If the molar specific heat at constant pressure for a polyatomic non-linear gas is x and the molar specific heat at constant volume for a diatomic gas is y, find the value of `x//y`.

Explanation

To find the value of \( \frac{x}{y} \), where \( x \) is the molar specific heat at constant pressure for a polyatomic non-linear gas and \( y \) is the molar specific heat at constant volume for a diatomic gas, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between specific heats**:
The molar specific heat at constant pressure \( C_p \) is related to the molar specific heat at constant volume \( C_v \) by the equation:
\[ C_p = C_v + R \]
where \( R \) is the universal gas constant.

2. **Define variables**:
Let \( C_{v1} \) be the molar specific heat at constant volume for the polyatomic gas. Therefore, for the polyatomic gas:
\[ x = C_{v1} + R \]

3. **Determine \( C_v \) for the diatomic gas**:
For a diatomic gas, the molar specific heat at constant volume \( C_{v2} \) is given by:
\[ C_{v2} = \frac{F}{2}R \]
where \( F \) is the degrees of freedom. For a diatomic gas, \( F = 5 \), so:
\[ C_{v2} = \frac{5}{2}R = y \]

4. **Determine \( C_v \) for the polyatomic gas**:
For a polyatomic non-linear gas, the degrees of freedom \( F_1 \) is typically 6. Thus:
\[ C_{v1} = \frac{F_1}{2}R = \frac{6}{2}R = 3R \]

5. **Substitute back to find \( x \)**:
Now substituting \( C_{v1} \) into the equation for \( x \):
\[ x = C_{v1} + R = 3R + R = 4R \]

6. **Calculate the ratio \( \frac{x}{y} \)**:
Now we can find the ratio:
\[ \frac{x}{y} = \frac{4R}{\frac{5}{2}R} = \frac{4R \cdot 2}{5R} = \frac{8}{5} \]

### Final Answer:
\[ \frac{x}{y} = \frac{8}{5} \]