Question #f4c7f

1 Answer
SCooke
Jun 19, 2017

#2.80 xx 10^23# molecules.

Explanation:

You have to be careful with the “N” ( normal temperature (20’C)) and STP ( standard 0’C, in degrees Kelvin (273.15) for the equation). One is for convenience, the other is for accuracy. The “22.4L/mol” figure is handy and useful, but based on the conditions at STP, not NTP.

Start with the Ideal Gas Law: #P*V = (n*R*T) #
Rearranging the Ideal Gas Law equation we have #V = (n*R*T)/P#
where R = 0.0820575 L-atm/K-mol , P = 1.0 atm, n = 1 mol and T = 0’C = 273.15’K (STP)

#V = (n*R*T)/P# ; #V = (1.0*0.0821*273.15)/1.0#
#V = (1.0*0.0821*273.15)/1.0 = 22.4L#

At NTP this would be

#V = (1.0*0.0821*293.15)/1.0 = 24.1L#!
That’s a 10% difference from the STP value!

Directly calculated for this problem, we have:

#n = (P*V)/(R*T)# ; #n = (1.0*11.2)/(0.0821*293.15) = 0.465 mol#

Applying Avogadro's Number to this we obtain: #0.465 mol * 6.022 xx 10^23 = 2.80 xx 10^23# molecules.

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