If the pressure and the volume of certain quantity of ideal gas are halved, then its temperature

Options

A. is doubled
B. becomes one-fourth
C. remains constant
D. become four times

Explanation

To solve the problem, we will use the ideal gas law, which states that:

\[ PV = nRT \]

Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles of gas (constant in this case)
- \( R \) = Universal gas constant
- \( T \) = Temperature

Given that both the pressure and volume of a certain quantity of ideal gas are halved, we can denote the initial conditions as \( P_1 \), \( V_1 \), and \( T_1 \), and the final conditions as \( P_2 \), \( V_2 \), and \( T_2 \).

### Step-by-Step Solution:

1. **Initial and Final Conditions**:
- Initial Pressure: \( P_1 \)
- Initial Volume: \( V_1 \)
- Initial Temperature: \( T_1 \)
- Final Pressure: \( P_2 = \frac{P_1}{2} \)
- Final Volume: \( V_2 = \frac{V_1}{2} \)

2. **Using the Ideal Gas Law**:
According to the ideal gas law, we can write the equations for the initial and final states:
\[ P_1 V_1 = n R T_1 \quad \text{(1)} \]
\[ P_2 V_2 = n R T_2 \quad \text{(2)} \]

3. **Substituting Final Conditions into the Equation**:
Substitute \( P_2 \) and \( V_2 \) into equation (2):
\[ \left(\frac{P_1}{2}\right) \left(\frac{V_1}{2}\right) = n R T_2 \]

4. **Simplifying the Equation**:
This simplifies to:
\[ \frac{P_1 V_1}{4} = n R T_2 \quad \text{(3)} \]

5. **Relating Equations (1) and (3)**:
From equation (1), we know:
\[ P_1 V_1 = n R T_1 \]
Now, substitute this into equation (3):
\[ \frac{n R T_1}{4} = n R T_2 \]

6. **Canceling Common Terms**:
Since \( n R \) is common in both sides, we can cancel it out:
\[ \frac{T_1}{4} = T_2 \]

7. **Final Result**:
Rearranging gives us:
\[ T_2 = \frac{T_1}{4} \]

Thus, if the pressure and volume of a certain quantity of ideal gas are halved, the temperature will be one-fourth of the initial temperature.

### Final Answer:
The temperature \( T_2 \) is \( \frac{T_1}{4} \).