I have seen multiple questions which talk about a tank with a partition in the middle and different ideal gases on either side at some different value of pressure,volume and temperature and then they say that when the partition is removed find final temperature of the mixture of gases....what has been told to me is that we can conserve internal energy writing * nCv(T1) initial of gas 1 + nCv(T2)initial of gas 2= nCv(Tfinal) of mixture. My problem with this is that these are not fundamental internal energies...infact it was explicitly mentioned by my teachers that this formula only represents change in internal energy and the actual value of internal energy can never be calculated since there are too many energy parameters and we can never sum all of them up accurately...so how is this approach to the solution correct becoz it always manages to give the right answer..is it something like the net energy of system shouldn't change so the net changes in each side should be equal to the entire change?? Is there any other approach to this problem...I was also thinking of mole conservation as an approach but was not able to execute it
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$\begingroup$ Are you talking about a mixture of two ideal gases? $\endgroup$– Chet MillerCommented Aug 11, 2019 at 16:30
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$\begingroup$ Yes.. initially separated by a partition and later the partition is removed $\endgroup$– Schwarz KugelblitzCommented Aug 11, 2019 at 16:31
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$\begingroup$ @SchwarzKugelblitz I have revised and simplified my answer. Hope it works better for you. $\endgroup$– Bob DCommented Aug 12, 2019 at 16:22
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$\begingroup$ @SchwarzKugelblitz So does it now answer your question acceptably? $\endgroup$– Bob DCommented Aug 12, 2019 at 20:16
2 Answers
Your teachers are correct that we normally talk about the change in internal energy of an ideal gas according to
$$\Delta U=nC_{v}\Delta T$$
It’s true you generally can't determine the absolute value of internal energy because of the many contributions (kinetic plus potential) to it at the molecular level. But it really doesn't matter if you are only interested in the change in internal energy. In reality the above equation is doing just that when you expand it as follows
$$\Delta U=nC_{v}\Delta T=nC_{v}(T_{2}-T_{1})=nC_{v}T_{2}-nC_{v}T_{1}=U_{2}-U_{1}$$
Where
$U_2=nC_{v}T_{2}$ is the internal energy of the gas at temperature $T_{2}$
$U_1=nC_{v}T_{1}$ is the internal energy of the gas at temperature $T_1$.
This demonstrates that it doesn't matter if you assign an absolute value to the internal energy if you are only interested in a change in internal energy. An example of where this is done is in the steam tables for water. Internal energy is assigned a value of zero at the triple point for water. Then the absolute values of the liquid and gaseous components of internal energy in the table are based on that. The application of these values is only to changes in internal energy.
Hope this helps.
In engineering, we regard an ideal gas as a substance which exhibits the limiting behavior of a real gas in the region of low gas density. As such, it displays the following key characteristics:
Its equation of state is Pv=RT, where v is the molar volume
Its molar internal energy u is a function only of temperature T, and not molar volume or pressure. This temperature-dependence is not typically linear, such that its molar heat capacity $c_v=\frac{du}{dT}$ is not constant. And so, itsmolar internal energy cannot be represented simply as u=c_vT. Instead, it must be represented as an integral with respect to temperature: $$u(T)=u(T_{ref})+\int_{T_{ref}}^T{c_v(T')dT'}$$where T' is a dummy variable of integration, $c_v(T)$ is the molar heat capacity in the limit of ideal gas behavior, and $u(T_{ref})$ is the absolute molar internal energy at a specified reference temperature $T_{ref}$. The value of the molar internal energy at this specified reference temperature does not actually need to be known because it will cancel out of all calculations that we make of practical interest. Over a limited range of temperatures in the vicinity of $T_{ref}$, we can often assume that the molar heat capacity is nearly constant, and we can write $$u(T)\approx u(T_{ref})+\bar{c_v}(T-T_{ref})$$
- In a mixture of ideal gases, each gas of the mixture behaves as if it is the only gas present in terms of its contribution to the overall mixture internal energy. So we can write for a mixture of ideal gas components, $$U=n_1u_1(T)+n_2u_2(T)+...$$where $n_i$ is the number of moles of species i in the mixture and $u_i(T)$ is the molar internal energy of pure species i at the same temperature T.
Based on these considerations, your conservation of internal energy equation for your tank-with-a-partiation problem can be written as follows: $$U=n_1[u_1(T_{ref_1})+\bar{c}_{v1}(T_1-T_{ref_1})]+n_2[u_1(T_{ref_2})+\bar{c}_{v_2}(T_2-T_{ref_2})]=n_1[u_1(T_{ref_1})+\bar{c}_{v1}(T_f-T_{ref_1})]+n_2[u_1(T_{ref_2})+\bar{c}_{v2}(T_f-T_{ref_2})]$$where $T_f$ is the final equilibrium temperature. What do you get when you solve this for for the final temperature, and does it involve the internal energies of the two species at their reference temperatures?
