Free expansion of Real Gases - Dieterici EoS, Change in Temperature

[curious_mind]

Homework Statement

A box is divided into two equal halves with a partition. The volume of the entire box is $V$. One partition contains the Real gas satisfying Dieterici Equation of state at temperature $T_0$. Take Dieterici equation of state in this case as $PV e^{\frac{a}{RTV}} = nRT$
where $a$ is a constant. Now, the partition is removed instantaneously, and the gas is allowed to expand to fill the full volume of the box and come to equilibrium. Calculate the temperature of gas at the final stage when it is at equilibrium.

Relevant Equations

First internal energy equation, using Maxwell's relation $\left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial P}{\partial T} \right)_V - P$.

I proceeded in the usual manner in which we take $dU = 0$ in the case of free expansion because there is no heat transfer in the box, as well as no work is done.

We can write, taking $U$ as the function of $V$ and $T$, $$ dU(V,T) = \left( \dfrac{\partial U}{\partial V} \right)_T ~ dV + \left( \dfrac{\partial U}{\partial T} \right)_V ~ dT $$ Also, $\left( \dfrac{\partial U}{\partial T} \right)_V = C_V$ is specific heat at constant volume.

Therefore, $0 = C_V ~ dT + \left( \dfrac{\partial U}{\partial V} \right)_T ~ dT$.

Using the "First internal energy equation" given above, I calculated that (for Dieterici EoS, as given in the question) $\left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial P}{\partial T} \right)_V - P = \dfrac{Pa}{RTV}$.
[I have double verified my calculation, but still I request to verify that again.]

Using given equation of state, $P = \dfrac{nRT}{V} e^{-\frac{a}{RTV}}$. So, $\left( \dfrac{\partial U}{\partial V} \right)_T = \dfrac{na}{V^2} e^{-\frac{a}{RTV}}$

Therefore, I obtain the expression $0 = C_V ~ dT + \left( \dfrac{na}{V^2} e^{-\frac{a}{RTV}} \right) dV$. Writing it in differential equation form,
$$ \dfrac{dV}{dT} = - \dfrac{C_V V^2}{na} e^{\frac{a}{RTV}} $$.
Now, I am not sure how to solve this differential equation, where I have to take limits for volume $\frac{V}{2}$ to $V$ and temperature from $T_0$ to temperature to be obtained $T$.

Problem with the obtained differential equation is that, I am not getting it in a seperable form. Also, I know very little differential equation theory which could be helpful in solving this.I also tried Wolfram Mathemtica DSolve function, but it is also getting very large expressions.

The question is given here https://snboseplc.com/thermodynamics-tifr/ - Question number 22. The answer given here is $T-\left(\frac{na}{C_V}\right)\ln{2}$. How do I obtain this answer ? Is there any mistake in my approach, or there can be other approach ?
Even if the answer is not accurate, how do I obtain the simplified answer ? How do people manage to get temperature change in Engineering thermodynamics to deal with such problems?Any help would be appreciable. Thanks.

 

[Chestermiller]

Homework Statement:: A box is divided into two equal halves with a partition. The volume of the entire box is $V$. One partition contains the Real gas satisfying Dieterici Equation of state at temperature $T_0$. Take Dieterici equation of state in this case as $PV e^{\frac{a}{RTV}} = nRT$
where $a$ is a constant. Now, the partition is removed instantaneously, and the gas is allowed to expand to fill the full volume of the box and come to equilibrium. Calculate the temperature of gas at the final stage when it is at equilibrium.
Relevant Equations:: First internal energy equation, using Maxwell's relation $\left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial P}{\partial T} \right)_V - P$.

I proceeded in the usual manner in which we take $dU = 0$ in the case of free expansion because there is no heat transfer in the box, as well as no work is done.

We can write, taking $U$ as the function of $V$ and $T$, $$ dU(V,T) = \left( \dfrac{\partial U}{\partial V} \right)_T ~ dV + \left( \dfrac{\partial U}{\partial T} \right)_V ~ dT $$ Also, $\left( \dfrac{\partial U}{\partial T} \right)_V = C_V$ is specific heat at constant volume.

Therefore, $0 = C_V ~ dT + \left( \dfrac{\partial U}{\partial V} \right)_T ~ dT$.

Using the "First internal energy equation" given above, I calculated that (for Dieterici EoS, as given in the question) $\left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial P}{\partial T} \right)_V - P = \dfrac{Pa}{RTV}$.
[I have double verified my calculation, but still I request to verify that again.]

Using given equation of state, $P = \dfrac{nRT}{V} e^{-\frac{a}{RTV}}$. So, $\left( \dfrac{\partial U}{\partial V} \right)_T = \dfrac{na}{V^2} e^{-\frac{a}{RTV}}$

I agree with your results up to this point. But, next, there is a problem with the n. If you are working on a per-mole basis, then the n shouldn't be in there. Otherwise you should be writing your next equation as $$dU=nC_vdT+ \left( \dfrac{na}{V^2} e^{-\frac{a}{RTV}} \right) dV$$or, $$\frac{1}{n}dU=C_vdT+ \left( \dfrac{a}{V^2} e^{-\frac{a}{RTV}} \right) dV$$
The next step is to recognize from this equation that $C_v$ is a function of V. Do you know how to show this?

To get what you want, namely the change in temperature for the specified volume change at constant internal energy, you will not be able to integrate the equation directly. You will have to apply Hess's Law in a 3-step path. Can you figure out how to do this?

 

[Chestermiller]

$$nC_v=\left(\frac{\partial U}{\partial T}\right)_v$$so $$\left(\frac{\partial nC_v}{\partial V}\right)_T=\frac{\partial^2 U}{\partial T\partial V}$$Also, $$\left(\frac{\partial U }{\partial V}\right)_T= \frac{na}{V^2} e^{-\frac{a}{RTV}} $$From this equation, what do you get for $\frac{\partial^2 U}{\partial T\partial V}$?

 

[curious_mind]

From this equation, what do you get for $\frac{\partial^2 U}{\partial T\partial V}$ ?

Ok I understood why $C_v$ has to be the function of $V$
$\dfrac{\partial^2 U}{\partial T \partial V} = \dfrac{\partial C_v}{\partial V} = \dfrac{a^2}{R T^2 V^3} e^{-\frac{a}{RTV}}$
Thanks. So there is no way I can integrate out the equation.

Is this the Hess's law, I need to apply ? https://en.m.wikipedia.org/wiki/Hess's_law

I am not getting your hint, do I need to apply Enthalpy state function instead of internal energy ?

 

[Chestermiller]

Ok I understood why $C_v$ has to be the function of $V$
$\dfrac{\partial^2 U}{\partial T \partial V} = \dfrac{\partial C_v}{\partial V} = \dfrac{a^2}{R T^2 V^3} e^{-\frac{a}{RTV}}$
Thanks. So there is no way I can integrate out the equation.

No, unless you do it numerically.

Is this the Hess's law, I need to apply ? https://en.m.wikipedia.org/wiki/Hess's_law

I am not getting your hint, do I need to apply Enthalpy state function instead of internal energy ?

Hess's law says that U is a function of state, so we can integrate du along any convenient path from the initial state to the final state, and then solve for the value of T for which $\Delta U = 0$. It turns out that there is a particular path for which the equation for dU can be integrated analytically.

The method capitalizes on the fact that, at very high molar volumes, we approach ideal gas behavior, where the heat capacity Cv approaches the ideal gas heat capacity, which is a function only of temperature (and not volume). That is $$C_v(T,\infty)=C_v^{IG}(T)$$Not only that but, for most gases, the ideal gas limiting heat capacity as a function of temperature is available in the literature.

Our method consists of integrateing du over three separate segments, two of which are isothermal and the other is constant volume:

Segment 1: $\Delta U_1=U(T_0,\infty)-U(T_0,V_0)$
Segment 2: $\Delta U_2=U(T,\infty)-U(T_0,\infty)=\int_{T_0}^T{C_v^{IG}(T')dT'}$
Segment 3: $\Delta U_3=U(T,2V_0)-U(T,\infty))$
So, $$\Delta U=\Delta U_1+\Delta U_2+\Delta U_3)=U(T,2V_0)-U(T_0,V_0)$$

See what you get when you integrate dU along this convenient path.

Questions?