Question

If x moles of a monoatomic are mixed with y moles of
non-linear triatomic gas and mixture behaves like
diatomic gas, then (neglect vibrational modes)
x=2y
2x=y
3X=y
X=3y​

Answer 1

Answer:

x=3y+x=2y+3x=y+y+=8 it is a correct answer

Answer 2

Given info : x moles of a monoatomic gas are mixed with y moles of non - linear triatomic gas and mixture behaves like diatomic gas.

To find : the relation between x and y is ..

solution :

for monoatomic gas, Cv = 3R/2

for diatomic gas, Cv = 5R/2

for non linear triatomic gas, Cv = 6R/2

Cv of mixture = (n₁C₁ + n₂C₂)/(n₁ + n₂)

here n₁ = x , n₂ = y , C₁ = 3R/2 , C₂ = 6R/2

so, Cv of mixture = (x × 3R/2 + y × 6R/2)/(x + y)

but Cv of mixture = 5R/2 [ ∵ mixture behaves like diatomic gas ]

so, 5R/2 = (x × 3R/2 + y × 6R/2)/(x + y)

⇒5 = (3x + 6y)/(x + y)

⇒5x + 5y = 3x + 6y

⇒2x = y

Therefore the relation between x and y is 2x = y.