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 and the relationship between pressure, volume, and temperature. The relevant equation for this scenario is derived from Gay-Lussac's Law, which states that for a fixed amount of gas at constant volume, the pressure of the gas is directly proportional to its absolute temperature.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The gas is at Standard Temperature and Pressure (STP), which means:
- Initial Temperature (T1) = 0°C = 273 K
- Initial Pressure (P1) = 1 atm

2. **Determine Final Temperature**:
- If the temperature is doubled, then:
- Final Temperature (T2) = 2 × T1 = 2 × 273 K = 546 K

3. **Use Gay-Lussac's Law**:
- According to Gay-Lussac's Law, the relationship between pressure and temperature at constant volume is given by:
\[ \frac{P1}{T1} = \frac{P2}{T2} \]
- Rearranging the equation to find P2 gives:
\[ P2 = P1 \times \frac{T2}{T1} \]

4. **Substitute Known Values**:
- Substitute the known values into the equation:
\[ P2 = 1 \, \text{atm} \times \frac{546 \, \text{K}}{273 \, \text{K}} \]

5. **Calculate P2**:
- Simplifying the equation:
\[ P2 = 1 \, \text{atm} \times 2 = 2 \, \text{atm} \]

6. **Final Answer**:
- The final pressure of the gas when the temperature is doubled at constant volume is **2 atm**.