`4` moles of `H_(2) at 500K` is mixed with `2` moles of `He` at `400k`. The mixture attains a temperature `T` and volume `V`. Now the mixture is compressed adiabatically to a volume `V'` and temperature `T'`. If `(T')/(T) = ((V)/(V'))^(n)`, find the value of `13n`.

Options

A. `4`
B. `6`
C. `5`
D. `13`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Determine the specific heat capacities (Cv) for each gas.
- For hydrogen (H₂), which is a diatomic gas, the specific heat capacity at constant volume (Cv) is given by:
\[ C_{v,\text{H}_2} = \frac{5R}{2} \]
- For helium (He), which is a monatomic gas, the specific heat capacity at constant volume (Cv) is given by:
\[ C_{v,\text{He}} = \frac{3R}{2} \]

### Step 2: Calculate the total Cv for the mixture.
- The total number of moles in the mixture is:
\[ n_{\text{total}} = n_{\text{H}_2} + n_{\text{He}} = 4 + 2 = 6 \]
- The total Cv for the mixture can be calculated as:
\[ C_{v,\text{mixture}} = \frac{n_{\text{H}_2} \cdot C_{v,\text{H}_2} + n_{\text{He}} \cdot C_{v,\text{He}}}{n_{\text{total}}} \]
\[ C_{v,\text{mixture}} = \frac{4 \cdot \frac{5R}{2} + 2 \cdot \frac{3R}{2}}{6} \]
\[ = \frac{10R + 3R}{6} = \frac{13R}{6} \]

### Step 3: Find the value of γ for the mixture.
- The value of γ (gamma) is related to Cv by:
\[ \gamma = \frac{C_p}{C_v} \]
- Since \( C_p = C_v + R \), we have:
\[ C_{p,\text{mixture}} = C_{v,\text{mixture}} + R = \frac{13R}{6} + R = \frac{13R}{6} + \frac{6R}{6} = \frac{19R}{6} \]
- Thus, we can find γ:
\[ \gamma_{\text{mixture}} = \frac{C_{p,\text{mixture}}}{C_{v,\text{mixture}}} = \frac{\frac{19R}{6}}{\frac{13R}{6}} = \frac{19}{13} \]

### Step 4: Use the adiabatic relation.
- The adiabatic relation is given by:
\[ T V^{\gamma - 1} = \text{constant} \]
- For initial and final states, we can write:
\[ T V^{\gamma - 1} = T' V'^{\gamma - 1} \]
- This leads to:
\[ \frac{T'}{T} = \left(\frac{V}{V'}\right)^{\gamma - 1} \]
- Substituting \( \gamma - 1 = \frac{19}{13} - 1 = \frac{6}{13} \):
\[ \frac{T'}{T} = \left(\frac{V}{V'}\right)^{\frac{6}{13}} \]

### Step 5: Compare with the given relation.
- The problem states:
\[ \frac{T'}{T} = \left(\frac{V}{V'}\right)^{n} \]
- By comparing both equations, we find:
\[ n = \frac{6}{13} \]

### Step 6: Calculate \( 13n \).
- Now, we can find \( 13n \):
\[ 13n = 13 \cdot \frac{6}{13} = 6 \]

### Final Answer:
Thus, the value of \( 13n \) is \( \boxed{6} \).

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