0.7 kg Ar tank 1800 kPa, 40℃ volume

0.7 kg Ar tank 1800 kPa, 40℃ volume

 

 

A mass of 0.7 kg of argon is maintained

at 1800 kPa and 40℃ in a tank.

(Assume that the argon is an ideal gas.

Molar mass of argon is 39.948 kg/kmol and

the universal gas constant is 8.314 kJ/kmol•K.)

Determine the volume of the tank. m3

 

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700 g / (39.948 g/mol) = 17.5228 mol Ar

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

8.314 kJ/kmol•K

= 8.314 J/mol•K

= 8.314 Pa•m3/mol•K

( 참고 https://ywpop.tistory.com/1988 )

 

 

 

PV = nRT 로부터,

V를 계산하면,

V = nRT / P

= (17.5228 mol) (8.314 Pa•m3/mol•K) (273.15 + 40 K) / (1800×1000 Pa)

= (17.5228) (8.314) (273.15 + 40) / (1800×1000)

= 0.025345 m3

= 25.345 L

 

 

 

답: 0.025345 m3 또는 25.345 L

 

 

 

 

[참고] 0.5 kg Ar tank 1400 kPa, 40℃ volume

 

V = nRT / P

= (500 / 39.948 mol) (8.314 Pa•m3/mol•K) (273.15 + 40 K) / (1400×1000 Pa)

= (500 / 39.948) (8.314) (273.15 + 40) / (1400×1000)

= 0.023276 m3

= 23.276 L