The volume of a gas increases by a factor of 2 while the pressure decrease by a factor of 3 Given that the number of moles is unaffected, the factor by which the temperature changes is `:`

Options

A. `(3)/(2)`
B. `3 xx 2`
C. `(2)/(3)`
D. `(1)/(2) xx 3`

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
- \( R \) = ideal gas constant
- \( T \) = temperature

### Step-by-Step Solution:

1. **Identify the Initial and Final Conditions:**
- Let \( P_1 \) be the initial pressure.
- Let \( V_1 \) be the initial volume.
- Let \( T_1 \) be the initial temperature.
- According to the problem:
- The volume increases by a factor of 2: \( V_2 = 2V_1 \)
- The pressure decreases by a factor of 3: \( P_2 = \frac{1}{3} P_1 \)
- The number of moles \( n \) remains constant.

2. **Set Up the Ideal Gas Law for Initial and Final States:**
- For the initial state:
\[ P_1 V_1 = nRT_1 \]
- For the final state:
\[ P_2 V_2 = nRT_2 \]

3. **Substituting the Final Conditions into the Ideal Gas Law:**
- Substitute \( P_2 \) and \( V_2 \) into the final state equation:
\[ \left(\frac{1}{3} P_1\right) (2V_1) = nRT_2 \]

4. **Simplifying the Equation:**
- This simplifies to:
\[ \frac{2}{3} P_1 V_1 = nRT_2 \]

5. **Relate the Two States Using the Ideal Gas Law:**
- From the initial state, we have:
\[ P_1 V_1 = nRT_1 \]
- Now we can relate the two equations:
\[ \frac{2}{3} P_1 V_1 = nRT_2 \]
- Dividing both sides by \( P_1 V_1 \):
\[ \frac{2}{3} = \frac{nRT_2}{P_1 V_1} \]

6. **Substituting for \( nRT_1 \):**
- Since \( P_1 V_1 = nRT_1 \), we can substitute:
\[ \frac{2}{3} = \frac{T_2}{T_1} \]

7. **Finding the Factor of Temperature Change:**
- Rearranging gives:
\[ T_2 = \frac{2}{3} T_1 \]

### Conclusion:
The factor by which the temperature changes is \( \frac{2}{3} \).