Question

Question #caa3e

Answer 1

All of these problems depend on the choice of gas constant `R`.....

Explanation:

And typically, chemists, measure volume in `"litres"`, and pressure in `mm*Hg`, where `"760 mm Hg"-=1*atm`. (NB, you do not use a mercury column to measure pressures > `1*atm`.

And thus (at least in my opinion) the gas constant that is generally most useful is `R=0.0821*L*atm*K^-1*mol^-1`.

The old standby, `P=(nRT)/V`. Which equation is this?

`=((25*g)/(44.01*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx298*K)/(0.500*L)=27.8*atm` (which is rather high pressure!).

We might have expected the high pressure, because the molar volume at `1*atm` is approx. `25*L`, and here we have compressed the gas into a much smaller volume.