If both the temperature and the volume of an ideal gas are doubled, the pressure

Options

A. Increases by a factor of 4
B. Is also doubled
C. Remains unchanged
D. Is diminished by a factor of 4

Answer

To solve the problem, we will use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

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

### Step-by-Step Solution:

1. **Initial Conditions**:
Let's denote the initial pressure, volume, and temperature as \( P \), \( V \), and \( T \) respectively. According to the Ideal Gas Law, we can write:
\[ PV = nRT \]

2. **New Conditions**:
If both the temperature and volume are doubled, the new volume \( V' \) and new temperature \( T' \) can be expressed as:
\[ V' = 2V \]
\[ T' = 2T \]

3. **Applying the Ideal Gas Law to New Conditions**:
The new pressure \( P' \) can be expressed using the Ideal Gas Law as follows:
\[ P'V' = nRT' \]

4. **Substituting New Values**:
Substitute \( V' \) and \( T' \) into the equation:
\[ P'(2V) = nR(2T) \]
Simplifying this gives:
\[ 2P'V = 2nRT \]

5. **Cancelling Common Factors**:
We can divide both sides of the equation by 2:
\[ P'V = nRT \]

6. **Comparing with Initial Conditions**:
From the initial conditions, we know that \( PV = nRT \). Thus, we can equate:
\[ P'V = PV \]

7. **Solving for New Pressure**:
Since the left-hand side and right-hand side are equal, we can conclude:
\[ P' = P \]

### Conclusion:
The pressure remains unchanged when both the temperature and the volume of an ideal gas are doubled. Therefore, the answer is:
\[ \text{Pressure remains the same (P' = P).} \]