Question

A sample of argon has a volume of 205 `cm^3` when it's temperatures is -44 degrees C. and its pressure is 712 mm of Hg. What would be the volume of the argon at STP?

Answer 1

We use the combined gas law, which states for a GIVEN quantity of gas..........`(P_1V_1)/T_1=(P_2V_2)/T_2`; and we eventually get `V_2~=230*cm^3`

Explanation:

`T` is always expressed as `"temperatura assoluta"`, which sets absolute zero at `0*K-=-273.15` `""^@C`. And depending on your syllabus, `STP` specifies a temperature of `273.15*K`, and a pressure of `1*atm`. And we know that `1*atm` will support a column of mercury that is `760*mm` high.........

We want `V_2=(P_1xxV_1xxT_2)/(P_2xxT_1)`.....

And already we see from the quotient we ARE going to get an answer with `"VOLUME units........"` as required.

`V_2=(712*mm*Hgxx205*cm^3xx273.15*K)/(760*mm*Hgxx229.15*K)`

`~=230*cm^3`...........