# Sublimation: invariant heat or internal energy?

1. Nov 29, 2016

### Vitor Pimenta

1. The problem statement, all variables and given/known data

Below, two experiments (1 and 2) are described, in which the same quantity of solid carbon dioxide is completely sublimated, at 25ºC:
1. The process is carried out in a hermetically sealed container, non-deformable with rigid walls;
2. The process is carried out in a cilinder with a piston, whose mass is negligible and moves without friction
The problem then makes statements about $\Delta U$ (change in internal energy in each process), $q$ (heat exchanged in each process) and $w$ (work done on the system of CO2 for each process), looking for the wrong statements.

My problem is with the following statements:

a) $\Delta {U_1} = \Delta {U_2}$
b) ${q_1} = {q_2}$
The false statement (according to the textbook) should be (a) , but I think (b) is the wrong one.

2. Relevant equations

• First Law of Thermodynamics: $\Delta U = q + w$
• Calorimetry equation for sublimation: $q = n{C_s}$ (in which $n$ is the number of moles of gas and ${C_s}$ is the molar latent heat of sublimation)

3. The attempt at a solution

About process (1) , no work is done by the gas being generated (after all the container is rigid), so first law of thermodynamic reduces to $\Delta {U_1} = {q_1}$.

Now, about process (2), work is done (since the piston can be moved by the gas being generated inside the cilinder), and the same law can only be written fully: $\Delta {U_2} = {q_2} + {w_2}$

I further assume that, considering CO2 as ideal gas, the change in internal energy will be the same for both processes, since they start from the same state (solid, 25ºC) and end with gas at the same temperature (25ºC).

This way of thinking, however, would lead to ${q_1} = {q_2} + {w_2}$ , that is, the realization that heat exchanged by process (1) is not the same as by process (2). I reached the opposite conclusion of the exercise´s intended answer (since I think both changes in internal energies are the same and that the heat exchanged are different).

Lastly, I included calorimetry equation because that seems to be an argument for reaching the exercise´s intended answers: Assuming ${C_s}$ depends uniquely on temperature, its value would be the same for both processes (1 and 2), something that would lead to the conclusion that ${q_1}={q_2}$ (both processes employ the same $n$ of gas).

2. Nov 29, 2016

### Charles Link

Both systems start out the same but they don't specify if both systems finish up with the same volume. They both finish up with T=25 degrees Centigrade. For an ideal gas, which perhaps can be assumed here, $U=U(T)$ independent of volume. Let's assume $\Delta U_1=\Delta U_2$,(i.e. the two containers at 25 degrees Centigrade have equal $U$ even if their volumes are different). The piston does a considerable amount of work in case (2) ($w_2=p_o \Delta V_2$ where $p_o$ is the ambient air pressure), so that $q_2=\Delta U_2+w_2 >q_1=\Delta U_1=\Delta U_2$ so that $q_2>q_1$. $\\$ (The first law of thermodynamics correctly reads $q=\Delta U+w$ if you define $w$ as the work done by the system (which I think is the usual convention)). $\\$ I am in agreement with you that (b) is the false statement. $\\$ editing...This is the case regardless of whether the final volumes are the same or not in the two cases. $\\$ Additional note: I think the calorimetry equation is somewhat incomplete and does not justify the result $q_1=q_2$. The heat of sublimation I found to be 6030 cal/mole in one source at T=195 K, but a quick estimate of $w_2=p_o \Delta V$ (V=22.4 liters at S.T.P. (T=0 degrees C) with $p_o=1 \, atm$ that $w_2=530 \, cal/mole$ because $R$ =.08206 liters-atm(mole-deg K)=1.987 cal/(mole-deg K) This makes 22.4 l-atm=[(22.4)(1.987)/(.08206)} calories. At T=25 C, the volume will be even slightly greater than 22.4 liters.

Last edited: Nov 29, 2016
3. Nov 30, 2016

### Charles Link

Additional comment to the above: The heat of vaporization/sublimation is usually given as $\Delta H$. $\$ $\Delta U$ for the energy of sublimation is normally smaller than this $\Delta H$ because $\Delta H=\Delta U +p_o \Delta V$ For a quick estimate of $\Delta U$ I think $\Delta H$ is often used, but for more precise calculations, the distinction should be made between $\Delta U$ and $\Delta H$. $\\$ One other additional item in the above is the $\Delta U$ process is complicated by having the solid at T=195 K and the gaseous state at T= 25 C. To do any precise computations of $\Delta U$, the $U(298 K)-U(195 K)$ for the gaseous state would also need to be computed. The $\Delta U= 6030 \, cal$ and $w_2= 530 \, cal$ were simply computed as rough estimates, and were not intended to be of high precision.

4. Nov 30, 2016

### Vitor Pimenta

Thanks for the response, I see you agreed with me that changes in internal energy should be the same for both IF the gas is treated as ideal. However, I could not understand what would be responsible for the "incompleteness" you saw in the calorimetry equation. Would it be that, instead of describing $q$, the equation describes ${{\Delta}H}$ ? (so that it would no longer justify ${q_1}={q_2}$ ?)

5. Nov 30, 2016

### Charles Link

The calorimetry equation is usually a $\Delta H$, but it is likely to be the $\Delta H$ for the specific temperature at which the phase change occurs. In this sense it is a "q", but it is an incomplete "q". If the gas of your problem above finished at T=195 K in both cases, $q_2=\Delta H$ and $q_1=\Delta U=\Delta H-p_o \Delta V$. The calorimetry equation is incomplete if it doesn't give more specific details on parameters such as the external pressure, and also the final temperature, etc. Otherwise, you can apply it for rough estimates, but in the above example, the calorimetry number might be $c_s=6030 cal/mole$ and it could be off by 530 cal/mole because of the pressure conditions $(w_2)$, and even additional amounts that include the specific heat of the gas ($C_v$) in going from T=195 K to T=298 K. I did rough estimates of $w_2$ to show that $q_1$ and $q_2=q_1+w_2$ could differ significantly. If $w_2 < 50 cal/mole$, (with $q_2=6030 \, cal/mole$), then it would have been a reasonably good approximation to say $q_1=q_2$. Clearly it is not a very good approximation to say $q_1=q_2$, but $\Delta U_1=\Delta U_2$ to a reasonably good approximation.

6. Nov 30, 2016

### Vitor Pimenta

I think I get it now. You were arguing why you think the calorimetry equation is incomplete and (just the same as I think) you consider that the latent heat is not completely determined only by temperature, but also needs other parameters like external pressure.

That is exactly why I question the textbook´s answer: Equating the changes in internal energy seems much more "secure" than equating heats.

Thanks for taking the time ! I guess I was not crazy after all...

7. Nov 30, 2016

### Staff: Mentor

If the final gas were an ideal gas, then, assuming that both solids started out at the same state, the final internal energies would be equal if the final temperatures are the same. This is true even though the final pressure in case 1 will be higher, since internal energy is a function only of temperature for an ideal gas. So, $\Delta U_1=\Delta U_2$.

If we again assume that the final gas is an ideal gas, then we have the following sequence of logic:
In case 1, the change in internal energy is equal to the heat added: $$q_1=\Delta U_1$$ In case 2, the heat added is equal to the change in enthalpy: $$q_2=\Delta H_2$$ Now, $$\Delta H_2=H_{2f}-U_i-P_2V_i$$ where $H_{2f}$ is the final enthalpy in case 2 (which is also the final enthalpy of case 1), $U_i$ is the initial internal energy in cases 1 and 2, $V_i$ is the initial volume of the solid in cases 1 and 2, and $P_2$ is the constant pressure in case 2 (and the initial pressure in case 1). But, $$H_{2f}=U_{2f}+P_2V_{2f}$$Therefore, $$\Delta H_2=U_{2f}-U_i+P_2V_{2f}-P_2V_i=\Delta U+P_2(V_{2f}-V_i)$$ where $$\Delta U=\Delta U_1=\Delta U_2$$Therefore, $$q_2=q_1+P_2(V_{2f}-V_i)$$

Now, these results are all derived assuming that the CO2 vapor in the final state at 25 C (77 F) behaves like an ideal gas. But that is not really the case. Here is a Pressure-Enthalpy diagram for CO2 which clearly indicates that the behavior is not ideal.

So, how do you think that this changes things?

8. Dec 1, 2016

### Vitor Pimenta

Obviously, assuming equality between changes in internal energy is no longer valid. Even so, affirming equality of heat exchanged based on calorimetry also seems wrong (because latent heat of sublimation probably depends on pressure as well).

An alternative I am thinking of is to treat CO2 as a Van der Waals gas, since it would then have internal energy of the form:

$$U(T,V,N) = U_\mathrm{ideal}(T,N) - \frac{aN^2}{V}$$

Even like this, however, we would need the volume of the container for the first process, in order to compare both processes.

9. Dec 1, 2016

### Staff: Mentor

It has nothing to do with the effect of pressure on the heat of sublimation. Even if, once sublimated, CO2 were an ideal gas, the two amounts of heat would not be the same.
The pressure-enthalpy graph I presented provides sufficient information to quantify the difference in the internal energy changes. But, certainly, the final internal energies $U_{1f}(T, P_1)$ and $U_{2f}(T,P_2)$ are not equal if CO2 is non-ideal. So, assuming the two systems start in the same state, the change in internal energy for the two processes will be different.
You are expected to assume that two initial volumes, pressures, and temperatures are the same in the two cases.

If you would like to perform a quantitative calculation using the pressure-enthalpy diagram I presented, I can show you how to do it.

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