What is the volume of a container that holds 25.0 g of carbon gas at STP?

1 Answer
Jul 1, 2016

Answer:

The container has a volume of #46.6 L#.

Explanation:

Since there are only one set of conditions and we are at STP, we must use the ideal gas law equation

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  • P represents pressure (may have units of atm, depending on the units of the universal gas constnat)
  • V represents volume (must have units of liters)
  • n represents the number of moles
  • R is the proportionality constant (has units of #(Lxxatm)/ (molxxK)#)
  • T represents the temperature, which must be in Kelvins.

Next, list your known and unknown variables. Our only unknown is the volume of #C(g)#. Our known variables are P,n,R, and T.

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 #(Lxxatm)/ (molxxK)#

The only issue is the mass of #C(g)#, we need to convert it into moles of #C(g)# in order to use the ideal gas law.

#25.0cancel"g" xx (1molC)/(12.01cancel"g")# = 2.08 mol #C#

Now all we have to do is rearrange the equation and solve for V like so:
#V = (nxxRxxT)/P#
#V = (2.08molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)#
#V = 46.6 L#