Proof gas loses 1/273 of its volume - Charle's Law

[ghostanime2001]

Homework Statement

Show that any gas loses 1/273 of its volume at 0 °C, when it is cooled by 1 Celsius degree.

$V_{1}=?$

$V_{2}=?$

$T_{1}=0 °C=273 K$

$T_{2}=272 K$

Homework Equations

I'm assuming I have to use Charle's Law AND starting with any starting arbitrary volume:

$\dfrac{V_{1}}{T_{1}}=\dfrac{V_{2}}{T_{2}}$

The Attempt at a Solution

So substituting in all the values above and solving for V2 I get:

$\dfrac{V_{1}}{273}=\dfrac{V_{2}}{272}$

$272V_{1}=273V_{2}$

$\dfrac{272}{273}V_{1}=V_{2}$

$\dfrac{273-1}{273}V_{1}=V_{2}$

$\left(\dfrac{273}{273}-\dfrac{1}{273}\right)V_{1}=V_{2}$

$\left(1-\dfrac{1}{273}\right)V_{1}=V_{2}$

$V_{1}-\dfrac{1}{273}V_{1}=V_{2}$

That's as far as I can get and I guess it makes sense? If I start with any $V_{1}$ i subtract that with 1/273th of the initial volume and i get the second volume. Assuming I start with

$V_{1}=1 L$, the second volume $V_{2}$ would be 0.996 L. The second set of experimental conditions are colder so I expect the volume to decrease following Charle's Law.

BUT! am I supposed to show explicitly that $\dfrac{1}{273}=V_{2}$ ? How do I show $\dfrac{1}{273}=V_{2}$ explicitly?

 

[Borek]

You are almost done. Hint: loses 1/273 volume doesn't mean V₂=1/273. Try to express the loss in terms of V₁-V₂, or V₂/V₁.

 

[technician]

Volume is proportional to T in Kelvins.
The volume at 0C (273K) = Vo, the volume at -273C (0K) = 0
Change in volume ? Change in temp ?

 

[Barakn]

the volume at -273C (0K) = 0
Change in volume ? Change in temp ?

The volume at 0K = 0 only holds for ideal gases (hint: nothing is ideal).

 

[technician]

removed

 

[Borek]

This discussion if off topic, and in no way helpful for the OP. Question doesn't need any information about gas behavior at 0 deg K to be solved.