Question

If one mole of a monoatomic. gas is mixed with one mole of a triatomic gas the value of $\gamma$ for the mixture is

• A. 1.40
• B. 1.44
• C. 1.53
• D. 3.07

Answer 1

Answer: (B)

For monoatomic gas $C_{V_1}=\frac{3R}{2}$
For triatomic gas $C_{V_2}=3R$
For the mixture of 1 mole of monoatomic gas and 1 mole of triatomic gas,
$C_V=\frac{1\left(\frac{3R}{2}\right)+1(3R)}{2}=\frac{9R}{4}$
$C_P=\frac{9R}{4}+R=\frac{13R}{4}$
$\Rightarrow {\gamma }_{\text{mix}}=\frac{C_P}{C_V}=\frac{13}{9}=1.44$