2
$\begingroup$

The gas that liquifies first, when cooled from 500 K to its critical temperature given in parenthesis is
a) $\ce{CO2}$ (304.1 K)
b) $\ce{NH3}$ (405.5 K)
c) $\ce{O2}$ (154.3 K)
d) $\ce{N2}$ (126.0 K)

Answer: b

I know that the critical temperature is the highest temperature at which the gas can be liquified and above which the gas can not be liquified whatever high pressures are used.

Also, $$\frac{p_\mathrm c V_\mathrm c}{T_\mathrm c} =nR =\frac{m}{M}R$$

i.e., $T_\mathrm c$ is directly proportional to $M$, when all the other quantities are constant.

So, based on $M$ $\ce{NH3}\lt\ce{N2}\lt\ce{O2}\lt\ce{CO2}$, so b should be the answer.

This is how I have solved this problem. Please tell me if I am wrong and if so, then please tell me the correct way to answer this.

$\endgroup$
3
$\begingroup$

Good question. I think you got the right answer but by using some pretty suspect logic.

First, there isn't enough information to solve the problem in general because pressure isn't specified. Is the cooling of the gas isobaric? Or adiabatic? Or ... ? Let's assume its isobaric, i.e., it happens at constant pressure. If so, what's the pressure? If the pressure is below 5.1 atm, then $\ce{CO2}$ will never condense into a liquid, instead it would transition directly to a solid phase (dry ice). Ammonia's triple point is roughly 0.06 atm. But your question says "cooled to the critical temperature", which implies not below the critical temperature, which means the only way any of them will liquefy is if the pressure is at or above the critical pressure. So let's assume the pressure is isobaric and is at least 11.8 MPa, the critical pressure of ammonia, the gas with the highest critical pressure among the possibilities. But if we do that, the question is easy. The gas that liquefies first is the gas with the highest critical temperature, which would be ammonia.

Relaxing this assumption makes the question much harder. Complex calculations would be required. The theory of corresponding states would likely be useful.

Your logic is OK, but in your equation, $w$, the weight, and $V_c$, the critical volume, also vary. So the statement that

$T_c$ is directly proportional to $M$

is not true, because the other factors are not constant.$%edit$

$\endgroup$
3
  • 1
    $\begingroup$ By definition, critical temperature is the highest temp at which gas can be liquefied, and its proportional to ratio of a and b so, higher the a , more intermolecular forces strength, easier it is to liquify, since sizes here are pretty much same {b} and can't cause difference of 100K, it must be big difference in intermolecular forces, again, Tc being proportional to M is wrong like Curt specified, if it was a different case of sizes and pressure not being enough for liquification, lots of complications would be there $\endgroup$
    – Mrigank
    Feb 9 '16 at 16:50
  • $\begingroup$ The Q is flawed and is being considered for deletion. Do you have objections to deletion, and/or can you see a way to rescue the OPs post through edits? $\endgroup$
    – Buck Thorn
    Jun 3 '21 at 12:59
  • $\begingroup$ Why is it being considered for deletion? What exactly is the flaw? $\endgroup$
    – Curt F.
    Jun 3 '21 at 15:43
1
$\begingroup$

The method you have used doesn't work for all cases. The equation is valid only for ideal gases. When some form of inter-molecular forces come into picture, the critical temperature doesn't depend on the mass alone. For example, $\ce{H2O}$ and $\ce{NH3}$ have similar molecular masses, but the critical temperature of $\ce{H2O}$ is higher because of stronger hydrogen bonding (~650 K). However, a better approximation of critical temperature would be

$T_c=\frac {8a}{27bR}$ which accounts for the inter-molecular forces and the size of the gas molecules.

$\endgroup$
0
$\begingroup$

You have assumed incorrectly that all other quantities like critical pressure and critical volume are the same for the gases. In reality it is possible to solve for the critical temperature in terms of the van der Waals constants $a$ and $b$ only so that $$T_c=\frac {8a}{27Rb}$$ Now qualitatively you can deduce that since for $\ce{NH_3}$ intermolecular forces are strong, so it's $a$ is higher than the others, and it is difficult to guess this in case of a generic problem.

$\endgroup$
0
-1
$\begingroup$

You should check the type of bonding whether it is hydrogen bonding:London dispersion.... Simply check out if all are organic go for... mass directly proportional to. Tc And if other then this prefer those which possess strong hydrogen bonding. Like water possess more strong hydrogen bond than NH3.... So Tc for water is more it is easy to liquefy!

$\endgroup$

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.