Question

At constant temperature, a `3.8*L` volume of gas at `765*mm*Hg`, was expanded to give a pressure of `0.500*atm`. What is the new volume?

Answer 1

`P_1V_1=P_2V_2` under conditions of constant temperature.

Explanation:

We need to know that `1*atm` will support a column of mercury that is `760*mm` high, and thus a mercury column may used as a highly visual measurement of pressure:

`1*atm-=760*mm*Hg` (note that given the schemozzle that occurs when you spill mercury in the lab, you really should not put mercury under a pressure of MORE than 1 atmosphere). Here.......

`P_1=(765*mm*Hg)/(760*mm*Hg*atm^-1)=1.01*atm.`

We solve for `V_2=(P_1V_1)/P_2=(1.01*atmxx3.8*L)/(0.500*atm)=7.60*L`