Find the mass of carbon dioxide contained in 0.85-m^3 tank at 2068.44 kpaa, 37.8°C?

2 Answers
Mar 23, 2018

Well, we use the old Ideal Gas equation...and find the appropriate value of #R#...these should be listed as supplementary material in an exam....

Explanation:

#n="mass"/"molar mass"=(PV)/(RT)=(2068.44*kPaxx850*L)/(8.314*L*kPa*K^-1*mol^-1xx311.0*K)=680.0*mol#

Note that a #m^3# is a LARGE volume, and corresponds to #1000*L#, or #1000*dm^3#...

And, if I have dun my sums right, this gives a mass of #680.0*molxx44.01*g*mol^-1xx10^-3*kg*g^-1=29.93*kg#...take that environment....

Mar 24, 2018

see a solution process below;

Explanation:

#"Volume",V=0.85m³#

#"Pressure",P=2068.44kpa#

#P= 2068.44 × 10³Pa#

#"Temperature", T =37.8°C#

#T= 37.8 + 273 = 310.8K#

#R ("Constant")= 8.31Jmol¯¹K¯¹#

#"No. of Moles", n =? #

Using;

#PV = nRT#

#n = (PV)/(RT)#

#n= (2068.44×10³×0.85)/(310.8 × 8.31)#

#n = 680.74mol#

#"Molar mass of" color(white)xCO₂=44g#

#1"mole" = 44g color(white)x"of"color(white)x CO₂#

#680.74"mole" = x#

#x = 680.74 × 44#

#x = 29,952.45g#

#x = 29.95245kg#

#x = 29.95kg (2 d.p)#