How is the Ideal Gas Law Derived from Temperature and Pressure Relationships?

[Kathhhriine]

Homework Statement

In the question, it was first asked to state the relationship between p1, p2, t1 and t', and v1, v2, t' and t2. This was found to be p1/t1=p2/t' and v1/t'=v2/t2. Further, it is asked to use these relationships to deduce that for an ideal gas PV=KT where K=constant, this is the part i dont understand.
For further understanding of the question, the question states this: "The ideal gas is held in a cylinder by a moveable piston. The pressure of the gas is p1, its volume is V1 and its kelvintemperature is T1. The pressure, volume and temperature are changed to p2, V2and T2 respectively." from p1 t', the volume is kept constant, giving p2, t' v1, and from v1, t', pressure is kept constant, giving v2, t2 and p2.

Relevant Equations

PV=KT
p1/t1=p2/t' and v1/t'=v2/t2

I figured that T' is a common factor for both relationships and from there deduceted that T'=p2xt1/p1=v1xt2/v2. However, I don't understand how that can be further manipulated to PV=KT.

 

[Chestermiller]

Since, at constant volume, pressure is proportional to temperature, you can write $$P=f(V)T$$ and since at constant pressure, volume is proportional to temperature, you can write $$V=g(P)T$$ So, if you divide one equation by the other, you get $$\frac{P}{V}=\frac{f(V)}{g(P)}$$or$$Pg(P)=Vf(V)$$So the left hand side is a function only of pressure and the right hand side is a function only of volume. Mathematically, what is the only way this can happen?

 

[ehild]

p1/t1=p2/t' and v1/t'=v2/t2

I figured that T' is a common factor for both relationships and from there deduceted that T'=p2xt1/p1=v1xt2/v2. However, I don't understand how that can be further manipulated to PV=KT.

You have p2xt1/p1=v1xt2/v2 . Divide both sides of the equation by (t1t2) and muliply by (p1v2).