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

### Step-by-Step Solution:

1. **Initial State**:
Let's denote the initial pressure, volume, and temperature as \( P_1 \), \( V_1 \), and \( T_1 \) respectively. According to the Ideal Gas Law:
\[ P_1 V_1 = nRT_1 \]

2. **Final State**:
According to the problem, both the temperature and volume of the gas are doubled. Therefore, the final volume \( V_2 \) and final temperature \( T_2 \) can be expressed as:
\[ V_2 = 2V_1 \]
\[ T_2 = 2T_1 \]

3. **Applying Ideal Gas Law to Final State**:
For the final state, we can write the Ideal Gas Law as:
\[ P_2 V_2 = nRT_2 \]
Substituting the expressions for \( V_2 \) and \( T_2 \):
\[ P_2 (2V_1) = nR(2T_1) \]

4. **Simplifying the Equation**:
We can simplify this equation:
\[ 2P_2 V_1 = 2nRT_1 \]
Dividing both sides by 2:
\[ P_2 V_1 = nRT_1 \]

5. **Relating Final and Initial Pressures**:
From the initial state, we know:
\[ P_1 V_1 = nRT_1 \]
Therefore, we can equate:
\[ P_2 V_1 = P_1 V_1 \]
Since \( V_1 \) is not zero, we can divide both sides by \( V_1 \):
\[ P_2 = P_1 \]

6. **Conclusion**:
This shows that the final pressure \( P_2 \) remains equal to the initial pressure \( P_1 \). Thus, the pressure of the gas remains unchanged when both the temperature and volume are doubled.

### Final Answer:
The pressure remains unchanged.

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