N2 78.08% O2 20.94% CO2 0.05% Ar 0.93% 25℃ 1.00 atm 1.00 L

N2 78.08% O2 20.94% CO2 0.05% Ar 0.93% 25℃ 1.00 atm 1.00 L

air sample N2 78.08% O2 20.94% CO2 0.05% Ar 0.93% 25℃ 1.00 atm 1.00 L

 

 

A sample of air contains 78.08% nitrogen, 20.94% oxygen,

0.05% carbon dioxide, and 0.93% argon, by volume.

How many molecules of each gas are present in 1.00 L

of the sample at 25℃ and 1.00 atm?

 

 

25℃, 1.00 atm에서 시료 1.00 L에는

각 기체가 몇 분자씩 들어있는가?

 

---------------------------------------------------

 

PV = nRT 로부터,

( 참고 https://ywpop.tistory.com/3097 )

 

공기 시료의 몰수를 계산하면,

n = PV / RT

= [(1.00) (1.00)] / [(0.08206) (273.15 + 25)]

= 0.04087 mol air

 

 

 

V ∝ n 이므로,

( 참고 https://ywpop.tistory.com/1979 )

 

(0.04087 mol) (78.08/100) (6.022×10^23 개/mol)

= (0.04087) (78.08/100) (6.022×10^23)

= 1.92×10^22 개 N2

 

 

 

(0.04087) (20.94/100) (6.022×10^23)

= 5.15×10^21 개 O2

 

 

 

(0.04087) (0.05/100) (6.022×10^23)

= 1.23×10^19 개 CO2

 

 

 

(0.04087) (0.93/100) (6.022×10^23)

= 2.29×10^20 개 Ar