In joule thomson process no heat is exchanged and no work is done. why does the gas temperature change? what is the reason(s) of the temp. reduction of the most gases?
You are mixing the two types (or should I say two descriptions) of Joule-Thomson. In ordinary life, the expanding gas does work and the internal energy (temperature) is reduced. You may try this by going to your favorite auto garage and using their air line to spin screwdrivers in the air while noticing the air gun body gets cold.
For an ideal gas (we don't have any, but some are close), no work is done and the temperature stays constant. Helium comes very close to doing this.
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"Well, I've wrestled with reality for 35 years, Doctor, and I'm happy to state I finally won out over it." Elwood P. Dowd
Have you ever wondered why there's a picture of you in all the area garages saying "Call the police if this guy asks to use the air hose."?
But, as long as you're causing trouble anyway, if you ever get the use of anybody's liquid nitrogen tank, try it with that. You can get some tremendous expansion on hot days!
__________________
"Well, I've wrestled with reality for 35 years, Doctor, and I'm happy to state I finally won out over it." Elwood P. Dowd
thanks, but my purpose is not ordinary life as you mentioned. as you know, based on the thermodynamic, in the joule thomson process, like a JT Valve in the gas industry, pressure drop of gas flow causes reduction in its temp.(except H2, NE, HE). in this process, there is no heat/ work exchange. the enthalpy is constant.
I'm willing to be clarified the concept(mechanism or reason) of the temp. change.
OK, I did tell you. For real gasses, there IS work. Do you have a standard thermodynamics text available, or is there a site you like? I'm willing to walk you through the explanation, but I'm not going to debate established facts.
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"Well, I've wrestled with reality for 35 years, Doctor, and I'm happy to state I finally won out over it." Elwood P. Dowd
Its not that complicated. You have the basic law: PxV/T is constant. when P changes both V&T change.
Your law is true for ideal gases, (except that V&T don't both have to change) but as has already been mentioned the JT process does not work on ideal gases. Enthalpy is constant in the JT process, not the quantity PV/T. Joule-Thomson predicts that dP/dT is proportional to V times a temperature dependent factor. You can't get this result from the combined gas law.
Can you post an example where the law dosn't work? Its been working for a very long here on Earth so I must be missing something. Very puzzling so a simple example would be very helpful. thanks
Can you post an example where the law dosn't work? Its been working for
a very long here on Earth so I must be missing something. Very puzzling
so a simple example would be very helpful. thanks
It doesn't work in the case of real gases- the only place on earth where it works exactly is in freshman chemistry texts. A simple example is the Joule-Thomson process. You're welcome.
The JT process is an engineering trick that is based on the ideal gas law. The JT trick has built in power losses as has any machine. The ideal gas law is as it says an ideal without the built in losses. If you want to see how efficient your JT trick is use the ideal gas law to establish how an ideal machine would be 100% efficient and compare that with what you get from the device using the JT process which will always be less than 100% (most are much less than the ideal). Its not a very good process if efficiency is a priority(I personally never liked the JT process)
I agree with your words. But, My question seeks the conceptof the matter. I don't mean it justified by the formula, PV=nRT, as done by greendog. Let me please to say my question in details as follows,
Consider a JT( throttling) process, adiabatic free expansion, in which a fluid passes a restriction ( like a valve),
If the fluid is a real gas, the temperature decreases(or increase) due to drop in pressure. why or how does this temperature reduction(or rise) take place?
If the fluid is an ideal gas, the temperature doesn't change. why or how does the temperature remain constant?
I can tell you why gas does what it does. Send me £500 in small bills (the ones that have Kris's picture on them instead of that Liz person) and I'll explain about stuffed cabbage and pickled beans.
__________________
"Well, I've wrestled with reality for 35 years, Doctor, and I'm happy to state I finally won out over it." Elwood P. Dowd
There is no similarity between
these two subject(wet and temp. change). Please think about the query carefully.
For more clarification, Lets
review the concept of the temperature by thermodynamic. By definition,
temperature is the degree of hotness or coldness of a material. Or, temperature
is a measurement of the average kinetic energy of the molecules in an object or
system. The greater the heat absorbed by a material, the more rapidly the atoms
within the material begin to move( or vibration), and thus the greater the rise
in temperature. So, changing in temperature of a material denotes the energy
change in it. Now, we should come back to our discussion, seek an answer for
this question, how does the energy of
a real gas, undergoes a throttling process( adiabatic free expansion),
change? What about a ideal gas?
By definition the energy is unchanged in both gases(real or ideal). The use of English language words might complicate matters. So I'm want to apoligize for any confusion I caused by asking why water is wet. Some things just are what they are and there is no way to know why that is.