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 of how the temperature of an ideal gas changes when both its pressure and volume are halved, we can use the ideal gas law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure of the gas
- \( V \) = volume of the gas
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant
- \( T \) = temperature of the gas (in Kelvin)

### Step-by-Step Solution:

1. **Initial Conditions**:
Let the initial pressure, volume, and temperature of the gas be \( P_1 \), \( V_1 \), and \( T_1 \) respectively. According to the ideal gas law, we have:
\[ P_1 V_1 = n R T_1 \]

2. **New Conditions**:
If the pressure and volume are both halved, the new pressure \( P_2 \) and volume \( V_2 \) can be expressed as:
\[ P_2 = \frac{P_1}{2}, \quad V_2 = \frac{V_1}{2} \]

3. **Applying the Ideal Gas Law Again**:
For the new conditions, we can write:
\[ P_2 V_2 = n R T_2 \]
Substituting the new values of pressure and volume:
\[ \left(\frac{P_1}{2}\right) \left(\frac{V_1}{2}\right) = n R T_2 \]
Simplifying this, we get:
\[ \frac{P_1 V_1}{4} = n R T_2 \]

4. **Relating the Two Equations**:
From the initial conditions, we know that \( P_1 V_1 = n R T_1 \). We can substitute this into our equation for the new conditions:
\[ \frac{n R T_1}{4} = n R T_2 \]

5. **Canceling Common Terms**:
Since \( n R \) is common on both sides, we can cancel it out (assuming \( n \) and \( R \) are not zero):
\[ \frac{T_1}{4} = T_2 \]

6. **Final Result**:
Thus, the new temperature \( T_2 \) is:
\[ T_2 = \frac{T_1}{4} \]

### Conclusion:
If the pressure and volume of a certain quantity of ideal gas are halved, then its temperature becomes one-fourth of the initial temperature.

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