If a sample of gas is intially at 1.8 atm, 22.0 l, and 26.4 c, what will be the volume if the pressure is reduced by 0.8 atm and the temperature is lowered to 20.3 c?

pv=nrT Initial state (1.8atm) (22.0 l) = n (0.082057) (26.4+273.15); r=.082057, and converting C to K Solving for n = (1.8) (22) / (.082057 * (26.4+273.15) moles n = 1.611 moles in initial state Now we solve for new volume pv=nrT (.8atm) v = (1.611) (.082057) (20.3+273.15) v = (1.611) (.082057) (20.3+273.15) /.8 v=48.49 l