25℃ total pressure 1.00 atm 0.586 g He water 23.8 torr

25℃ total pressure 1.00 atm 0.586 g He water 23.8 torr

 

 

Helium is collected over water at 25℃ and 1.00 atm total pressure.

What total volume of gas must be collected to obtain 0.586 g helium?

(At 25℃ the vapor pressure of water is 23.8 torr.)

 

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P_total = P_He + P_water

( 참고 https://ywpop.tistory.com/3490 )

 

 

 

P_water = (23.8 torr) (1 atm / 760 torr) = 23.8/760 atm

 

 

 

P_He = P_total – P_water

= 1.00 – (23.8/760)

= 0.9687 atm

 

 

 

He의 몰질량 = 4.00 g/mol

0.586 g / (4.00 g/mol) = 0.1465 mol He

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

PV = nRT

( 참고 https://ywpop.tistory.com/3097 )

 

V = nRT / P

= (0.1465) (0.08206) (273.15+25) / (0.9687)

= 3.70 L

 

 

 

 

[키워드] 헬륨 수상포집 기준문서, 헬륨 수상치환 기준문서, He 수상포집 기준문서, He 수상치환 기준문서, 25℃ 1.00 atm He water He 0.586 g volume 23.8 torr