For a fixed amount of an ideal gas present at STP, if temperature is doubled keeping volume same then the final pressure of

Options

A. 1 atm
B. 2 atm
C. 3 atm
D. 4 atm

Answer

To solve the problem, we will use the Ideal Gas Law, which is represented by the equation:

\[ PV = nRT \]

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

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- At STP (Standard Temperature and Pressure), the initial pressure \( P_1 \) is 1 atm, the initial temperature \( T_1 \) is 273 K, and the volume \( V \) is constant.

2. **Understand the Changes**:
- The problem states that the temperature is doubled while keeping the volume the same. Therefore, the new temperature \( T_2 \) will be:
\[ T_2 = 2T_1 = 2 \times 273 \, \text{K} = 546 \, \text{K} \]

3. **Using the Ideal Gas Law**:
- Since the volume \( V \) and the number of moles \( n \) remain constant, we can express the relationship between pressure and temperature:
\[ P \propto T \]
This means that pressure is directly proportional to temperature when volume is constant.

4. **Calculate Final Pressure**:
- If the initial pressure is \( P_1 = 1 \, \text{atm} \) and the initial temperature is \( T_1 = 273 \, \text{K} \), then when the temperature is doubled to \( T_2 = 546 \, \text{K} \), the final pressure \( P_2 \) can be calculated as:
\[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \]
Substituting the known values:
\[ \frac{P_2}{1 \, \text{atm}} = \frac{546 \, \text{K}}{273 \, \text{K}} = 2 \]
Therefore:
\[ P_2 = 2 \times 1 \, \text{atm} = 2 \, \text{atm} \]

### Final Answer:
The final pressure of the gas when the temperature is doubled while keeping the volume constant is **2 atm**.

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