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In another question, I asked about the theoretical justification for the experiment that I will describe in what follows.

The experiment was done by Rossini and Frandsen in 1932 at the National Bureau of Standards.

Here is the apparatus used

enter image description here

What we have here is $n$ moles of gas in a container $B$ at pressure $P$.

When the valve is open, gas escapes to the atmosphere through the long coil surrounding the container $B$.

There is a water bath around the container whose temperature can be maintained constant at exactly the same value as that of the surrounding atmosphere.

The experiment is as follows

  1. The valve is opened slightly and gas flows slowly through the long coil and out into the air.

  2. At the same time, the temperature of the gas, the container, the coil, and the water is maintained constant by an electric heating coil immersed in the water.

  3. The electrical energy supplied to the water is, therefore, the heat $Q$ absorbed by the gas during the expansion.

Thus, we have a gas expanding in volume at constant temperature.

The work done by the gas is

$$W=-P_0(nv_0-V)\tag{1}$$

where $P_0$ is atmospheric pressure, $v_0$ is the molar volume at atmospheric temperature and pressure, $V$ is the volume of the container $B$, and $nv_0$ is larger than $V$.

If $u(P,T)$ is the molar internal-energy at pressure $P$ and temperature $T$ and if $u(P_0,T)$ is the molar internal energy at atmospheric pressure and the same tempeature, then, from the first law, the change of molar internal-energy can be expressed in terms of the measured quantities $Q$ and $W$ as

$$u(P,T)-u(P_0,T) = \frac{Q+W}{n}\tag{2}$$

provided that corrections have been made to take account of the energy changes due to the contraction of the walls of the container.

My first question is about (2). Shouldn't it be $u(P_0,T)-u(P,T)$?

In this way, the change of molar internal-energy $\Delta u$ was measured for various values of the initial pressure $P$ at constant temperature $T$. The values of $\Delta u$ were plotted against the corresponding pressure $P$, as shown in Fig. 5-3.

enter image description here

My main question is about the interpretation of these results contained in the following snippet

Since $u(P_0,T)$ is constant, the slope of the resulting curve is equal to $(\partial u,\partial P)_T$ at any value of $P$. Within the pressure range of 1 to 40 standard atmospheres, the experimental points fall on a straight line, meaning that $(\partial u/\partial > P)_T$ has the same value at every pressure; that is $(\partial u/\partial P)_T$ is independent of the pressure, depending only on the temperature. Thus,

$$\left ( \frac{\partial u}{\partial P}\right )_T = f(T)\tag{3}$$

Rossini and Frandsen's experiments with air, oxygen, and mixtures o oxygen and carbon dioxide led to the conclusion that the internal energy of a real gas is a function of both temperature and pressure. They found no pressure or temperature range in which the quantity $(\partial u/\partial P)_T$ was equal to zero. In other words, their real gases did not reach the low-pressure limit of the ideal gas.

Questions

1)

Since $u(P_0,T)$ is constant, the slope of the resulting curve is equal to $(\partial u,\partial P)_T$ at any value of $P$.

I think this claim comes from

$$dU=\left (\frac{\partial u}{\partial T}\right )_P dT+\left (\frac{\partial u}{\partial P}\right )_T dP$$

But since $dT=0$ we have

$$dU=\left (\frac{\partial u}{\partial P}\right )_T dP$$

However, if we are always measuring $u(P,T)-u(P_0,T)$, (and actually as I said before the way I understand it the final pressure is $P_0$ and so the difference should be the other way around) then aren't we measuring some kind of average slope instead of the actual slope at at a point?

2.

My second question has to do I believe with this other question which is about why we would want to do this experiment in the first place and the theoretical basis for it.

Rossini and Frandsen's experiments with air, oxygen, and mixtures o oxygen and carbon dioxide led to the conclusion that the internal energy of a real gas is a function of both temperature and pressure.

This seems to imply that since they found $(\partial u/\partial P)_T\neq 0$ at all pressures, then it is not possible to have an adiabatic free expansion in which the temperature stays constant.

Is this the correct interpretation?

3.

In other words, their real gases did not reach the low-pressure limit of the ideal gas.

What does this means exactly?

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1 Answer 1

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One form of the general virial equation of state for a real gas is $$\frac{Pv}{RT}=1+B^*(T)P+C^*(T)P^2+...$$where v is the molar volume. At low pressures, this approaches $$\frac{Pv}{RT}=1+B^*(T)P$$ For a real gas, $$\left(\frac{\partial u}{\partial P}\right)_T=\left[-P+T\left(\frac{\partial P}{\partial T}\right)_v \right]\left(\frac{\partial v}{\partial P}\right)_T$$At low pressure, this approaches $$\left(\frac{\partial u}{\partial P}\right)_T=-RT^2\frac{dB^*}{dT}$$which is a function only of temperature.

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  • $\begingroup$ Where does the third equation come from? $\endgroup$
    – xoux
    Nov 12, 2023 at 19:27
  • $\begingroup$ It come from $du=Tds-Pdv$, $ds=\frac{C_v}{T}dT+\left(\frac{\partial s}{\partial v}\right)_Tdv$, and $\left(\frac{\partial s}{\partial v}\right)_T=\left(\frac{\partial P}{\partial T}\right)_v$ $\endgroup$ Nov 12, 2023 at 21:35
  • $\begingroup$ I really don't follow. We have $dU=(\partial U/\partial T)_V dT+(\partial U/\partial V)_T dV=T\frac{C_V}{T} dT +(\partial U/\partial V)_T dV=T\frac{C_V}{T}dT+T(\partial P/\partial T)_V dV +(\partial U/\partial V)_T dV - T(\partial P/\partial T)_V dV=Tds+dV((\partial U/\partial V)_T - T(\partial P/\partial T)_V)$. According to what you wrote we should have then $(\partial U/\partial V)_T - T(\partial P/\partial T)_V = -P$. Not sure where that comes from. $\endgroup$
    – xoux
    Nov 12, 2023 at 22:37
  • $\begingroup$ The derivation is in every thermo book. Have you not learned about entropy yet? $\endgroup$ Nov 12, 2023 at 22:39
  • $\begingroup$ No, entropy is a few chapters ahead still. $\endgroup$
    – xoux
    Nov 12, 2023 at 22:41

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