Question

What is the new volume of a gas at `655*mm*Hg` pressure AND volume of `955*L` that is compressed to `1.34*atm`?

Answer 1

Well, old `"Boyle's Law"` holds that `P_1V_1=P_2V_2`...

Explanation:

...i.e. `Pprop1/V`

And we know that `1*atm` will support a column of mercury that is `760*mm` high...

`760*mm*Hg-=1*atm`

(I hope this has been demonstrated to you in the lab...I cannot demonstrate this in my lab because of the health and safety mafia.)

So we solve for `V_2=(P_1V_1)/P_2=((655*mm*Hg)/(760*mm*Hg*atm^-1)xx955*L)/(1.34*atm)`

Clearly, we will get an answer in LITRES, why `"clearly"`?

What do you make the volume.....?