Question

What volume would a sample of gas that has a volume of `1.63*L` at `793*"mm Hg"` occupy at `755*"mm Hg"`?

Answer 1

Again it is highly non-standard (and dangerous) to quote a pressure in `mm*Hg` for pressures OVER 1 atmosphere.

Explanation:

We know that `1*atm-=760*mm*Hg`...or that one atmosphere will support a column of mercury that is `760*mm` high...pressures higher than one atmosphere ARE NEVER quoted in `mm*Hg` (except in these scenarios where some punter who has NEVER used a mercury manometer mistakenly proposes an illegitimate measurement).

And so `793*"Torr"-=(793*"Torr")/(760*"Torr"*atm^-1)=1.04*atm`

And given `P_1V_1=P_2V_2`...

`V_2=(P_1V_1)/P_2=((755*"Torr")/(760*"Torr"*atm^-1))/(1.04*atm)xx1.63*L=1.56*L`..

Reasonably, the volume has reduced slightly. But your chemistry teacher has no business asking questions with these whack units.