jun 12, 2018 at 19:47
  • $\begingroup$ So I was now testing this and obtained the equation $$T_2=(3/2RT_1 - \frac{na}{3V_1^3}+\frac{na}{3V_2^3})\frac{2}{3R}$$, which leads me to an absurd result of -803133... What am I doing wrong now? $\endgroup$
    – WatsonHolmes
    Commented Jun 12, 2018 at 20:21
  • $\begingroup$ What is the integral of $dV/V^2$? $\endgroup$
    – Chet Miller
    Commented Jun 12, 2018 at 20:33
  • $\begingroup$ oh god... sorry I'm tired $\endgroup$
    – WatsonHolmes
    Commented Jun 12, 2018 at 20:37

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