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. `3xx2`
C. `(2)/(3)`
D. `(1)/(2)xx3`

Explanation

To solve the problem, we will use the Ideal Gas Law, which states:

\[ PV = nRT \]

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

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Let the initial pressure be \( P_1 \).
- Let the initial volume be \( V_1 \).
- Let the initial temperature be \( T_1 \).
- The number of moles \( n \) and the gas constant \( R \) remain constant.

From the Ideal Gas Law for the initial state:
\[ P_1 V_1 = nRT_1 \quad \text{(Equation 1)} \]

2. **Determine the New Conditions:**
- The volume of the gas increases by a factor of 2:
\[ V_2 = 2V_1 \]
- The pressure decreases by a factor of 3:
\[ P_2 = \frac{P_1}{3} \]

3. **Write the Equation for the New Conditions:**
Using the Ideal Gas Law for the new state:
\[ P_2 V_2 = nRT_2 \quad \text{(Equation 2)} \]

4. **Substitute the New Values into Equation 2:**
Substitute \( P_2 \) and \( V_2 \) into Equation 2:
\[ \left(\frac{P_1}{3}\right)(2V_1) = nRT_2 \]
Simplifying this gives:
\[ \frac{2P_1 V_1}{3} = nRT_2 \]

5. **Relate the Two Equations:**
Now, we can relate Equation 1 and the modified Equation 2:
From Equation 1:
\[ P_1 V_1 = nRT_1 \]
Substitute this into the modified Equation 2:
\[ \frac{2}{3} (nRT_1) = nRT_2 \]

6. **Cancel Out Common Terms:**
Since \( nR \) is common in both sides, we can cancel it out:
\[ \frac{2}{3} T_1 = T_2 \]

7. **Find the Factor by Which Temperature Changes:**
Rearranging gives:
\[ T_2 = \frac{2}{3} T_1 \]
Thus, the factor by which the temperature changes is:
\[ \frac{T_2}{T_1} = \frac{2}{3} \]

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