Question 242: Thermodynamics & KTG

Question

Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :

Accepted Answer

To calculate the molar specific heat at constant volume of a gas mixture, we can use a weighted average based on the molar specific heats of the individual gases and their respective amounts (moles).
Let's call $C_{V,m}$ the molar specific heat at constant volume of the mixture, $n_1$ the number of moles of the monoatomic gas, $n_2$ the number of moles of the diatomic gas, $C_{V,m1}$ the molar specific heat at constant volume of the monoatomic gas, and $C_{V,m2}$ the molar specific heat at constant volume of the diatomic gas.
For a monoatomic ideal gas, the molar specific heat at constant volume is:
$$ C_{V,m1} = \frac{3}{2}R $$
For a diatomic ideal gas, if we assume the gas is rigid and does not exhibit vibrational modes, the molar specific heat at constant volume is:
$$ C_{V,m2} = \frac{5}{2}R $$
The weighted average of the molar specific heat for the mixture is:
$$ C_{V,m} = \frac{(n_1 \cdot C_{V,m1} + n_2 \cdot C_{V,m2})}{n_1 + n_2} $$
Putting the values into the equation, we get:
$$ C_{V,m} = \frac{(2 \cdot \frac{3}{2}R + 6 \cdot \frac{5}{2}R)}{2 + 6} $$
$$ C_{V,m} = \frac{(3R + 15R)}{8} $$
$$ C_{V,m} = \frac{18R}{8} $$
$$ C_{V,m} = \frac{9}{4}R $$
Therefore, the molar specific heat of the mixture at constant volume is $\frac{9}{4}R$. The correct answer is:
Option D
$$ \frac{9}{4}R $$

Answer

$\frac{3}{2} \mathrm{R}$

Answer

$\frac{7}{4} \mathrm{R}$

Answer

$\frac{5}{2} \mathrm{R}$

Answer

$\frac{9}{4} \mathrm{R}$

Thermodynamics & KTG

Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :

Choose the correct answer: