돌턴의 분압법칙과 수상치환. 500.0 mL of a gas at 20.0℃ and 720.0 mmHg

돌턴의 분압법칙과 수상치환. 500.0 mL of a gas at 20.0℃ and 720.0 mmHg

 

 

500.0 mL of a gas was collected over water at 20.0℃ and 720.0 mmHg.

What is its volume at STP? The vapor pressure of water is 17.5 mmHg at 20.0℃.

 

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P1•V1 / T1 = P2•V2 / T2

 

P1 = 720.0 mmHg – 17.5 mmHg = 702.5 mmHg

V1 = 500.0 mL

T1 = 20.0℃ + 273 = 293.0 K

 

P2 = 760 mmHg

V2 = ? mL

T2 = 0℃ = 273.0 K

 

V2 = P1•V1•T2 / T1•P2

= [(702.5 mmHg)(500.0 mL)(273.0 K)] / [(293.0 K)(760 mmHg)]

= 430.6 mL

 

 

답: 430.6 mL