Two moles of an ideal gas with `(C_(P))/(C_(V))= (5)/(3)` are mixed with 3 moles of another ideal gas with `(C_(P))/(C_(V))= (4)/(3)`. The value of `(C_(P))/(C_(V))` for the mixture is

Options

A. 1.45
B. `1.50`
C. 1.47
D. 1.42

Answer

To find the value of \((C_{P})/(C_{V})\) for the mixture of two ideal gases, we can follow these steps:

### Step 1: Identify the given values
We have two gases:
- Gas 1: \(C_{P}/C_{V} = 5/3\) with \(n_1 = 2\) moles
- Gas 2: \(C_{P}/C_{V} = 4/3\) with \(n_2 = 3\) moles

### Step 2: Calculate \(C_{V}\) for each gas
Using the relation:
\[ \frac{C_{P}}{C_{V}} = 1 + \frac{R}{C_{V}} \]
For Gas 1:
\[ 1 + \frac{R}{C_{V1}} = \frac{5}{3} \]
\[ \frac{R}{C_{V1}} = \frac{5}{3} - 1 = \frac{2}{3} \]
Thus,
\[ C_{V1} = \frac{3R}{2} \]

For Gas 2:
\[ 1 + \frac{R}{C_{V2}} = \frac{4}{3} \]
\[ \frac{R}{C_{V2}} = \frac{4}{3} - 1 = \frac{1}{3} \]
Thus,
\[ C_{V2} = 3R \]

### Step 3: Calculate the total \(C_{V}\) for the mixture
The total \(C_{V}\) for the mixture can be calculated using the formula:
\[ C_{V \text{ mixture}} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2} \]
Substituting the values:
\[ C_{V \text{ mixture}} = \frac{2 \cdot \frac{3R}{2} + 3 \cdot 3R}{2 + 3} \]
Calculating this:
\[ = \frac{3R + 9R}{5} = \frac{12R}{5} \]

### Step 4: Calculate \((C_{P})/(C_{V})\) for the mixture
Using the relation:
\[ \frac{C_{P}}{C_{V \text{ mixture}}} = 1 + \frac{R}{C_{V \text{ mixture}}} \]
Substituting \(C_{V \text{ mixture}} = \frac{12R}{5}\):
\[ \frac{C_{P}}{C_{V \text{ mixture}}} = 1 + \frac{R}{\frac{12R}{5}} = 1 + \frac{5}{12} \]
Calculating this:
\[ = 1 + 0.4167 = 1.4167 \approx 1.42 \]

### Final Answer
Thus, the value of \((C_{P})/(C_{V})\) for the mixture is approximately \(1.42\).

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