Question

The label on an aerosol spray can contains a warning that the can should not be heated to over \(130^{\circ} \mathrm{F}\) because of the danger of explosion due to the pressure increase as it is heated. Calculate the potential volume of the gas contained in a \(500 .-\mathrm{mL}\) aerosol can when it is heated from \(25^{\circ} \mathrm{C}\) to \(54^{\circ} \mathrm{C}\) (approximately \(130^{\circ} \mathrm{F}\) ), assuming a constant pressure.

Short answer

The potential volume of the gas contained in a 500 mL aerosol can when it is heated from 25°C to 54°C, assuming constant pressure, is approximately \(549.61 \mathrm{mL}\).

Step-by-step solution

Convert temperatures to Kelvin

To use the Ideal Gas Law, we need temperatures in Kelvin. To convert temperatures from Celsius to Kelvin, we add 273.15. Initial temperature (T1) = 25°C + 273.15 = 298.15 K Final temperature (T2) = 54°C + 273.15 = 327.15 K

Use the constant pressure condition

Since the pressure is constant, we can use the relationship between initial and final volumes and temperatures: \( V_1/T_1 = V_2/T_2 \) Where: V1 = initial volume = 500 mL T1 = initial temperature = 298.15 K V2 = final volume (which we want to find) T2 = final temperature = 327.15 K

Solve for the final volume (V2)

Now we can plug in the values for V1, T1, and T2 into the equation \( V_1/T_1 = V_2/T_2 \) to solve for the final volume (V2). \( \frac{500}{298.15} = \frac{V_2}{327.15} \) Now solve for V2: \( V_2 = \frac{500 \times 327.15}{298.15} \) \( V_2 \approx 549.61 \mathrm{mL} \)

Round the result

Round the final volume to two decimal places: Potential volume of the gas after heating = 549.61 mL So, when the aerosol can is heated from 25°C to 54°C, the potential volume of the gas contained in the can increases to approximately 549.61 mL.

Key concepts

Temperature Conversion

Imagine we're cooking a dish that calls for precise temperature settings, but our kitchen thermometer displays a different temperature scale than our recipe. We need to convert temperatures to cook our meal correctly. A similar conversion is crucial in scientific calculations, particularly when working with the Ideal Gas Law.

Converting from Celsius to Kelvin is a fundamental step because the Ideal Gas Law, a cornerstone of chemistry and physics, requires temperatures to be in Kelvin. The Kelvin scale is an absolute temperature scale, meaning it starts at absolute zero, which is the theoretical point where particles have minimal thermal motion.

To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. This is because the difference between the freezing point of water at standard pressure on both scales is 273.15 units: \[ 0^{\text{°}}\text{C} + 273.15 = 273.15\text{K} \]
And for each degree you climb in Celsius, you climb one Kelvin as well, maintaining the direct relationship between the two scales. This is why, when solving our exercise, we added 273.15 to the given Celsius temperatures before using the Ideal Gas Law.

Kelvin Scale

The Kelvin scale holds a key role in the scientific world, being the base unit of temperature in the International System of Units (SI). Unlike Celsius or Fahrenheit, the Kelvin scale is not referred to in degrees; we simply say 'kelvins'.

This scale is essential because it provides a uniform standard for scientists globally, allowing accurate and consistent measurements and calculations in thermodynamics and kinetic theories, among other fields.

The significance of Kelvin in our exercise lies in its direct proportionality to the kinetic energy of particles in a gas. The Ideal Gas Law connects temperature, volume, and pressure of a gas, but the temperature must be in an absolute scale for the law to hold true. Hence, our need for Kelvin conversions.

Gas Volume Calculation

Consider playing with a balloon. As you warm it with your hands, it expands. This physical reality is part of what scientists describe with the Ideal Gas Law, where the volume of a gas is directly proportional to its temperature when pressure is held constant.

Gas volume calculation is a practical application of this law, and it follows a straightforward formula when dealing with constant pressure: \( V_1/T_1 = V_2/T_2 \)

Here, \(V_1\) and \(V_2\) are the initial and final volumes, while \(T_1\) and \(T_2\) are the initial and final temperatures (in Kelvin). To solve for the unknown volume, you multiply the initial volume by the ratio of the final to the initial temperature. This is precisely what we did in the exercise to find the potential volume of gas after heating the aerosol can.

The importance of converting to Kelvin becomes evident here. If we had used Celsius or Fahrenheit, we could have ended up with negative temperatures after conversion, which doesn't work when using ratios in our calculations. Kelvin provides a seamless scale where all temperatures are positive, making our math for volume calculation both possible and meaningful.