Question

When one mole of monoatomic gas is mixed with one mole of a diatomic gas, Then the equivalent value of $\gamma$ 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 gases :

$C_V=\frac{3}{2}R,C_P=\frac{5}{2}R$

For diatomic gases :

$C_V=\frac{5}{2}R$ $C_P=\frac{7}{2}R$

then, for the mixture of both gases

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

$C_V=\frac{8}{4}RT$

$C_V=2RT$

$C_P=\frac{\frac{5}{2}RT+\frac{7}{2}}{2}RT$

$C_P=\frac{12}{4}RT$

$C_P=3RT$

finally we know

$r=\frac{C_P}{C_V}=\frac{3RT}{2RT}=1.5$