320 mg CH4 175 mg Ar 225 mg Ne mixture 300 K 8.87 kPa

320 mg CH4 175 mg Ar 225 mg Ne mixture 300 K 8.87 kPa

 

 

A gas mixture consists of 320 mg of methane,

175 mg of argon, and 225 mg of neon.

The partial pressure of neon at 300 K is 8.87 kPa.

Calculate (i) the volume and (ii) the total pressure of the mixture.

 

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각 성분의 몰수를 계산하면,

 

CH4의 몰질량 = 16.04 g/mol 이므로,

0.320 g / (16.04 g/mol) = 0.01995 mol CH4

( 참고: n=W/M https://ywpop.tistory.com/7738 )

 

0.175 / 39.95 = 0.00438 mol Ar

 

0.225 / 20.18 = 0.01115 mol Ne

 

 

 

전체 몰수 = 0.01995 + 0.00438 + 0.01115 = 0.03548 mol

 

 

 

1 atm = 101.325 kPa 이므로,

8.87 kPa × (1 atm / 101.325 kPa) = 0.08754 atm

 

 

 

P_Ne = X_Ne × P_tot

( 참고 https://ywpop.tistory.com/48 )

 

P_tot = P_Ne / X_Ne

= 0.08754 atm / (0.01115 / 0.03548)

= 0.279 atm

---> 혼합물의 전체 압력

 

 

 

PV = nRT

( 참고 https://ywpop.tistory.com/3097 )

 

V = nRT / P

= (0.03548) (0.08206) (300) / (0.279)

= 3.13 L

---> 혼합물의 전체 부피