Question

For a gas Cp/R= 3, then gas may be
(1) Monatomic
(2) Diatomic
(3) Polyatomic
(4) Mixture of monatomic and diatomic​

Answer 1

Answer:

Option (4) Mixture of monoatomic and diatomic

Explanation:

We know that

$C_p=C_v+R$

For monoatomic gas

$C_v=\frac{3R}{2}$

Thus, $C_p=\frac{3R}{2}+R$

$\implies C_p=\frac{5R}{2}$

And for diatomic gas

$C_v=\frac{5R}{2}$

Thus, $C_p=\frac{5R}{2}+R$

$\implies C_p=\frac{7R}{2}$

Since the given $C_p$ lies between the these two, the gas may be a mixture of monoatomic and diatomic gas.

Hope the answer is helpful.

Answer 2

Answer:

4

Explanation:

for a monatomic gas Cp=5R/2 ,Cv=3R/2

                                             Cp/Cv=1.67

for diatomic gas Cp/Cv=1.4

given Cp=3R

Cv=Cp-R=2R

Cp/Cv=3R/2R=1.5

this value lies between monatomic and diatomic gas