He gas Henry 25℃ 3.8×10^(-4) M/atm 293 mmHg 10.00 L water

He gas Henry 25℃ 3.8×10^(-4) M/atm 293 mmHg 10.00 L water

 

 

물에서 헬륨 기체의 용해도에 대한 Henry의 법칙 상수는

25℃에서 3.8×10^(-4) M/atm이다.

 

 

The Henry’s law constant for the solubility of helium gas in water

is 3.8×10^(-4) M/atm at 25℃.

a) Express the constant for the solubility of helium gas in M/mmHg.

b) If the partial pressure of He at 25℃ is 293 mmHg,

what is the concentration of dissolved He in mol/L at 25℃?

c) What volume of helium gas can be dissolved in 10.00 L of water

at 293 mmHg and 25℃? (Ignore the partial pressure of water.)

 

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a) Express the constant for the solubility of helium gas in M/mmHg.

 

1 atm = 760 mmHg 이므로,

(3.8×10^(-4) M/atm) (1 atm / 760 mmHg)

= (3.8×10^(-4)) (1 / 760)

= 5.0×10^(-7) M/mmHg

 

 

 

 

b) If the partial pressure of He at 25℃ is 293 mmHg,

what is the concentration of dissolved He in mol/L at 25℃?

 

S_g = k P_g

( 참고 https://ywpop.tistory.com/1911 )

 

= (5.0×10^(-7) M/mmHg) (293 mmHg)

= (5.0×10^(-7)) (293)

= 0.0001465 M

= 1.47×10^(-4) M

= 1.47×10^(-4) mol/L

 

 

 

 

c) What volume of helium gas can be dissolved in 10.00 L of water

at 293 mmHg and 25℃? (Ignore the partial pressure of water.)

 

(1.47×10^(-4) mol/L) (10 L) = 0.00147 mol He

 

 

 

PV = nRT 로부터, V를 계산하면,

( 참고: 이상 기체 방정식 https://ywpop.tistory.com/3097 )

 

V = nRT / P

= [(0.00147) (0.08206) (273.15+25)] / (293/760)

= 0.0933 L

(= 93.3 mL)

 

 

 

 

[키워드] 헬륨의 헨리 법칙 기준문서, He의 Henry 법칙 기준문서