If the temperature of an ideal gas in a sealed, rigid container is increased to 1.5 times the initial value (in K), the density of gas

Options

A. becomes 1.5 times the initial value
B. becomes 2.5 times the initial value
C. becomes 2.25 times the initial value
D. remains same

Answer

To solve the problem, we need to analyze how the density of an ideal gas changes when its temperature is increased while keeping the volume constant.

### Step-by-Step Solution:

1. **Understanding Density**:
Density (\( \rho \)) is defined as the mass (\( m \)) of the gas divided by its volume (\( V \)):
\[ \rho = \frac{m}{V} \]

2. **Conditions Given**:
- The gas is in a sealed, rigid container, which means the volume (\( V \)) does not change.
- The temperature of the gas is increased to 1.5 times its initial value (\( T \rightarrow 1.5T \)).

3. **Ideal Gas Law**:
The ideal gas law states:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

4. **Constant Mass and Volume**:
Since the container is sealed, the mass of the gas remains constant, and since it is rigid, the volume also remains constant. Therefore, the number of moles (\( n \)) remains constant.

5. **Effect of Temperature on Pressure**:
When the temperature increases, according to the ideal gas law, if the volume is constant, the pressure must increase. However, this does not affect the mass or volume of the gas.

6. **Density Calculation**:
Since both the mass and volume of the gas remain unchanged, the density will also remain unchanged:
\[ \text{New Density} = \frac{m}{V} = \text{Initial Density} \]

7. **Conclusion**:
The density of the gas will remain the same even after the temperature is increased.

### Final Answer:
The density of the gas remains unchanged.

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