If 1.93 * 10^-3 mol of argon occupies a 65.8-mL container at 20 degrees Celsius, what is the pressure (in torr)?

1 Answer
Jul 6, 2017

Answer:

#P~=0.706*atm-=??"Torr"#

Explanation:

From the Ideal Gas equation, which you must be getting sick of.......

#P=(nRT)/V=(1.93xx10^-3*molxx0.0821*(L*atm)/(K*mol)xx293*K)/(65.8*mLxx10^-3*L*mL^-1)#

#-=0.706*atm#

And here the trick is to know that #1000*mL-=1*L#, i.e. #1*mL-=10^-3*L#.

Anyway, you know the relationship between #"atmospheres"# and #mm*Hg# so have at it.............