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Topic: Ideal Gas Equation  (Read 3207 times)

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Offline CSG

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Ideal Gas Equation
« on: March 19, 2010, 10:23:22 AM »

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


Offline sjb

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Re: Ideal Gas Equation
« Reply #1 on: March 19, 2010, 11:43:48 AM »
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 ?

Offline billnotgatez

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Re: Ideal Gas Equation
« Reply #2 on: March 28, 2010, 07:23:30 AM »
I am reusing part of a post I did before
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)
 T2 = 4T1 for this situation

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