If pressure and temperature of an ideal gas are doubled and volume is halved, the number of molecules of the gas

Options

A. A) remain constant
B. B) become half
C. C) become two times
D. D) become four times

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 (or number of molecules)
- \( R \) = Ideal gas constant
- \( T \) = Temperature

### Step-by-Step Solution:

1. **Write the Initial Ideal Gas Equation:**
\[ P_1 V_1 = n_1 R T_1 \]
Here, \( P_1 \), \( V_1 \), \( n_1 \), and \( T_1 \) are the initial pressure, volume, number of molecules, and temperature respectively.

2. **Apply the Changes to Pressure, Volume, and Temperature:**
According to the problem:
- Pressure is doubled: \( P_2 = 2P_1 \)
- Volume is halved: \( V_2 = \frac{1}{2} V_1 \)
- Temperature is doubled: \( T_2 = 2T_1 \)

3. **Write the New Ideal Gas Equation:**
\[ P_2 V_2 = n_2 R T_2 \]
Substituting the changes:
\[ (2P_1) \left(\frac{1}{2} V_1\right) = n_2 R (2T_1) \]

4. **Simplify the New Equation:**
\[ P_1 V_1 = n_2 R (2T_1) \]
Now, we can equate this with the initial equation:
\[ n_1 R T_1 = n_2 R (2T_1) \]

5. **Cancel Out Common Terms:**
Since \( R \) and \( T_1 \) are common in both sides, we can cancel them:
\[ n_1 = 2n_2 \]

6. **Solve for the New Number of Molecules:**
Rearranging gives:
\[ n_2 = \frac{n_1}{2} \]
This means the number of molecules \( n_2 \) becomes half of the initial number of molecules \( n_1 \).

### Conclusion:
The number of molecules of the gas becomes half when the pressure and temperature are doubled, and the volume is halved. Therefore, the correct answer is option B: the number of molecules becomes half.

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