For an ideal gas, `Cv//Cp` is

Options

A. `lt1`
B. `gt1`
C. `=1`
D. `ge1`

Answer

To solve the question regarding the ratio \( \frac{C_v}{C_p} \) for an ideal gas, we can follow these steps:

### Step 1: Understand the definitions
- \( C_v \) is the molar specific heat at constant volume.
- \( C_p \) is the molar specific heat at constant pressure.

### Step 2: Write the formulas for \( C_v \) and \( C_p \)
For an ideal gas, the formulas are:
- \( C_v = \frac{F}{2} R \)
- \( C_p = \frac{F}{2} R + R \)

Where \( F \) is the degrees of freedom of the gas and \( R \) is the gas constant.

### Step 3: Substitute the formulas into the ratio
Now, we can express the ratio \( \frac{C_p}{C_v} \):
\[ \frac{C_p}{C_v} = \frac{\frac{F}{2} R + R}{\frac{F}{2} R} \]

### Step 4: Simplify the expression
This can be simplified as follows:
\[ \frac{C_p}{C_v} = \frac{\frac{F}{2} R + R}{\frac{F}{2} R} = \frac{\frac{F + 2}{2}}{\frac{F}{2}} = \frac{F + 2}{F} \]
Thus, we have:
\[ \frac{C_p}{C_v} = 1 + \frac{2}{F} \]

### Step 5: Analyze the degrees of freedom
The degrees of freedom \( F \) varies depending on the type of gas:
- For a monatomic gas, \( F = 3 \)
- For a diatomic gas, \( F = 5 \)
- For a triatomic gas, \( F = 6 \)

### Step 6: Conclude the relationship
Since \( F \) is always positive, \( \frac{C_p}{C_v} \) is always greater than 1. Therefore, the reciprocal \( \frac{C_v}{C_p} \) must be less than 1:
\[ \frac{C_v}{C_p} < 1 \]

### Final Answer
Thus, for an ideal gas, \( \frac{C_v}{C_p} \) is less than 1.

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