Question

One mole each of a monoatomic, diatomic and triatomic gases are mixed, $C_p/C_v$ for the mixture is

• A. $1.40$
• B. $1.428$
• C. $1.67$
• D. none of these

Answer 1

The correct option is B $1.428$

As we know

$C_v$

for monoatomic gas is $3R/2$

for diatomic gas is $5R/2$

for triatomic gas is $7R/2$

so for their mixture

$C_v=\frac{[\frac{3}{2}R+\frac{5}{2}R+\frac{6}{2}R]}{3}=\frac{7}{3}$

As $C_p-C_v=R$

$C_p=\frac{[\frac{5}{2}R+\frac{7}{2}R+\frac{8}{2}R]}{3}=\frac{10}{3}$

$\therefore \frac{C_p}{C_v}=\frac{10}{7}=1.428$