A closed container of volume 0.02m^(3)contains a mixture of neon and argon gases at a temperature of 27^(0) C and pressure of 1 xx 10^(5) N//m^(2)The total mass of the mixture is 28gm. If the gram molecular weights of neon and argon are 20 and 40 respectively, find the masses of the individual gases in the container, assuming them to be ideal.

A closed container of volume 0.02m^(3)contains a mixture of neon and a

1.	A closed container of volume 0.02m^(3)contains a mixture of neon and argon gases at a temperature of 27^(0) C and pressure of 1 xx 10^(5) N//m^(2)The total mass of the mixture is 28gm. If the gram molecular weights of neon and argon are 20 and 40 respectively, find the masses of the individual gases in the container, assuming them to be ideal.
Answer» SOLUTION :If the mass of neon is .m., the mass of argon wiU be (28- m) , so `n_(Ne) = (m)/(20) and n_(Ar) = ((28 - m))/(40)` `therefore n = n_(Ne) + n_(Ar) = (m)/(20) + ((28 - m))/(40) = (28 + m)/(40)` ..... (1) n = `(PV)/(RT) = (1 xx 10^(5) xx 0.02 )/(8.314 xx 300) = 0.8 `... (2) So from EQUATION (1) and (2), `(28 + m )`40 = 0.8 , i.e., m = 4 gm so `m_(Ne) = `4 gm and `m_(Ar)` = 28 - 4 = 24 gm