sorry, another question - Tilted Forum Project Discussion Community

danny_boy · 01-13-2006, 07:57 AM

Temp of earth's surface drops 5 degrees C for every 1000m of elevation above the surface. If the ground temp is 15 degrees C and pressure is 760mmHg, at what elevation will the pressure be 380mmHg. Assume ideal gas behavior.

Im taking an engineering course to broaden my horizons, but feel lost here.

Ideal gas law is Pv=nRT, so I know I need a relationship between temp and pressure, and then relate that to distance (elevation).

I need to find what the temperature is, when the pressure decrease by 1/2, and am unsure if I merely need to substitute values into a known equation, or formulate a new relationship. Any advice would be appreciated.

Thanks,

AngelicVampire · 01-13-2006, 10:09 AM

Ok, the first thing is that this relates A change in Pressure to a change in temperature:

So if we assume that the volume of air and the number of moles of the gas are fixed (which is not totally true but close enough for my example). So we now have:

Pk = T, we have the pressure and a now constant k, so simply run the numbers through to get the new temperature for that pressure. Then simply add 1000/5degrees to get your answer.

Thats what I'd do anyway.

Pip · 01-13-2006, 10:12 AM

Since P and T are directly proportional, you don't even have to bother finding out k.

danny_boy · 01-13-2006, 03:24 PM

Ok,

So, after using pK=t, K is 0.0197, and solving for the new temp at 380mmHg, the new temp is 7.5 degrees C.

If the temp drops 5 degrees for every 1000m of elevation, then the temp went from 15 degrees down to 7.5 degrees. Setting up the equation:

5/1000 = 7.5/X

X is =1500m

Have I done this correctly ?

Thanks.

AngelicVampire · 01-13-2006, 09:26 PM

Looks ok, of course you should note that pressure drops as height increases so the volume of the gas will expand so k is not truly a constant. However other than that thats the sort of way I would go at that problem.

01-13-2006, 07:57 AM	#1 (permalink)
danny_boy Crazy	sorry, another question Temp of earth's surface drops 5 degrees C for every 1000m of elevation above the surface. If the ground temp is 15 degrees C and pressure is 760mmHg, at what elevation will the pressure be 380mmHg. Assume ideal gas behavior. Im taking an engineering course to broaden my horizons, but feel lost here. Ideal gas law is Pv=nRT, so I know I need a relationship between temp and pressure, and then relate that to distance (elevation). I need to find what the temperature is, when the pressure decrease by 1/2, and am unsure if I merely need to substitute values into a known equation, or formulate a new relationship. Any advice would be appreciated. Thanks,

01-13-2006, 10:12 AM	#3 (permalink)
Pip Likes Hats Location: Stockholm, Sweden	Since P and T are directly proportional, you don't even have to bother finding out k. Last edited by Pip; 01-13-2006 at 10:14 AM.. Reason: Too slow :D

01-13-2006, 10:09 AM	#2 (permalink)
AngelicVampire Insane	Ok, the first thing is that this relates A change in Pressure to a change in temperature: So if we assume that the volume of air and the number of moles of the gas are fixed (which is not totally true but close enough for my example). So we now have: Pk = T, we have the pressure and a now constant k, so simply run the numbers through to get the new temperature for that pressure. Then simply add 1000/5degrees to get your answer. Thats what I'd do anyway.

01-13-2006, 03:24 PM	#4 (permalink)
danny_boy Crazy	Ok, So, after using pK=t, K is 0.0197, and solving for the new temp at 380mmHg, the new temp is 7.5 degrees C. If the temp drops 5 degrees for every 1000m of elevation, then the temp went from 15 degrees down to 7.5 degrees. Setting up the equation: 5/1000 = 7.5/X X is =1500m Have I done this correctly ? Thanks.

01-13-2006, 09:26 PM	#5 (permalink)
AngelicVampire Insane	Looks ok, of course you should note that pressure drops as height increases so the volume of the gas will expand so k is not truly a constant. However other than that thats the sort of way I would go at that problem.