Question

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

Answer 1

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:

`(P_1)/(T_1) = (P_2)/(T_2)`

`(750.0)/(323.0) = (P_2)/(273.15)`

`323.0*P_2 = 750.0*273.15`

`323.0\:P_2 = 204862.5`

`P_2 = (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! =)

Answer 2

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.


`(P)/(T) = (P)/(T)`

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


`(750.0)/(323.0) = (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