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 solve the problem, we need to find the ratio \( \frac{x}{y} \), 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.

### Step 1: Determine \( x \) for the polyatomic non-linear gas

The molar specific heat at constant pressure \( C_p \) for a polyatomic non-linear gas can be calculated using the formula:

\[ C_p = \frac{F}{2} R + R \]

where \( F \) is the degree of freedom of the gas and \( R \) is the universal gas constant.

For a polyatomic non-linear gas, the degree of freedom \( F \) is typically 6. Therefore, we can substitute \( F = 6 \) into the equation:

\[ C_p = \frac{6}{2} R + R = 3R + R = 4R \]

Thus, we have:

\[ x = 4R \]

### Step 2: Determine \( y \) for the diatomic gas

The molar specific heat at constant volume \( C_v \) for a diatomic gas can be calculated using the formula:

\[ C_v = \frac{F}{2} R \]

For a diatomic gas, the degree of freedom \( F \) is 5. Therefore, we can substitute \( F = 5 \) into the equation:

\[ C_v = \frac{5}{2} R \]

Thus, we have:

\[ y = \frac{5}{2} R \]

### Step 3: Calculate the ratio \( \frac{x}{y} \)

Now we can find the ratio \( \frac{x}{y} \):

\[ \frac{x}{y} = \frac{4R}{\frac{5}{2} R} \]

When we simplify this expression, we can cancel \( R \):

\[ \frac{x}{y} = \frac{4}{\frac{5}{2}} = 4 \cdot \frac{2}{5} = \frac{8}{5} \]

### Step 4: Final Answer

Thus, the value of \( \frac{x}{y} \) is:

\[ \frac{x}{y} = \frac{8}{5} = 1.6 \]