Question

\(\Lambda\) cooking gas cylinder can resist a prcssurc of \(15 \mathrm{~atm}\). At \(27^{\circ} \mathrm{C}\) the pressurc of the gas is \(12 \mathrm{~atm}\). Calculate the minimum temperature in \({ }^{\circ} \mathrm{C}\) ' above which it burst (1) \(375^{\circ} \mathrm{C}\) (2) \(1002^{\circ} \mathrm{C}\) (3) \(102^{\circ} \mathrm{C}\) (4) \(300^{\circ} \mathrm{C}\)

Short answer

Option (3) 102°C.

Step-by-step solution

Understand the Given Data

Identify the given data: The maximum pressure the cylinder can withstand is 15 atm, the initial pressure is 12 atm, and the initial temperature is 27°C, which needs to be converted to Kelvin for calculations.

Convert Initial Temperature to Kelvin

Convert the given temperature from Celsius to Kelvin using the formula: \[ T(K) = T(°C) + 273.15 \]So, the initial temperature is: \[ 27 + 273.15 = 300.15 \, K \]

Use the Ideal Gas Law (Combined Form)

According to the combined gas law: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]Where, \( P_1 = 12 \) atm (initial pressure)\( T_1 = 300.15 \) K (initial temperature)\( P_2 = 15 \) atm (final pressure)

Solve for Final Temperature in Kelvin

Rearrange the equation to solve for the final temperature \( T_2 \): \[ T_2 = T_1 \cdot \frac{P_2}{P_1} \]Substituting the values: \[ T_2 = 300.15 \times \frac{15}{12} = 375.1875 \, K \]

Convert the Final Temperature Back to Celsius

Convert the final temperature back to Celsius: \[ T(°C) = T(K) - 273.15 \]Thus, \[ 375.1875 - 273.15 = 102.0375 \approx 102 \, °C \]

Determine the Correct Option

The closest option to the calculated temperature is 102°C. Therefore, the correct answer is option (3) 102°C.

Key concepts

Pressure

Pressure is a fundamental concept in the study of gases. It is defined as the force exerted per unit area. In the context of the Ideal Gas Law and the Combined Gas Law, pressure is a critical variable. In this exercise, we deal with pressures in atmospheres (atm). To recap, one atmosphere is the pressure exerted by the weight of the atmosphere at sea level. When a gas cylinder can withstand a maximum pressure of 15 atm but currently holds a pressure of 12 atm, it is essential to understand how this pressure can increase with temperature.

In the ideal gas context, pressure can change by adjusting the temperature, volume, or the amount of gas. Since the volume and the amount of gas remain constant in this problem, any increase in temperature will directly result in an increase in pressure due to the gas molecules moving faster and colliding more frequently with the cylinder walls.

Key points to remember about pressure in gas laws:

• It is often measured in atmospheres (atm) or pascals (Pa).
• A higher pressure means gas molecules are more tightly packed and collision frequency is higher.
• Pressure is related to temperature, volume, and the number of gas molecules.

Temperature Conversion

Temperature conversion, especially from Celsius to Kelvin, is a vital step in gas law problems. The Kelvin scale is directly related to the kinetic energy of gas molecules, making it more suitable for calculations.

To convert Celsius to Kelvin, use the formula:

i.e. \[ T(K) = T(°C) + 273.15 \]

This formula simply shifts the Celsius scale up by 273.15. So if you start with 27°C, converting to Kelvin gives:

27 °C + 273.15 = 300.15 K.

Always keep in mind:

• The Kelvin scale starts at absolute zero (-273.15°C or 0 K), which means no molecular motion.
• Kelvin is the preferred unit in gas law equations because it avoids negative values that can make calculations difficult.

Lastly, converting back to Celsius is equally important to match the desired units in the final answer:

\[ T(°C) = T(K) - 273.15 \].

Combined Gas Law

The Combined Gas Law integrates the relationships between pressure, temperature, and volume. For this exercise, the volume and the number of gas particles are constant. This law is derived from Boyle's Law, Charles's Law, and Gay-Lussac's Law combined.

The formula used here is:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
where P represents pressure and T represents temperature (in Kelvin).

This law helps us find the new temperature or pressure when one of these variables changes while volume remains constant. For example:

Given:

• Initial pressure (\( P_1 \)) is 12 atm
• Initial temperature (\( T_1 \)) is 300.15 K (converted from 27°C)
• Final pressure (\( P_2 \)) is 15 atm

We need to determine the final temperature \( T_2 \):

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

By substituting the values:

\[ 375.1875 K = 300.15 K \cdot \frac{15}{12} \]

This updated temperature in Kelvin must then be converted back to Celsius to match the options given:

\[ T(°C) = T(K) - 273.15 \]
This gives us 102°C as the final answer.

Always remember:

• The Combined Gas Law is practical for finding unknown variables when dealing with gas samples where volume is constant.
• It is helpful to keep track of units and make sure temperatures are in Kelvin for consistency.