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if a gas is cooled from 323.0 K to 273.15 k and the volume is kept constant what final pressure would result if the original pressure was 750.0 mm Hg

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IncredibleCaitlyn
Alright, so, this seems to be dealing with gas laws. 
 

You have temperature, volume (Though volume wont matter in this equation becasue it would cancel itself out) and pressure. 

[tex] \frac{P}{T} = \frac{P}{T} [/tex]

Your first fraction is what you start with, your second is what you end with 

[tex] \frac{750.0}{323.0} = \frac{x}{273.13} [/tex]

Now in order to find the unknown pressure, you will multiple the 273.13 on both sides (On the first side to get x alone and then on the next becasue what is done to one side must be done to the other) 

x = 634.2 mm Hg




dexteright02

Hello!

if a gas is cooled from 323.0 K to 273.15 k and the volume is kept constant what final pressure would result if the original pressure was 750.0 mmHg

We have the following information:

P1 (initial pressure) = 750.0 mmHg

T1 (initial temperature) = 323.0 K

P2 (final pressure) = ? (in mmHg)

T2 (final temperature) = 273.15 K

According to the Law of Charles and Gay-Lussac in the study of gases, we have an isochoric (or isovolumetric) transformation when its volume remains constant or equal, then we will have the following formula:

[tex]\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}[/tex]

[tex]\dfrac{750.0}{323.0} = \dfrac{P_2}{273.15}[/tex]

[tex]323.0*P_2 = 750.0*273.15[/tex]

[tex]323.0\:P_2 = 204862.5[/tex]

[tex]P_2 = \dfrac{204862.5}{323.0}[/tex]

[tex]P_2 = 634.249226... \to \boxed{\boxed{P_2 \approx 634.2\:mmHg}}\end{array}}\qquad\checkmark[/tex]

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I Hope this helps, greetings ... Dexteright02! =)

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