An ideal gas has pressure `p_(0)` in container of volume `V_(0)`.The number of molecules of the gas is doubled without changing the temperature. What is the new value of pressure times volume?

Options

A. `(p_(0)V_(0))/4`
B. `(p_(0)V_(0))/2`
C. `p_(0)V_(0)`
D. `2p_(0)V_(0)`

Explanation

To solve the problem, we will use the ideal gas law, which states that:

\[ PV = nRT \]

where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles of gas,
- \( R \) is the universal gas constant,
- \( T \) is the temperature.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The initial pressure is \( P_0 \).
- The initial volume is \( V_0 \).
- The initial number of moles is \( n \).
- The initial temperature is \( T \).

From the ideal gas law, we have:
\[ P_0 V_0 = nRT \]

2. **Doubling the Number of Molecules**:
- If the number of molecules is doubled, the new number of moles \( n' \) becomes:
\[ n' = 2n \]

3. **Temperature Remains Constant**:
- The temperature remains constant, so \( T' = T \).
- The gas constant \( R \) also remains unchanged, so \( R' = R \).

4. **New Pressure and Volume**:
- We want to find the new value of pressure times volume, \( P'V' \).
- According to the ideal gas law for the new conditions, we have:
\[ P'V' = n'RT' \]
Substituting the new values:
\[ P'V' = (2n)R(T) = 2(nRT) \]

5. **Relate to Initial Conditions**:
- Since \( nRT = P_0 V_0 \) from the initial conditions, we can substitute:
\[ P'V' = 2(P_0 V_0) \]

6. **Final Result**:
- Therefore, the new value of pressure times volume is:
\[ P'V' = 2P_0 V_0 \]

### Conclusion:
The new value of pressure times volume is twice the initial value, which can be expressed as:
\[ P'V' = 2P_0 V_0 \]