Answer to Question #157826 in Physical Chemistry for sally

Question

A sample of oxygen gas is collected over water at 25°C where the vapor pressure at this

temperature is 23.8 mmHg. The wet gas occupies a volume of 7.28 L at a total pressure

of 1.25 bar. If all the water is removed, what volume will the dry oxygen occupied at a

pressure of 1.07 atm and a temperature of 37°C?

Answer 1

P_tot=P(water)+P(oxygen)

P(oxygen)=1.26625atm-0.0352223=1.23atm

PV=nRT

n=PV/RT

=1.23*7.28/(0.082057*298)

n=0.361mol oxygen

PV=nRT

V=nRT/P

=0.361*0.082057*310/1.23

V=8.586l