The temperature of a gas in a closed container is `27^(@)`C. If the temperature is raised to `327^(@)`C the pressure exerted is :

Options

A. reduced to half
B. doubled
C. reduced to one-third
D. cannot be calculated from the given information.

Explanation

To solve the problem, we will use the ideal gas law, which states that for a given amount of gas at constant volume, the pressure of the gas is directly proportional to its absolute temperature. This can be expressed mathematically as:

\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

Where:
- \( P_1 \) is the initial pressure,
- \( T_1 \) is the initial temperature in Kelvin,
- \( P_2 \) is the final pressure,
- \( T_2 \) is the final temperature in Kelvin.

### Step 1: Convert the temperatures from Celsius to Kelvin
- The initial temperature \( T_1 \) is given as \( 27^\circ C \).
- To convert to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]

- The final temperature \( T_2 \) is given as \( 327^\circ C \).
- To convert to Kelvin:
\[ T_2 = 327 + 273 = 600 \, K \]

### Step 2: Set up the equation using the ideal gas law
Using the relationship \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), we can rearrange it to find \( P_2 \):
\[ P_2 = P_1 \cdot \frac{T_2}{T_1} \]

### Step 3: Substitute the known values
Assuming the initial pressure \( P_1 \) is some arbitrary value (let's say \( P_1 = P \)), we can substitute the values for \( T_1 \) and \( T_2 \):
\[ P_2 = P \cdot \frac{600}{300} \]

### Step 4: Simplify the equation
\[ P_2 = P \cdot 2 \]

### Conclusion
The final pressure \( P_2 \) is double the initial pressure \( P_1 \):
\[ P_2 = 2P_1 \]