2
$\begingroup$

Can we justify that "For sublimation of a solid at 1 atm $\Delta U>0$ at low temperature and $\Delta U<0$ at high temperature?"

I read several answers and came to know that $U=f(T)$ is valid only for ideal gases at $\pu{1 atm}$ pressure where $T$ is temperature.

Internal energy is the sum of kinetic energy of molecules and chemical potential energy of molecules.

For phase transition even if the temperature remains constant, its internal energy changes.

I tried to justify by considering that while sublimation as solid changes to gases its degrees of freedom increases and we may say internal energy is a function of degrees of freedom so as temperature increases so $U_2 > U_1$ so $\Delta U >0$ but this contradicts what we have to prove.

Also similarly what can we say about internal energy for other phase transitions ?

enter image description here

Answer given is: ABCD

source:Physical Chemistry (Vol. II) (Class XI) Chemistry Module For JEE (Main+Advanced) Paperback – 1 January 2018 by Resonance Eduventures Limited

$\endgroup$
2
$\begingroup$

Can we justify that "For sublimation of a solid at 1 atm $\Delta U>0$ at low temperature and $\Delta U<0$ at high temperature?"

No.

$\Delta U>0$, always, for sublimation, because of the energy needed to separate the atoms or molecules in changing from the solid to the gas phase.

As for the enthalpy,

$$H = U +PV \Rightarrow \Delta H = \Delta U+\Delta (PV)=\Delta U+ P\Delta V+ V \Delta P$$

Thus at constant pressure:

$$\Delta H = \Delta U+ P\Delta V \approx \Delta U+ nRT$$

[Here I've used the ideal gas law to approximate the volume of the gas, and ignored the volume of the solid, which (at 1 atm, room temperature) is about 3 orders of magnitude smaller.]

I.e., $\Delta H$ will be even more positive than $\Delta U$, because of the pV-work required to make space for the gas.

The reason why sublimation becomes more favorable as the temperature increases is because sublimation has a positive $\Delta S$, and $\Delta G = \Delta H - T \Delta S$. But, irrespective of temperature, at any temperature and pressure at which the solid can exist, $\Delta U_{solid->gas} >0 $.

$\Delta U >0$ for liquid->gas and solid->liquid phase transitions as well (with the possible exception of the solid->liquid phase transitions for helium-3 and helium-4 at extremely low temperatures which, at least according to https://en.wikipedia.org/wiki/Enthalpy_of_fusion, have $\Delta H < 0$; but solid helium doesn't exist at the 1 atm pressure specified by the OP).

*Yes, you might be able to come up some extreme hypothetical mechanical system where the solid is under such great pressure that the intermolecular forces are so far into the repulsive part of their potentials that $\Delta U_{solid->gas} <0 $. But that's clearly not what the OP had in mind, since s/he specified 1 atm. And, in addition, such an arrangement would require that only the solid, but not the gas, be at that extreme pressure. So, strictly speaking, that $\Delta U$ would not be for the sublimation alone, it would be for the sublimation plus the pressure change.

$\endgroup$
3
  • $\begingroup$ btw I got this question in my textbook which was a multiple choice question which said "for sublimation of a solid at 1 atm,which of the following may be correct" and options $\Delta U >0$ at low temperature,$\Delta U < 0$ at high temperature were given correct,so did they necessarily meant extreme hypothetical conditions? $\endgroup$ – user69608 Aug 14 '20 at 13:31
  • $\begingroup$ @user69608 Can you give a complete reference to the textbook, and a screenshot of the problem and answer? $\endgroup$ – theorist Aug 14 '20 at 16:16
  • $\begingroup$ edited the question $\endgroup$ – user69608 Aug 14 '20 at 16:42
1
$\begingroup$

Can we justify that "For sublimation of a solid at 1 atm $\Delta U>0$ at low temperature and $\Delta U<0$ at high temperature?"

A justification can be suggested by writing

$$\Delta_{\mathrm{sub}} U = \Delta _{\mathrm{sub}} H - RTn$$

(but in retrospect it has a flaw, see below). If you assume the enthalpy of sublimation is approximately constant (weakly dependent on T) then it is clear that to sublimate a constant amount of substance n requires an input of energy at low T (in the form of input heat $q_p = \Delta _{\mathrm{sub}} H$ to break bonds in the solid lattice) but results in a net reduction in internal energy of the system (due to work done on expansion) at sufficiently high T.

As hinted at in another answer, however, this justification has one flaw*. The flaw is that (in the constant pressure scenario) heat input provides energy not only to break bonds, but also to drive the expansion of the gas (perform work). In the isochoric (constant volume) scenario there is no work and $\Delta U = q_V$. In the isobaric (constant pressure) scenario the sum of the additional energy input as heat (>0) to perform expansion work and energy loss due to the expansion work (<0) cancel, so these changes do not alter $\Delta U$.

It should be noted that $\Delta _{\mathrm{sub}} H>0$ (sublimation is endothermic).

It is interesting to note that in the case of vaporization the enthalpy drops to zero as T nears the critical point. Similarly the expansion work required drops to zero near the critical point.

*I thought there was an additional flaw, but haven't figured it out entirely.

$\endgroup$
19
  • $\begingroup$ What if for a solid at high temperature $\Delta H$ is sufficiently greater than $RTn$ then $\Delta U$ would come positive which contradicts.? $\endgroup$ – user69608 Aug 8 '20 at 18:02
  • $\begingroup$ Can you answer the other questions in the post? $\endgroup$ – user69608 Aug 8 '20 at 18:03
  • 1
    $\begingroup$ @BuckThorn There are two issues here. The first is $\Delta H$ vs $\Delta U$: This is a sublimation, so it's at constant T. Imagine an endothermic process (reaction, phase change, doesn't matter) at constant T & V. q_v = $\Delta U$. Now suppose we switch to constant p, and further suppose the reactant is a solid and the product is a gas. More heat (q_p) will need to flow into the reactant mixture (equal to $p \Delta V$) because now we additionally need to do the work to expand the system. Look at the math: $\Delta H = \Delta U + p \Delta V$. The pV term is positive, not negative. $\endgroup$ – theorist Aug 10 '20 at 21:16
  • 1
    $\begingroup$ But the first issue ($\Delta H$ vs $\Delta U$) is incidental to the question, which is what is happening with $\Delta U$. Think about it physically -- you are pulling molecules apart. $\Delta U$ can't be negative, irrespective of whether you have to add or subtract pV to get $\Delta H$ (though you do have to add pV: $\Delta H = \Delta U + p \Delta V$). The pV term only affects $\Delta H$. It has no effect on $Delta U$. $\endgroup$ – theorist Aug 10 '20 at 21:18
  • 1
    $\begingroup$ We're doing this at constant temperature, so any pV work required by doing this at constant p is cancelled out by increased heat flow into the system, leaving $\Delta U$ unaffected by the pV work. That increased heat flow due to pV work at constant p accounts for nearly all the difference between $\Delta H$ and $\Delta U$*. [*There will be a slight effect on$\Delta U$ from the p-dependence of the internal energy of the solid, but that is small.] And you can carry out sublimation at constant V by putting a solid into an evacuated container. $\endgroup$ – theorist Aug 10 '20 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.