The pressure of an ideal gas of diatomic molecules is doubled by halving the volume. The ratio of the new internal energy to the old, both measured relative to the internal energy at 0 K, is

Options

A. `1//4`
B. `1//2`
C. `1`
D. `2`

Answer

To solve the problem, we need to analyze the relationship between the internal energy of an ideal gas and its state variables, specifically when the pressure is doubled and the volume is halved.

### Step-by-Step Solution:

1. **Understanding Internal Energy of an Ideal Gas**:
The internal energy \( U \) of an ideal gas is given by the formula:
\[ U = \frac{F}{2} nRT \]
where \( F \) is the degrees of freedom, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature.

2. **Identifying the Initial State**:
Let the initial pressure be \( P \), the initial volume be \( V \), and the initial temperature be \( T_1 \). The initial internal energy \( U_1 \) can be expressed as:
\[ U_1 = \frac{F}{2} nRT_1 \]

3. **Identifying the New State**:
According to the problem, the pressure is doubled and the volume is halved. Therefore, the new pressure \( P' \) and new volume \( V' \) are:
\[ P' = 2P \quad \text{and} \quad V' = \frac{V}{2} \]

4. **Using the Ideal Gas Law**:
The ideal gas law states:
\[ PV = nRT \]
For the initial state:
\[ P \cdot V = nRT_1 \]
For the new state:
\[ P' \cdot V' = nRT_2 \]
Substituting the new pressure and volume:
\[ 2P \cdot \frac{V}{2} = nRT_2 \]
This simplifies to:
\[ PV = nRT_2 \]
Therefore, we have:
\[ T_2 = T_1 \]

5. **Calculating the New Internal Energy**:
Since the temperature remains the same, the new internal energy \( U_2 \) is:
\[ U_2 = \frac{F}{2} nRT_2 = \frac{F}{2} nRT_1 = U_1 \]

6. **Finding the Ratio of New Internal Energy to Old Internal Energy**:
The ratio of the new internal energy to the old internal energy is:
\[ \frac{U_2}{U_1} = \frac{U_1}{U_1} = 1 \]

### Final Answer:
The ratio of the new internal energy to the old internal energy is:
\[ \boxed{1} \]