5 moles of a gas in a closed vessel was heated from 300 K to 600 K. The pressure of the gas doubled. The number of moles of the gas will be

Options

A. 5
B. 2.5
C. 10
D. 20

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

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Initial number of moles (\( n_1 \)) = 5 moles
- Initial temperature (\( T_1 \)) = 300 K
- Final temperature (\( T_2 \)) = 600 K
- Initial pressure (\( P_1 \)) = P (unknown)
- Final pressure (\( P_2 \)) = 2P (since the pressure doubles)

2. **Write the Ideal Gas Law for Initial and Final States**:
- For the initial state:
\[ P_1 V = n_1 R T_1 \]
- For the final state:
\[ P_2 V = n_2 R T_2 \]

3. **Substitute Known Values**:
- Substitute \( n_1 = 5 \) moles, \( T_1 = 300 \) K, \( T_2 = 600 \) K, and \( P_2 = 2P_1 \):
- Initial equation:
\[ P_1 V = 5 R (300) \]
- Final equation:
\[ 2P_1 V = n_2 R (600) \]

4. **Divide the Two Equations**:
- Dividing the final equation by the initial equation:
\[ \frac{2P_1 V}{P_1 V} = \frac{n_2 R (600)}{5 R (300)} \]
- Simplifying gives:
\[ 2 = \frac{n_2 \cdot 600}{5 \cdot 300} \]

5. **Simplify the Right Side**:
- The right side simplifies to:
\[ 2 = \frac{n_2 \cdot 600}{1500} \]
- This further simplifies to:
\[ 2 = \frac{n_2}{2.5} \]

6. **Solve for \( n_2 \)**:
- Rearranging gives:
\[ n_2 = 2 \cdot 2.5 = 5 \]

7. **Conclusion**:
- The number of moles of the gas remains the same, which is \( n_2 = 5 \) moles.

### Final Answer:
The number of moles of the gas will be **5 moles**.