- #1
ghostanime2001
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Homework Statement
Show that any gas loses 1/273 of its volume at 0 °C, when it is cooled by 1 Celsius degree.
[itex]V_{1}=?[/itex]
[itex]V_{2}=?[/itex]
[itex]T_{1}=0 °C=273 K[/itex]
[itex]T_{2}=272 K[/itex]
Homework Equations
I'm assuming I have to use Charle's Law AND starting with any starting arbitrary volume:
[itex]\dfrac{V_{1}}{T_{1}}=\dfrac{V_{2}}{T_{2}}[/itex]
The Attempt at a Solution
So substituting in all the values above and solving for V2 I get:
[itex]\dfrac{V_{1}}{273}=\dfrac{V_{2}}{272}[/itex]
[itex]272V_{1}=273V_{2}[/itex]
[itex]\dfrac{272}{273}V_{1}=V_{2}[/itex]
[itex]\dfrac{273-1}{273}V_{1}=V_{2}[/itex]
[itex]\left(\dfrac{273}{273}-\dfrac{1}{273}\right)V_{1}=V_{2}[/itex]
[itex]\left(1-\dfrac{1}{273}\right)V_{1}=V_{2}[/itex]
[itex]V_{1}-\dfrac{1}{273}V_{1}=V_{2}[/itex]
That's as far as I can get and I guess it makes sense? If I start with any [itex]V_{1}[/itex] i subtract that with 1/273th of the initial volume and i get the second volume. Assuming I start with
[itex]V_{1}=1 L[/itex], the second volume [itex]V_{2}[/itex] 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 [itex]\dfrac{1}{273}=V_{2}[/itex] ? How do I show [itex]\dfrac{1}{273}=V_{2}[/itex] explicitly?