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Someone explain PV = nRT

Of course it makes perfect sense.. but when you compress a gas, its temperature increases. But according to that equation, if you halve volume, pressure should double and nRT should all remain constant. If temperature were to increase, pressure would have to increase less than double... This doesn't make sense. Am I missing osmething?
 
i don't know if i'm falling for some joke, but...

That last statement is wrong. If you halved volume and increased temperature, pressure would be more than double.

Because, when you increase temperature it wants to expand.
 
Originally posted by: bigalt
i don't know if i'm falling for some joke, but...

That last statement is wrong. If you halved volume and increased temperature, pressure would be more than double.

Because, when you increase temperature it wants to expand.

No. You aren't increasing temperature directly. But when you compress a gas it increases in temperature doesn't it? In an engine, compressing gasses in the piston increase in both temperature and pressure.
 
PV = nRT

Pretend you have a steel SCUBA tank filled with compressed air. For all intents and purposes, the volume is fixed. If you heat it up, the pressue will increase. If you cool it down the pressure will decrease.

Now transfer the gas to another SCUBA tank that is half the size. The presure inside this tank will be double....


blah blah blah...

==

I don't know dude. Its a fvcking math problem.. use your noodle...
 
Originally posted by: Mwilding
PV = nRT

Pretend you have a steel SCUBA tank filled with compressed air. For all intents and purposes, the volume is fixed. If you heat it up, the pressue will increase. If you cool it down the pressure will decrease.

Now transfer the gas to another SCUBA tank that is half the size. The presure inside this tank will be double....


blah blah blah...

==

I don't know dude. Its a fvcking math problem.. use your noodle...

I'm not talking about a sealed tank. I'm talking about a cylinder with a piston.
 
The pressure will likely more than double, as the electromagnetic force pushing the atoms apart decreases with the square of their distance from each other.
 
Originally posted by: ClueLis
The pressure will likely more than double, as the electromagnetic force pushing the atoms apart decreases with the square of their distance from each other.

Thanks. Are you sure though?
 
Originally posted by: WinkOsmosis
Originally posted by: ClueLis
The pressure will likely more than double, as the electromagnetic force pushing the atoms apart decreases with the square of their distance from each other.

Thanks. Are you sure though?

Not positive about the entire explanation, although I'm sure the bit about the electromagnetic force it true.
 
P=[n*.08206*(T in Kelvin)]/V
2.93atm = [.60 moles Ne * .08206 * 298K] / 5L
5.87atm = [.60 moles Ne * .08206 * 298K] / 2.5L

That's changing your volume. When you solve for temperature the equation becomes

T in Kelvin = P * V / nR

Thus:
298K = (2.93atm * 5L) / (.6 moles * .08206)
Compressing it yields
589K = (5.8atm * 5L) / (.6 moles * .08206)
An increase in temperature

What's difficult to understand?
 
Originally posted by: dtyn
P=[n*.08206*(T in Kelvin)]/V
2.93atm = [.60 moles Ne * .08206 * 298K] / 5L
5.87atm = [.60 moles Ne * .08206 * 298K] / 2.5L

That's changing your volume. When you solve for temperature the equation becomes

T in Kelvin = P * V / nR

Thus:
298K = (2.93atm * 5L) / (.6 moles * .08206)
Compressing it yields
589K = (5.8atm * 5L) / (.6 moles * .08206)
An increase in temperature

What's difficult to understand?

What's difficult for me to understand is how you figure compressing the gas allows it to stay at 5L.
 
Originally posted by: WinkOsmosis
Originally posted by: dtyn
P=[n*.08206*(T in Kelvin)]/V
2.93atm = [.60 moles Ne * .08206 * 298K] / 5L
5.87atm = [.60 moles Ne * .08206 * 298K] / 2.5L

That's changing your volume. When you solve for temperature the equation becomes

T in Kelvin = P * V / nR

Thus:
298K = (2.93atm * 5L) / (.6 moles * .08206)
Compressing it yields
589K = (5.8atm * 5L) / (.6 moles * .08206)
An increase in temperature

What's difficult to understand?

What's difficult for me to understand is how you figure compressing the gas allows it to stay at 5L.

increasing the pressure of a gas does not necessarily mean changing the volume

 
If temperature were to increase, pressure would have to increase less than double...

What is your specific question? I think you mean that if temp goes up, pressure increases too. That's true, of course. But we're looking at absolute temperature. If you have some air at 20ºC and you heat it to 40ºC the pressure does not double because 20ºC = 293K and 40ºC = 313K. Instead, if you heated the air from 20ºC to 313ºC (586K), then the pressure would indeed be double.

Convert everything to absolute temperature before calculating the T part of the equation.
 
Originally posted by: GeneValgene
Originally posted by: WinkOsmosis
Originally posted by: dtyn
P=[n*.08206*(T in Kelvin)]/V
2.93atm = [.60 moles Ne * .08206 * 298K] / 5L
5.87atm = [.60 moles Ne * .08206 * 298K] / 2.5L

That's changing your volume. When you solve for temperature the equation becomes

T in Kelvin = P * V / nR

Thus:
298K = (2.93atm * 5L) / (.6 moles * .08206)
Compressing it yields
589K = (5.8atm * 5L) / (.6 moles * .08206)
An increase in temperature

What's difficult to understand?

What's difficult for me to understand is how you figure compressing the gas allows it to stay at 5L.

increasing the pressure of a gas does not necessarily mean changing the volume

To increase the pressure without changing the volume, you would have to add more of the gas or heat it. I'm talking about heat as a result of compression
 
I understand your question now.

As you compress the gas, the atoms retain they're starting kinetic energy. As the volume decreases, more collisions occur releasing heat. The heat created is not enough to balance the equation, so the pressure increases also. If you compress the gas slowly, the temperature change is negligible. Do it quickly and it is VERY noticable.

anecdote: an inexperience SCUBA shop tech will open the valves all the way and will fill your tank to its rated pressure of 3000 PSI in less than a minute. They then give you a warm tank to take diving. When you get to the dive site, the tank has cooled and you have 2800PSI. An experienced tech will take 5-10 minutes to fill your tank and you will get to dive with 3000PSI
 
Originally posted by: Mwilding
I understand your question now.

As you compress the gas, the atoms retain they're starting kinetic energy. As the volume decreases, more collisions occur releasing heat. The heat created is not enough to balance the equation, so the pressure increases also. If you compress the gas slowly, the temperature change is negligible. Do it quickly and it is VERY noticable.

anecdote: an inexperience SCUBA shop tech will open the valves all the way and will fill your tank to its rated pressure of 3000 PSI in less than a minute. They then give you a warm tank to take diving. When you get to the dive site, the tank has cooled and you have 2800PSI. An experienced tech will take 5-10 minutes to fill your tank and you will get to dive with 3000PSI

That makes sense. I didn't realize speed had anything to do with it. I figured the reduced heating at slower speed would be a result of having more time to cool.
 
Originally posted by: WinkOsmosis
Originally posted by: dtyn
P=[n*.08206*(T in Kelvin)]/V
2.93atm = [.60 moles Ne * .08206 * 298K] / 5L
5.87atm = [.60 moles Ne * .08206 * 298K] / 2.5L

That's changing your volume. When you solve for temperature the equation becomes

T in Kelvin = P * V / nR

Thus:
298K = (2.93atm * 5L) / (.6 moles * .08206)
Compressing it yields
589K = (5.8atm * 5L) / (.6 moles * .08206)
An increase in temperature

What's difficult to understand?

What's difficult for me to understand is how you figure compressing the gas allows it to stay at 5L.

Sorry, forgot to change the volume. Take it down to 2.6L, you get 305K, temperature changes from compressing are very minute. Screw this, I'm a history major, no more thinking about science crap 🙂.
 
Originally posted by: Mwilding
I understand your question now.

As you compress the gas, the atoms retain they're starting kinetic energy. As the volume decreases, more collisions occur releasing heat. The heat created is not enough to balance the equation, so the pressure increases also. If you compress the gas slowly, the temperature change is negligible. Do it quickly and it is VERY noticable.

anecdote: an inexperience SCUBA shop tech will open the valves all the way and will fill your tank to its rated pressure of 3000 PSI in less than a minute. They then give you a warm tank to take diving. When you get to the dive site, the tank has cooled and you have 2800PSI. An experienced tech will take 5-10 minutes to fill your tank and you will get to dive with 3000PSI

Dang, I was all like "wow" when you got "for all intents and purposes" right above but then you went and did that. 🙁
 
Originally posted by: KingNothing
Originally posted by: Mwilding
I understand your question now.

As you compress the gas, the atoms retain they're starting kinetic energy. As the volume decreases, more collisions occur releasing heat. The heat created is not enough to balance the equation, so the pressure increases also. If you compress the gas slowly, the temperature change is negligible. Do it quickly and it is VERY noticable.

anecdote: an inexperience SCUBA shop tech will open the valves all the way and will fill your tank to its rated pressure of 3000 PSI in less than a minute. They then give you a warm tank to take diving. When you get to the dive site, the tank has cooled and you have 2800PSI. An experienced tech will take 5-10 minutes to fill your tank and you will get to dive with 3000PSI

Dang, I was all like "wow" when you got "for all intents and purposes" right above but then you went and did that. 🙁
lol - I noticed that when I read it, and didn't fix it...

😱
 
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