`2` mole ideal He gas and `3` mole ideal `H_(2)` gas at constant volume find out `C_(v)` of mixture

Options

A. `(21R)/(10)`
B. `(11R)/(10)`
C. `(21R)/(5)`
D. `(11R)/(5)`

Answer

To find the heat capacity at constant volume (\(C_v\)) of a mixture of gases, we can use the following steps:

### Step 1: Identify the moles of each gas
We have:
- \(n_{\text{He}} = 2\) moles of Helium gas
- \(n_{\text{H}_2} = 3\) moles of Hydrogen gas

### Step 2: Determine the \(C_v\) values for each gas
- For Helium (a monatomic gas), the heat capacity at constant volume is given by:
\[ C_{v, \text{He}} = \frac{3}{2}R \]
- For Hydrogen (a diatomic gas), the heat capacity at constant volume is given by:
\[ C_{v, \text{H}_2} = \frac{5}{2}R \]

### Step 3: Calculate the total \(C_v\) for the mixture
The formula for the heat capacity at constant volume of the mixture is:
\[ C_{v, \text{mixture}} = \frac{n_{\text{He}} \cdot C_{v, \text{He}} + n_{\text{H}_2} \cdot C_{v, \text{H}_2}}{n_{\text{He}} + n_{\text{H}_2}} \]

Substituting the values:
\[ C_{v, \text{mixture}} = \frac{2 \cdot \frac{3}{2}R + 3 \cdot \frac{5}{2}R}{2 + 3} \]

### Step 4: Simplify the equation
Calculating the numerator:
\[ = \frac{2 \cdot \frac{3}{2}R + 3 \cdot \frac{5}{2}R}{5} = \frac{3R + \frac{15}{2}R}{5} = \frac{3R + 7.5R}{5} = \frac{10.5R}{5} = \frac{21R}{10} \]

### Step 5: Final result
Thus, the heat capacity at constant volume of the mixture is:
\[ C_{v, \text{mixture}} = \frac{21R}{10} \]

### Answer
The final answer is:
\[ C_{v, \text{mixture}} = \frac{21R}{10} \]

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