What is the pressure of 2.50 moles of an ideal gas if it has a volume of 50.0 liters when the temperature is 300.0 K

Start with the ideal gas equation, PV=nRT PV=nRT and rearrange for pressure to get p = nRT V p = nRT V. You have all the necessary variables in their proper units, so plug em' into the equation to solve for pressure in units of atmospheres.

P = (2.5 mol) (300 K) (0.08206 L atm K - 1 mo l - 1 50.0 L = 1.23 atm P = (2.5 mol) (300 K) (0.08206 L atm K - 1 mo l - 1 50.0 L = 1.23 atm

All that needs to be done now is converting atmospheres to mm Hg Hg.

1.23 atm∗ 760 mm Hg 1 atm = 935 mm Hg 1.23 atm∗ 760 mm Hg 1 atm = 935 mm Hg.

That value makes sense, since the original pressure in atmospheres was above 1, the pressure in mm Hg Hg will be above 760.