Question #caa3e

1 Answer
Apr 27, 2017

Answer:

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.