Three moles of ideal gas A with `(C_(p))/(C_(v))=(4)/(3)` is mixed with two moles of another ideal gas B with `(C _(P))/(C_(v))=(5)/(3)` The `(C_(P))/(C_(v))` of mixture is (Assuming temperature is constant)

Options

A. `1.5`
B. `1.42`
C. `1.7`
D. `1.3`

Answer

To find the \( \frac{C_P}{C_V} \) (gamma) of the mixture of two ideal gases A and B, we can follow these steps:

### Step 1: Identify the given values
- For gas A:
- Number of moles, \( n_1 = 3 \)
- \( \frac{C_P}{C_V} = \gamma_1 = \frac{4}{3} \)

- For gas B:
- Number of moles, \( n_2 = 2 \)
- \( \frac{C_P}{C_V} = \gamma_2 = \frac{5}{3} \)

### Step 2: Use the formula for the \( \frac{C_P}{C_V} \) of the mixture
The formula for the \( \frac{C_P}{C_V} \) of the mixture is given by:

\[ \gamma_{mix} = \frac{n_1 \gamma_1 + n_2 \gamma_2}{n_1 + n_2} \]

### Step 3: Substitute the values into the formula
Substituting the values we have:

\[ \gamma_{mix} = \frac{3 \cdot \frac{4}{3} + 2 \cdot \frac{5}{3}}{3 + 2} \]

### Step 4: Calculate the numerator
Calculating the numerator:

\[ 3 \cdot \frac{4}{3} = 4 \]
\[ 2 \cdot \frac{5}{3} = \frac{10}{3} \]

Now adding these two results together:

\[ 4 + \frac{10}{3} = \frac{12}{3} + \frac{10}{3} = \frac{22}{3} \]

### Step 5: Calculate the denominator
The denominator is simply:

\[ n_1 + n_2 = 3 + 2 = 5 \]

### Step 6: Final calculation
Now we can substitute the values back into the equation:

\[ \gamma_{mix} = \frac{\frac{22}{3}}{5} = \frac{22}{15} \]

### Step 7: Simplify the result
Calculating \( \frac{22}{15} \):

\[ \frac{22}{15} \approx 1.4667 \]

### Conclusion
Thus, the \( \frac{C_P}{C_V} \) of the mixture is approximately \( 1.47 \).

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