A closed container of volume 0.02 m 3 contains a mixture of neon and argon gases, at a temperature of 27°C and pressure of 1 × 10 5 Nm –2 . The total mass of the mixture is 28 g. If the gram molecular weights of neon and argon are 20 and 40 respectively. Find the masses of the individual gases x and y in the container, assuming them to be ideal. (Universal gas constant R = 8.314 J/mole K) Give your answer as x + y.
Text Solution
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28 (m Ar = 24 + m Ne = 4)
PV = n Total RT
10 5 × 0.02 = n Total × 8.314 × 300
n Total = 0.8
+ 
= 0.8
m Ar + m Ne = 28
m Ar = 24 ; m Ne = 4
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