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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. 

\(\frac{P}{T} = \frac{P}{T}\)

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

\(\frac{750.0}{323.0} = \frac{x}{273.13}\)

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:

\(\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}\)

\(\dfrac{750.0}{323.0} = \dfrac{P_2}{273.15}\)

\(323.0*P_2 = 750.0*273.15\)

\(323.0\:P_2 = 204862.5\)

\(P_2 = \dfrac{204862.5}{323.0}\)

\(P_2 = 634.249226... \to \boxed{\boxed{P_2 \approx 634.2\:mmHg}}\end{array}}\qquad\checkmark\)

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

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