Question

When one mole of monatomic gas is mixed with one mole of a diatomic gas, then the equivalent value of $\lambda$ for the mixture will be (vibration mode neglected)

• A. $1.33$
• B. $1.40$
• C. $1.50$
• D. $1.6$

Answer 1

The correct option is C $1.50$

For monoatomic gas

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

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

For diatomic gas

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

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

After mixing new $C_v\&C_P$ are

$C_v=\frac{\frac{3}{2}RT+\frac{5}{2}RT}{2}=2RT$

$C_p=\frac{\frac{5}{2}RT+\frac{7}{2}RT}{2}=3RT$

$\lambda =\frac{C_p}{C_v}=\frac{3RT}{2RT}=1.5$