What will be the final pressure of an ideal gas present in a cylinder at 2 atm when the temperature of the gas is increased from `100^(@)C` to `500^(@)C`?

Options

A. 3.68 atm
B. 4.14 atm
C. 1.77 atm
D. 8.42 atm

Answer

To find the final pressure of an ideal gas when its temperature is increased, we can use the Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature when the volume is held constant.

### Step-by-Step Solution:

**Step 1: Convert the temperatures from Celsius to Kelvin.**
- The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273.15 \]
- For \( T_1 = 100°C \):
\[ T_1 = 100 + 273.15 = 373.15 \, K \]
- For \( T_2 = 500°C \):
\[ T_2 = 500 + 273.15 = 773.15 \, K \]

**Step 2: Use the Gay-Lussac's Law formula.**
- According to Gay-Lussac's Law:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
- Rearranging the formula to find \( P_2 \):
\[ P_2 = P_1 \times \frac{T_2}{T_1} \]

**Step 3: Substitute the known values into the equation.**
- Given \( P_1 = 2 \, atm \), \( T_1 = 373.15 \, K \), and \( T_2 = 773.15 \, K \):
\[ P_2 = 2 \, atm \times \frac{773.15 \, K}{373.15 \, K} \]

**Step 4: Calculate \( P_2 \).**
- Performing the calculation:
\[ P_2 = 2 \times \frac{773.15}{373.15} \approx 2 \times 2.07 \approx 4.14 \, atm \]

**Final Answer:**
- The final pressure \( P_2 \) is approximately \( 4.14 \, atm \).