Question

A `0.275*L` volume of an unspecified gas exerts a pressure of `732.6*mm*Hg` at a temperature of `-28.2` `""^@C`. What is the molar quantity of this gas?

Answer 1

Approx. `0.013*mol`

Explanation:

We need to know that `1*atm` will support a column of mercury `760*mm` high, and thus we can use the length of a mercury column to express pressure.

And thus `P=(732.6*mm*Hg)/(760*mm*Hg*atm^-1)=??*atm`

And we use the `"Ideal Gas Law"`, `n=(PV)/(RT)`

`n=(PV)/(RT)=((732.6*mm*Hg)/(760*mm*Hg*atm^-1)xx0.275*L)/(0.0821*L*atm*K^-1*mol^-1xx245*K)=`

`1.32xx10^-2*mol.`

Why did I change the temperature to `"degrees Kelvin"`?