Thermodynamics of a Constant Volume Gas Thermometer

[RJLiberator]

Homework Statement

A constant volume gas thermometer contains a gas whose equation of state is

$$(p+\frac{a}{V^2_m})(V_m-b)=RT$$
and another, of identical construction, contains a different gas which obeys the ideal gas law, $$pV_m = RT$$. The thermometers are calibrated at the ice and steam points. Show that they will give identical values for a temperature.
[Assume that the thermometers are constructed so that all the gas is at the temperature being measured.]

Homework Equations

V is constant, we must consider the variation of T with p.

The Attempt at a Solution

Since we have to consider the variation of T with p, I want to take the derivative $$\frac{dT}{dp}$$.

We start with the ideal gas law, and end up with:

$$\int \frac{pV_m}{R}dp=\int dT$$
This ends up being
$$T=\frac{V_m}{R}\frac{p^2}{2}$$

With the same strategy on the more complicated equation of state we see
$$\int( \frac{pV_m}{R}-\frac{pb}{r}+\frac{a}{RV_m}-\frac{ab}{RV^2_m}) dp=\int dT$$
This turns out to be
$$\frac{p^2V_m}{2R}-\frac{p^2b}{2R}+\frac{ap}{RV_m}-\frac{abp}{RV^2_m}=T$$

Here, we see there is a similar term $$\frac{p^2V_m}{2R}$$ on both sides.

I'm not sure if what I did was right...
Could I set the constants equal to some value to make the equations equal?

 

[TSny]

We start with the ideal gas law, and end up with:

$$\int \frac{pV_m}{R}dp=\int dT$$

The integrand on the left is incorrect.

Start with $pV_m = RT$ and take the differential of both sides.

 

[RJLiberator]

I'm sorry, I'm really struggling with taking the differential here.

The hint that I'm given is "Note that V is constant, then consider the variation of T with p." This leads me to believe I need to find dT/dp.

So, if I differentiate either side, I'm not sure what to differentiate with respects to.

We have $$pV_m=RT$$ If I differentiate so the left side becomes
$$V_m\frac{dp}{dT}$$ the right side becomes $$R\frac{dT}{dT}$$

which doesn't make sense?

 

[TSny]

The hint that I'm given is "Note that V is constant, then consider the variation of T with p." This leads me to believe I need to find dT/dp.

OK. That is a good approach.

So, if I differentiate either side, I'm not sure what to differentiate with respects to.

Note that dT/dp is a derivative with respect to p. So, If you want to find dT/dp then you would differentiate both sides with respect to p.

We have $$pV_m=RT$$ If I differentiate so the left side becomes
$$V_m\frac{dp}{dT}$$ the right side becomes $$R\frac{dT}{dT}$$

OK. Here you decided to take the derivative of both sides with respect to T, instead of p. That will also work.

which doesn't make sense?

I'm not sure why you feel that it doesn't make sense. Can you simplify $\frac{dT}{dT}$?

 

[RJLiberator]

dT/dT just equals 1, ah.

So we have $$R=V_m\frac{dp}{dT}$$

So far, correct?

 

[TSny]

dT/dT just equals 1, ah.

So we have $$R=V_m\frac{dp}{dT}$$

So far, correct?

Yes. Good.

 

[RJLiberator]

I decided to look at this problem by differentiating by dp.

So I know I have
$$V_m = R\frac{dT}{dp}$$

The other piece of information I'm given is:
$$(p+\frac{a}{V^2_m})(V_m-b)=RT$$

I've tried manipulating the second equation, I've tried substituting the first equation into the second equation, but nothing so far.

If I multiple the second equation by V^2 on both sides, simplify, and set it equal to 0, I get:

$$pV_m^3+V_m^2(-bp-RT)+aV_m-ab=0$$

The goal is to show that these two equations will give identical values for Temperature.

I guess I can try taking the derivative again

$$\frac{dp}{dp}V_m^3+V_m^2\frac{d(-bp-RT)}{dp}+\frac{d(aV_m-ab)}{dp}=0$$

=> $$V_m^3-bV_m^2=0$$
$$V_m-b=0$$
$$V_m=b$$

But this doesn't seem to help. Any ideas or any leads on if I'm doing anything right?

 

[TSny]

I decided to look at this problem by differentiating by dp.

So I know I have
$$V_m = R\frac{dT}{dp}$$

Good.

The other piece of information I'm given is:
$$(p+\frac{a}{V^2_m})(V_m-b)=RT$$

I've tried manipulating the second equation, I've tried substituting the first equation into the second equation, but nothing so far.

There is no reason for $V_m$ of the constant-volume ideal gas thermometer to be equal to $V_m$ for the constant-volume non-ideal gas thermometer. So, your substitutions are not valid.

Also, it's important to note that $T$ in the two equations of state represents the absolute thermodynamic temperature. It does not necessarily represent the temperature as measured by either of the constant volume gas thermometers. Imagine you are given a container of the ideal gas which has a pressure gauge. Using this as a constant-volume gas thermometer means that you are going to define a temperature scale, $t$, based on the pressure of the gas. Likewise for a container of the non-ideal gas. So, you could let $t_{\rm id}$ be the temperature as measured on the calibrated ideal gas thermometer and $t_{\rm n}$ be the temperature as measured on the calibrated non-ideal gas thermometer. You want to use the equations of state for the gases to prove that the two temperature scales, $t_{\rm id}$ and $t_{\rm n}$, will agree at all temperatures.

The absolute thermodynamic temperature $T$ is defined independently of the construction of a thermometer. So, $T$ is the same variable in the two equations of state.

From $V_m = R\frac{dT}{dp}$ for the ideal gas, see if you can show $\Delta P = \frac{R}{V_m} \Delta T$. (You can actually show this directly from the ideal gas equation of state without using derivatives.) Thus, for the ideal gas, the change in pressure is proportional to the change in thermodynamic temperature. $$\Delta P_{\rm id} = C_{\rm id} \Delta T$$ for some constant $C_{\rm id}$ and the subscript refers to the ideal gas. (As you can see, the value of the constant $C_{\rm id}$ is $\frac{R}{V_m}$. But, the value of $C_{\rm id}$ is not important. What's important is $\Delta P_{\rm id}$ is proportional to $\Delta T$ .)

See if you can show something similar for the non-ideal gas. Then see if you can use these results to get information about changes $\Delta t_{\rm id}$ and $\Delta t_{\rm n}$ of the temperature readings of the constant-volume gas thermometers constructed from the two containers of gas.

 

[RJLiberator]

Thank you, I am pretty sure with your words I completed this correctly.

We saw that
$$V_m = R\frac{dT}{dp}$$
Now, I can divide both sides by V_m and then multiply both sides by dp. From here, I can integrate to get

$$\Delta P_{\rm id} = C_{\rm id} \Delta T$$

Considering equation 2, we have in expanded form
$$p(V_m-b)+\frac{a}{V_m}-\frac{ab}{V_m^2}$$

Letting $$(V_m-b)= C_1 and \frac{a}{V_m}-\frac{ab}{V_m^2} = C_2$$
We have in total

$$RT = pC_1+C_2$$
Now, we can differentiate similar to the ideal gas and obtain
$$R\frac{dT}{dp}=\frac{dp}{dp}C_1+\frac{dC_2}{dp}$$

But this is simply the same strategy we saw in the ideal gas and gives us
$$\Delta T_{non}=\frac{C_1}{R} \Delta P_{non}$$

So, I'm left with the difference between the two equations being a difference of a constant, namely -R/b.

 

[Chestermiller]

All you need to show is that, at constant volume, p is a linear function of T. See if you can work the first equation into slope-intercept form.

 

[RJLiberator]

All you need to show is that, at constant volume, p is a linear function of T. See if you can work the first equation into slope-intercept form.

I believe I did this in my previous post:

$$RT = pC_1+C_2$$

Here, I can divide by R and the constants are still constants and we get a function in the form of y=mx+b.

 

[TSny]

OK. Good. So you have

$$\Delta T_{non}=\frac{C_1}{R} \Delta P_{non}$$

So, this tells you $\Delta P_{non}=C_{non} \Delta T$ for some constant $C_{non}$

And you also know $\Delta P_{id}=C_{id} \Delta T$.

Can you use this to help show that $t_{id} = t_{non}$?

[ @Chestermiller is coming at from a different angle. Go with that way if you prefer.]

 

[Chestermiller]

I believe I did this in my previous post:

$$RT = pC_1+C_2$$

Here, I can divide by R and the constants are still constants and we get a function in the form of y=mx+b.

Good. So, since it is a straight line relationship, you can also express it in point-slope form using the values at the melting point and boiling point. Let's see you do that.

 

[RJLiberator]

Solving for T and setting the equations equal to each other we can see that

$$\Delta P_{non} = C \Delta P_{id}$$

This shows us that the pressure from the ideal gas and non-ideal gas are compared with a constant of proportionality.

EDIT: One sec.

 

[RJLiberator]

Couldn't we say that the change in temperature = 100 from ice point to steam point for both of these equations, thus, the P_id and P_non are off by a certain constant of proportionality and thus the change in temperatures will always be equal to each other?

 

[TSny]

Couldn't we say that the change in temperature = 100 from ice point to steam point for both of these equations, thus, the P_id and P_non are off by a certain constant of proportionality and thus the change in temperatures will always be equal to each other?

The problem doesn't state that you are constructing a Celsius temperature scale. So, the change in temperature between the ice and steam points might not be 100. But it is some specific number.

I think your reasoning concerning the changes in temperature might be OK, but I'm not sure I quite follow it in detail. You are correct in post #14 that $\Delta P_{non} = C \Delta P_{id}$. It's getting from this to the statement that $t_{id} = t_{non}$ that I think needs clarification.

From the way in which a constant-volume gas temperature scale is constructed, what can you say about $\Delta P_{non}$ and $\Delta t_{non}$? Similarly for the ideal gas.

 

[RJLiberator]

In a constant-volume gas temperature scale, when there is an increase in temperature there is a corresponding increase in pressure of an ideal gas. Similarly with a decrease.

Now, in an ideal gas, we've shown that there is a difference between the change in P and the change in T and that difference is R/V_m product with the change in T.

Similarly, we've shown in the non-ideal gas, that there is a R/C_1 constant where C_1 = (V_m-b).

Now, we can think of this as a line on an xy graph. y=mx where y=the change in temperature, x = change in pressure and m = the different constants.
The constants won't matter as it won't have an effect on the change, so it becomes
the change in P = the change in T for both operations.

 

[TSny]

The temperature scale $t$, of a constant-volume gas thermometer is defined to be a linear function of $P$. So, $\Delta P_{id} = k_{id} \Delta t_{id}$ for some constant $k_{id}$. Similarly for the non-ideal gas. Combine this with $\Delta P_{non} = C \Delta P_{id}$ to deduce how $\Delta t_{id}$ and $\Delta t_{non}$ must be related.

 

[RJLiberator]

$$\Delta P_{non}=C_{non} \Delta T_{non}$$
$$\Delta P_{id} = k_{id} \Delta T_{id}$$
$$\Delta P_{non} = C \Delta P_{id}$$

By putting the first two equations into the third we get

$$C_{non} \Delta T_{non} = Ck_{id} \Delta T_{id}$$

The constants can be grouped to show

$$\Delta T_{non}= C \Delta T_{id}$$

This seems to show that T_non =/= T_ideal ?

 

[TSny]

$$\Delta T_{non}= C \Delta T_{id}$$

Good

This seems to show that T_non =/= T_ideal ?

No, it just shows that $t_{non}$ is linearly related to $t_{id}$. Finally, you can bring in calibration at the two fixed points to determine the values of the constants in the linear relation.

 

[TSny]

I imagine @Chestermiller has lost his patience with the tedious way that I dragged you through this problem.

A concise way to summarize the solution is

(a) Eliminating $T$ between the two equations of state shows that the pressures of the two gases are linearly related to one another.

(b) Since $t$ is defined to be a linear function of $P$ for each gas, statement (a) implies that $t_{non}$ is a linear function of $t_{id}$.

(c)Then, calibration at the fixed points requires $t_{non} = t_{id}$ at all temperatures.

 

[RJLiberator]

Thank you for walking me through the problem. I understand the problem much more because of your help.

 

[Chestermiller]

I imagine @Chestermiller has lost his patience with the tedious way that I dragged you through this problem.

Not really. We were just tired from watching football games all day and went to bed early.

@RJLiberator : What is the equation for the straight line (T vs p) that passes through the two points $(p_{273}, 273)$ and $(p_{373}, 373)$?

 

[TSny]

Not really. We were just tired from watching football games all day and went to bed early.

OK. (I see Michigan won!)

 

[Chestermiller]

OK. (I see Michigan won!)

Yeah! GO BLUE!

 

[Chestermiller]

$$T=273+100\frac{(p-p_{273})}{(p_{373}-p_{273})}$$