# What gas's temperature increases as it decompresses?

1. Feb 19, 2009

### noblethewhite

I'm not sure if this is where this question should go, but it can't really follow the template in the homework section. I've looked on search engines, phrasing this question differently but can't seem to find the answer. It would be the exception to Charles Law, I believe.

2. Feb 19, 2009

### Mapes

What makes you think that such a single gas exists? Or is a reaction involved?

3. Feb 19, 2009

### Bystander

See: "free expansion, and/or Joule-Thomson (also, Thompson), or Joule-Kelvin, or Kelvin-Joule coefficient/effect."

4. Feb 19, 2009

### Mapes

Ah, I was considering

$$\left(\frac{\partial T}{\partial V}\right)_{S,N}$$

which is never positive for a gas (a positive value corresponds to a gas heating during decompression). However, in the Joule-Thomson expansion it looks like we're considering

$$\left(\frac{\partial T}{\partial P}\right)_{H,N}$$

which can be negative at high temperatures (and a negative value corresponds to a gas heating during decompression). That's interesting, Bystander, I hadn't heard of the Joule-Thomson inversion temperature before, thanks.

5. Feb 20, 2009

### Mapes

I looked into this a little more because it's not intuitive (to me, at least) how the temperature of a gas could increase during "decompression." We could use that word to describe a Joule-Thomson experiment (constant and unequal pressures on either side of a porous plug), but it seems more accurate to say that we are pushing the gas irreversibly through the plug, and we are definitely doing work on the system. This, of course, has a different connotation than a free decompression, in which no positive work enters the system (and the gas may end up doing work on the environment, decreasing the temperature of the gas).

In the Joule-Thomson arrangement, we do work on the gas before it enters the plug, and recover that energy as work when the gas expands on the other side. In the case of an ideal gas, the amounts are equal and the temperature doesn't change. In the case of a real gas, atomic/molecular repulsion increases at high pressures and temperatures, and we must do more work on the upstream side to obtain a given pressure. This excess work ends up heating the gas.

6. Feb 20, 2009

### noblethewhite

I'm in a process technology course and the instructor asked this question as one of his bonus questions. There's suppose to be an actual gas that does this. I'm not really that far in my knowledge of physics yet but I'm guessing that would be the principle behind this gas's reaction to pressure.

Last edited: Feb 20, 2009
7. Feb 21, 2009

### Bystander

Lots of gases --- long's they're above their J-T inversion temperatures. Couple with J-T inversion Ts well below room T are technically "difficult" to liquify. Hint enough for you?

8. Feb 26, 2009

### noblethewhite

Well found out the answer. That is the principle behind it, but the answer he was looking for is hydrogen. Apparently in plants its a big worry if any pipe or container containing hydrogen leaks, b/c if it does the rapid drop in pressure will cause the hydrogen to heat up and ignite. So you can imgaine the problems that would cause.