Verifying Answer to PV/T at A: Is T2 > T1?

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

I have correctly calculated value of PV/T at A .

Assuming volume is constant , for a particular value of pressure P ( for example , at the lowest points of the two graphs ) PV/T value at T₂ is lower than at T₁ which means T₂ is greater than T₁ .

According to me answer is option a) .

But this is wrong . Book answer is option c) .

Could someone please verify the answer .

 

[Charles Link]

This, in some ways, is a somewhat specialized question. Even though I am quite familiar with the ideal gas law, this one is showing a very dramatic deviation from being ideal, so I googled a result which I think provides good qualitative info that might be useful: https://chem.libretexts.org/Textbook_Maps/General_Chemistry_Textbook_Maps/Map:_Chemistry:_The_Central_Science_(Brown_et_al.)/10:_Gases/10.9:_Real_Gases_-_Deviations_from_Ideal_Behavior It was sort of my guess that this (as shown in the "link" ) might be the case, but it's very hard to predict this sort of thing. $\\$ Editing: An analysis using the Van der Waal's equation showed the behavior of the blue curve (T=173 K) could occur at low temperatures. I think this was intended to be a qualitative exercise, but if you are interested, let me show the result: Van der Waal's equation is $(P+\frac{an^2}{V^2})(V-nb)=nRT$ . Expanding out the terms: $PV-nbP+\frac{an^2}{V}-\frac{abn^3}{V^2}=nRT$. Dividing by $nRT$ and rewriting: $\frac{PV}{nRT}=1+\frac{bP}{RT}-\frac{an}{VRT}+\frac{abn^2}{V^2 RT}$. Now $\frac{1}{V} \approx \frac{P}{nRT}$. Substituting this in gives: $\frac{PV}{nRT}=1+\frac{bP}{RT}-\frac{aP}{(RT)^2}+\frac{abP^2}{(RT)^3}$. (Notice, we now have an expression that will allow us to graph $\frac{PV}{nRT}$ vs. $P$ for various temperatures). At higher temperatures, the last two terms will be insignificant, and only the $1+\frac{bP}{RT}$ terms will be of significance. At low temperatures, the $-\frac{aP}{(RT)^2}$ term could cause an initial downward effect, (as $P$ is increased), greater (in absolute value) than the $\frac{bP}{RT}$ term, before the last term, that is second power in $P$, becomes large and causes the right side to be larger than 1. I think this analysis is beyond what they were expecting from the question. Without supplying additional info on $a$ and $b$, this analysis shows the result in the "link" is consistent, but it doesn't really predict it.

 

[Chestermiller]

 

[Jahnavi]

@Charles Link ,@Chestermiller Could you please help me understand the following lines from the textbook .

1) How can pressure be same throughout the system (liquid+vapour) ? Do they mean pressure at the interface ?

2) How is density same as molar volume ?

Thanks

 

[Charles Link]

@Charles Link ,@Chestermiller Could you please help me understand the following lines from the textbook .
View attachment 225026
1) How can pressure be same throughout the system (liquid+vapour) ? Do they mean pressure at the interface ?

2) How is density same as molar volume ?

Thanks

1) Yes, they mean pressure at the interface. They normally assume pressure is the same throughout the gas. (As you learned from studies of Archimedes' principle there are pressure gradients, especially in the liquid, and the pressure changes with depth). $\\$ In the case of a 3rd phase=solid, they generally would assume the liquid phase, (and the solid phase), isn't thick enough to create very much additional pressure. (In the phase diagrams, sometimes the pressure, (gas pressure=which is assumed to be the pressure everywhere in the system, even in the liquid and solid), can include pressures of 10 or 20 or even 100 atmospheres and more).$\\$ 2) Molar volume is the inverse of the density.

 

[Jahnavi]

Thanks for replying .

(gas pressure=which is assumed to be the pressure everywhere in the system, even in the liquid and solid),

Pressure is same throughout the gas , but different at various places in the liquid .No ?

What do you mean by "pressure everywhere in the system " ?

2) Molar volume is the inverse of the density.

How ?

 

[Charles Link]

Thanks for replying .
Pressure is same throughout the gas , but different at various places in the liquid .No ?

What do you mean by "pressure everywhere in the system " ?
How ?

1) The increase in the pressure from the depth in the liquid in a sample in a bottle or other container may be .05 atmospheres at most, and the change in gas pressure may be .01 atmospheres or less. On a scale where they are measuring things that may include pressures from 0 to 10 atmospheres, these changes are very small, e.g. the point on the graph at P=1.0 atmospheres is not affected significantly, and any differences caused by the effects of gravity to change the pressure can normally be ignored. $\\$ 2) Molar volume is the volume of 1 mole of material. If density is measured as $\delta$ (grams/cm^3), then $\frac{\delta}{M.W.}=\frac{1}{molar \, volume}$.

 

[Chestermiller]

molar volume = volume per mole
molar density = moles per volume

 

[Jahnavi]

OK .

1) The increase in the pressure from the depth in the liquid in a sample in a bottle or other container may be .05 atmospheres at most, and the change in gas pressure may be .01 atmospheres or less. On a scale where they are measuring things that may include pressures from 0 to 10 atmospheres, these changes are very small, e.g. the point on the graph at 1.0 atmospheres is not affected significantly, and any differences caused by the effects of gravity to change the pressure can normally be ignored. \\

Don't you think this is what the book meant when it says that the pressure is same throughout the system , and NOT that the pressure is same at the interface (even though it is ) to which you have agreed ?

 

[Charles Link]

OK .
Don't you think this is what the book meant when it says that the pressure is same throughout the system , and NOT that the pressure is same at the interface (even though it is ) to which you have agreed ?

In very precise terms, the pressure is the same at the interface, and will vary with height, due to gravity, throughout both the gas and liquid. Sometimes, it can be important to take these pressure differences into account. In calculations where the barometric pressure is computed as a function of altitude, it is of prime importance. $\\$ In most ordinary laboratory calculations, gravitational effects on the system are usually ignored, but a good researcher will point them out, and include them, if they have a significant effect on the system being considered.

 

[Jahnavi]

OK . Thanks !

@Charles Link , @Chestermiller Kindly help me understand another point given in the book especially the second and third lines .

 

[Charles Link]

OK . Thanks !

@Charles Link , @Chestermiller Kindly help me understand another point given in the book especially the second and third lines .

View attachment 225038

It would appear they are referring to a result that follows from Statistical and Thermal Physics, that the density of a gas as a function of energy will be of the form $\rho(E)=\rho_o \, e^{-E/(k_B T)}$ where $E=+mgh +(\frac{1}{2})mv^2$. This calculation is somewhat advanced. In addition, the average of $\frac{1}{2} mv^2=\frac{3}{2}k_B T$, so that you can compute the effect that the $mgh$ term will have on the density. $\\$ If $mgh<<k_B T$ , the height $h$ will have very little effect on the density. $\\$ Edit: The result of the above is basically that the density as a function of height $h$ will be $\rho(h)=\rho(0) \, e^{-mgh/(k_B T)}$. If $mgh <<k_BT$ for all $h$, then the density is nearly uniform. And if $mgh <<k_B T$, you can also say that $mgh << \frac{1}{2}mv^2$.$\\$ The book's author apparently is familiar with this type of calculation, but it is really a somewhat advanced upper level undergraduate topic. $\\$ ................................$\\$ And it may be worthwhile to relate this to Archimedes' result: force per unit volume from the pressure gradient must offset the gravitational force per unit volume, so that $-\nabla P=\rho g \, \hat{z}$. $\\$ Now $PV=nRT$ and $\rho=\frac{n \, M.W. }{V}$, so that $P=\frac{\rho RT }{ M.W.}$. $\\$ The result is $(\frac{RT}{M.W.}) \frac{d \rho}{dz}=-\rho g$. $\\$ Solving this differential equation, we get $\rho=\rho_o \, e^{-(M.W.) (g z)/(RT)}$ . Here the gas constant $R=N_o \, k_B$, where $N_o$ is Avagadro's number (particles/mole). Thereby, $\frac{ M.W. (in \, grams/mole)}{N_o}=m$, where $m$ is the mass of the gas molecule. $\\$ This gives $\rho=\rho_o \, e^{-mg z/(k_B T)}$. $\\$ These results from Archimedes and the ideal gas law are consistent with our formula above $\rho(h)=\rho(0) \, e^{-mgh/k_B T}$, which can also be written in the form $P(h)=P_o \, e^{-mgh/(k_BT)}$. Note that although the pressure has a somewhat complicated form as a function of height $h$, (more complicated than for a liquid), Archimedes principle still works with a gas as well: The buoyant force is equal to the weight of the gas that is displaced. $\\$ It may also be of interest, that for an uncompressible liquid, the pressure varies with depth $d$ as $P(d)=P(0) +\rho_o \, g \, d$. Gases are compressible, so that the equation describing pressure as a function of height is much different. $\\$ .............................. $\\$ And let's try some gas molecules at room temperature: $\\$ $k_B T=1.381 \cdot 10^{-23} (300) \approx 4.1 \cdot 10^{-21} \, joules$, $\\$ and $mgh$ for a height of $h= 1.0 \, meters$ , ($M.W.=30$), is $\\$ $mgh \approx 30 (1.67 \cdot 10^{-27}) (9.8)(1.0)=4.9 \cdot 10^{-25} \, joules$. $\\$ It looks like $k_BT>> mgh$. $\\$ The equation also gives the result that the atmospheric pressure should be $P(h) \approx 0.5 \, atm$ at $h=5800 \, meters$ altitude. $\\$ Edit: Note that the atmosphere is 78% $N_2$ with M.W.=28, and 21% $O_2$ with M.W.=32, so that using a number of M.W. $\approx$ 30 for the atmosphere should give us reasonably accurate results for the pressure as a function of altitude. (Temperature is also known to vary somewhat with altitude=we're just looking for approximate results for the pressure vs. altitude calculation).

 

[Jahnavi]

Thank you very much .

Real gases behave as ideal gases at low pressure AND high temperature .

I think I understand how high temperature implies that the molecules are moving fast enough such that their molecular interactions (potential energy ) could be neglected .

But I do not understand how low pressure is also a necessity .

I understand that at low densities , the volume of the gas molecules will be negligible as compared to the volume occupied by the gas (one of the assumptions for ideal gas ) .

Does a real gas at low pressure necessarily has low density ?

We get low pressure => low density if we use PV = nRT (keeping T constant ), but this itself is an ideal gas law , so we possible can't use it .

So how is it that a low pressure is required for a real gas to behave as an ideal gas AND does a real gas at low pressure necessarily imply low density ?

 

[Charles Link]

The ideal gas law is quite accurate for pressures at or below 1 atm at temperatures near $T=300 \, K$, so yes, we can use it, and we can be fairly certain the results are quite accurate. The equation is $PV=nRT$, where $n$ is the number of moles, and $R$ is the universal gas constant. Writing it as $P=\frac{n}{V} RT=\frac{\rho}{M.W.}RT$, where $\rho$ is the density, gives the result that the pressure is proportional to the density for a given temperature. $\\$ At high pressures and/or low temperatures is where the ideal gas law can become inaccurate and unreliable. Also, if the temperature of the gas is in a region of pressure and temperature where the gas can coexist in liquid form, (such as water vapor at room temperature), then the results could be somewhat inaccurate. For the application at hand, the ideal gas law is quite reliable.

 

[Jahnavi]

Thanks a lot