Problem 130 - Chemistry: An Atoms-Focused Approach

Question

The pressure in an aerosol can is 1.2 atm at \(27^{\circ} \mathrm{C}\). The can will withstand a pressure of 3.0 atm. Will it burst if heated in a campfire to \(450^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The initial temperature is 27 degrees Celsius, and the initial pressure is 1.2 atm. Answer: No, the can will not burst. The final pressure exerted on the can when heated to 450 degrees Celsius is 2.9 atm, which is less than the pressure the can can withstand (3.0 atm).

Step 1: Convert the temperatures to Kelvin

To convert from Celsius to Kelvin, add 273.15 to the temperature in Celsius. The initial temperature is 27 degrees Celsius, which is equivalent to 300.15 K. T1 = 27℃ = 300.15K The final temperature is given as 450 degrees Celsius, which is equivalent to 723.15K. T2 = 450℃ = 723.15K

Step 2: Use the Gay-Lussac's Law to calculate final pressure

The formula that gives the relationship between pressure and temperature while keeping volume and amount constant is P1/T1 = P2/T2, where P1 = initial pressure, T1 = initial temperature, P2 = final pressure, T2 = final temperature. In our case, P1 = 1.2 atm (Initial pressure) T1 = 300.15 K (Initial temperature in Kelvin) T2 = 723.15 K (Final Temperature in Kelvin) Substitute these values into the equation and solve for final pressure P2: P2 = (P1 * T2) / T1

Step 3: Perform the Calculation

Substitute the known values: P2 = (1.2 atm * 723.15 K) / 300.15 K Upon calculation, it is found that P2 is approximately 2.9 atm.

Step 4: Compare the final pressure to the pressure the can can withstand

The can can withstand a pressure of 3.0 atm. Compare this to the calculated final pressure (P2 = 2.9 atm). It is observed that the final pressure exerted on the can of 2.9 atm is less than the pressure the can can withstand of 3.0 atm. Thus, the can will not burst when heated in the campfire to 450 degrees Celsius.

Key Concepts

Gay-Lussac's Law

Understanding Gay-Lussac's Law is essential when discussing the relationship between pressure and temperature in gases. This law asserts that if the volume of a gas is held constant, the pressure exerted by the gas is directly proportional to its absolute temperature. In mathematical terms, Gay-Lussac's Law is expressed as \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), where:

• \( P_1 \) and \( P_2 \) are the initial and final pressures respectively.
• \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.

This means if the temperature of a gas increases, its pressure increases too, provided the volume doesn't change. Understanding this relationship helps predict pressure changes due to temperature variations. This can be especially important in real-life scenarios like the one in the exercise, with aerosol cans exposed to varying temperatures.

Pressure-Temperature Relationship

The pressure-temperature relationship described by Gay-Lussac's Law indicates how closely linked these two variables are for gases in a contained space. When the temperature of a gas increases, the molecules move faster. This increase in speed results in more frequent collisions with the walls of the container, leading to an increase in pressure. Conversely, lowering the temperature results in a decrease in molecular speed, collision frequency, and therefore, pressure. This principle helps us understand and predict the behavior of gases in different environmental conditions. For example, heating an aerosol can in a campfire as per the exercise can dramatically increase its pressure, risking a burst if the limits of the material are exceeded.

Kelvin Conversion

Converting temperatures into Kelvin is crucial when applying Gay-Lussac's Law, as this law relies on absolute temperatures. The Kelvin scale is an absolute temperature scale, starting from absolute zero (0 K), the lowest theoretically attainable temperature. Celsius temperatures must be converted to Kelvin by adding 273.15.For instance, in the exercise:

• The initial temperature of 27°C converts to 300.15 K by calculation: \( 27 + 273.15 = 300.15 \).
• Similarly, the final temperature is converted from 450°C to 723.15 K: \( 450 + 273.15 = 723.15 \).

These conversions make the calculations accurate and comply with the physical laws governing pressure and temperature relationships. Always remember, conversions are significant as they affect the precision and validity of the calculations involved in gas-related experiments.