[CSG]

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!

[sjb]

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 ?

[billnotgatez]

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