Question

An ideal gas in a closed rigid container is at 50$^\circ$C and pressure 3.23 kPa. If temperature is doubled, find new pressure in Pa :

• A. 3730 Pa
• B. 3230 Pa
• C. 6460 Pa
• D. 6430 Pa

Hint:

Always convert temperatures from Celsius to Kelvin before using gas laws. A common trap is to double the pressure directly if the temperature is doubled in Celsius. $50^\circ\text{C}$ to $100^\circ\text{C}$ is not a doubling of absolute temperature.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
An ideal gas is confined in a rigid container, which means its volume is constant ($V =$ constant). The initial temperature and pressure are given. The temperature is then "doubled". We need to find the final pressure in Pascals.
Step 2: Key Formula or Approach:
For a fixed mass of an ideal gas at constant volume, Gay-Lussac's Law applies:
$P \propto T$
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
Important: Temperature must be in Kelvin. "Doubled" refers to the Celsius temperature as per the question context, which means $T_2$ in Celsius is $2 \times 50^\circ\text{C} = 100^\circ\text{C}$.
Step 3: Detailed Explanation:
Given values:
Initial pressure, $P_{1} = 3.23 \text{ kPa} = 3230 \text{ Pa}$
Initial temperature, $T_{1} = 50^\circ\text{C} = 50 + 273 = 323 \text{ K}$
Final temperature in Celsius is double the initial value, so $t_2 = 2 \times 50^\circ\text{C} = 100^\circ\text{C}$.
Final temperature in Kelvin, $T_{2} = 100 + 273 = 373 \text{ K}$
Using Gay-Lussac's Law:
$\frac{P_{2}}{T_{2}} = \frac{P_{1}}{T_{1}}$
$P_{2} = P_{1} \times \left(\frac{T_{2}}{T_{1}}\right)$
$P_{2} = 3230 \text{ Pa} \times \left(\frac{373 \text{ K}}{323 \text{ K}}\right)$
$P_{2} = 10 \times 373 \text{ Pa}$
$P_{2} = 3730 \text{ Pa}$
Step 4: Final Answer:
The new pressure is 3730 Pa.