Idea gas law applied to throttling process

[FrankJ777]

Hi all.
I'm trying to become acquainted with thermal dynamics. I thought I had a fairly good handle on the idea gas law, PV=nRT, or PV=mRT, but I came across a problem involving the throttling of an idea gas, and the solution was totally not what I expected. Referring to picture where gas is forced through the valve I thought I tried using:

$\frac{P_{1}V_{1}}{T_{1}}$ = $\frac{P_{2}V_{2}}{T_{2}}$

Setting V$_{1}$ = V$_{2}$, then solving for T$_{2}$:

T$_{2}$ = $\frac{P_{2}T_{1}}{P_{1}}$

Of course I got the wrong answer. The solution explained that the throttling process produces no change in enthalpy for an idea gas, and therefore no temperature change. I'm trying to follow they're explanation, but what I'm more interested in understanding is why the application of the idea gas law is not valid for this situation. I would expect it to be valid in any situation with an idea gas, so obviously I'm misusing it. Could someone explain to me why it's not valid in this situation? Thanks a lot.

 

[Andrew Mason]

Hi all.
I'm trying to become acquainted with thermal dynamics. I thought I had a fairly good handle on the idea gas law, PV=nRT, or PV=mRT, but I came across a problem involving the throttling of an idea gas, and the solution was totally not what I expected. Referring to picture where gas is forced through the valve I thought I tried using:

$\frac{P_{1}V_{1}}{T_{1}}$ = $\frac{P_{2}V_{2}}{T_{2}}$

Setting V$_{1}$ = V$_{2}$, then solving for T$_{2}$:

T$_{2}$ = $\frac{P_{2}T_{1}}{P_{1}}$

Of course I got the wrong answer. The solution explained that the throttling process produces no change in enthalpy for an idea gas, and therefore no temperature change. I'm trying to follow they're explanation, but what I'm more interested in understanding is why the application of the idea gas law is not valid for this situation. I would expect it to be valid in any situation with an idea gas, so obviously I'm misusing it. Could someone explain to me why it's not valid in this situation? Thanks a lot.

The ideal gas law IS applicable to this situation, once everything settles down and reaches equilibrium. I am not sure why you say that volume does not change. The gas rapidly expands since there is no external pressure.

This is essentially an adiabatic free expansion. The gas expands but does no work (W=0) since there is no external pressure. Since it is adiabatic, Q = 0. So by applying the first law: Q = ΔU + W, one has to conclude that ΔU = nCvΔT = 0, which means ΔT=0.

AM

 

[FrankJ777]

OK. I was assuming that the volumes were the same based on the diagram. (In most throttling problems I see the diagrams are quite similar.) It seemed that judging by the diagrams, the volume, or at least the diameter of the pipe is assumed to be the same on both sides of the aperture. So I'm confused how the volume is different. I guess I'm confused in general what is meant by volume of the gas in that case. I know that we also use specific volume as a property of a gas. In this case I can see that the specific volume of the gas increases as it goes to the other side of the aperture, but I can't see how you determine that the Volume itself is greater for the gas on side 2 of the aperture.

 

[Andrew Mason]

OK. I was assuming that the volumes were the same based on the diagram. (In most throttling problems I see the diagrams are quite similar.) It seemed that judging by the diagrams, the volume, or at least the diameter of the pipe is assumed to be the same on both sides of the aperture. So I'm confused how the volume is different. I guess I'm confused in general what is meant by volume of the gas in that case. I know that we also use specific volume as a property of a gas. In this case I can see that the specific volume of the gas increases as it goes to the other side of the aperture, but I can't see how you determine that the Volume itself is greater for the gas on side 2 of the aperture.

The volume is determined the amount of space occupied by the gas when it returns to equilibrium. Throttling occurs when a gas expands rapidly through a small aperture into a vacuum. As it passes through the throttle valve the gas is in a dynamic state. When it regains equilibrium its volume has increased but because it expanded against 0 external pressure, no work was done on the surroundings. So there has been no change in internal energy. That is just a consequence of the first law for an adiabatic free expansion (Q=W=0).

Since T is constant for an ideal gas undergoing a free expansion (real gases will cool because the increase in volume requires work in increasing the separation of the molecules) the pressure is inversely proportional to volume.

AM

 

[FrankJ777]

OK. So does PV=mRT only apply at equilibrium? And for that matter is gas in the static state imply that it's in equilibrium, and visa verse?

 

[Andrew Mason]

OK. So does PV=mRT only apply at equilibrium? And for that matter is gas in the static state imply that it's in equilibrium, and visa verse?

PV=nRT is the equation of state for an ideal gas in equilibrium. Pressure, volume and temperature are not necessarily defined under dynamic conditions.

AM

 

[FrankJ777]

Thanks. That helps a lot. I'm taking the FE in a few weeks, and the only exposure to thermodynamics I've had was a couple of chapters in PHYS I and CHEM.

 

[Chestermiller]

What we're dealing with here is a continuous flow system operating at steady state. A constant rate of mass flow is passing through the system from the high pressure inlet pipe to the low pressure outlet pipe. To analyze this system, we need to use a modified form of the first law that applies to such as situation. I'm sure you must have been studying the first law in your thermo course, and have advanced to the point where continuous flow systems at steady state are being addressed. I'm not going to go through the derivation because it is covered in your textbook, but the final result is typically of the form:
$$h_{out}-h_{in}=\dot{Q}-\dot{W_S}$$
where $h_{out}$ is the enthalpy per unit mass of material exiting the system, $h_{in}$ is the enthalpy per unit mass of material entering the system, $\dot{Q}$ is the heat input per unit mass of material passing through the system, $\dot{W_S}$ is the so-called shaft work imposed per unit mass of material passing through the system.

In an adibatic throttling situation, $\dot{Q}$ and $\dot{W_S}$ are both equal to zero. Therefore, you have, $$h_{out}=h_{in}$$

Now, for your case, the enthalpy of your ideal gas is a function only of temperature. Therefore, you must have that:
$$T_{out}=T_{in}$$
The conclusion here is that, in the adiabatic throttling of an ideal gas, the change in temperature is equal to zero. This is the so-called Joule-Thompson effect. If you are operating at higher pressures, beyond the ideal gas region, the enthalpy of the gas is a weak function of the pressure, and there is a small change in temperature during throttling. This change in temperature can be calculated if you know the PVT behavior of your gas beyond the ideal gas region.

The ideal gas law still applies to the throttling situation you have described. Since the temperature does not change, you will have that:
$$P_2v_2=P_1v_1$$
where the lower case v's are the specific volumes of the gas at the two pressures and at temperature T.

I hope this helps.