N2 gas 1×10^5 atm 0.8 10 mol 298 K 5 atm

N2 gas 1×10^5 atm 0.8 10 mol 298 K 5 atm

 

 

The Henry’s law constant for the solubility

of N2 gas in water at 298 K is 1×10^5 atm.

The mole fraction of N2 in air is 0.8.

Calculate the number of moles of N2

​dissolved in 10 moles of water at 298 K and 5 atm.

 

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돌턴의 분압 법칙

P_a = X_a × P_tot

( 참고 https://ywpop.tistory.com/48 )

 

N2의 부분 압력, P_N2 = 0.8 × 5 atm = 4 atm

 

 

 

Henry의 법칙

P_g = k_H • X_g

( 참고 https://ywpop.tistory.com/1911 )

 

X_N2 = P_N2 / k_H

= 4 atm / (1×10^5 atm)

= 4×10^(-5)

 

 

 

X_N2 = n_N2 / (n_N2 + n_H2O)

( 참고: 몰분율 https://ywpop.tistory.com/2659 )

 

n_N2 <<< n_H2O 이므로,

(n_N2 + n_H2O) ≒ n_H2O

 

X_N2 = n_N2 / (n_N2 + n_H2O)

≒ n_N2 / n_H2O

 

 

 

n_N2 = X_N2 × n_H2O

= 4×10^(-5) × 10 mol H2O

= 4×10^(-4) mol N2

 

 

 

답: 4×10^(-4) mol

 

 

 

 

[키워드] 질소 헨리 법칙 기준문서, N2 헨리 법칙 기준문서