Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: CSG on March 19, 2010, 10:23:22 AM
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Hi,
Could anyone mathematically show how to answer the following question please?
"A gas sample occupies a Volume V1 at a pressure P1 and a Kelvin temperature T1. What would be the temperature of the gas, T2, if both its pressure and volume are doubled?"
The answer is T2=4T1
It makes sense for it to be 4, however I want to understand it in terms with PV/T = PV/T
Thanks!
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The ideal gas equation in full reads pV = nRT.
For a given sample of gas, n is constant, and R is really just a fudge factor to get the right values / units. Now what happens if p -> 2p, and V -> 2V ?
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I am reusing part of a post I did before
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Since this topic was posted a while ago, I guess I can use it without doing homework for someone.
Anyone can correct me if you see issues.
When analyzing the problem in general it looks like you can look at it as an ideal gas in situation 1 and situation 2.
Mathematically it can be represented by
Situation 1 ---- P1V1= n1RT1
Situation 2 ---- P2V2= n2RT2
It follows mathematically
Situation 1 ---- R = (P1V1) / (n1T1)
Situation 2 ---- R = (P2V2) / (n2T2)
Since R is the same in both situations
(P1V1) / (n1T1)= (P2V2) / (n2T2)
Now for the problem at hand
For this problem we can eliminate n1 and n2 from the equation since it is the same on both sides.
(P1V1) / (T1)= (P2V2) / (T2)
Algebraically we can change the formula to
(P1V1T2)= (P2V2T1)
Solving for T2
(T2) = (P2V2T1)/(P1V1)
Since P2 is 2P1 and V2 is 2V1 in this problem
(T2) = (2P12V1T1)/(P1V1)
Again algebraically we can eliminate the P1 and V1
(T2) = (2 . 2 . T1)
or
T2 = 4T1 for this situation