Question

An aerosol spray at room temperature, \(25^{\circ} \mathrm{C}\), is under a pressure of 10 atm at a volume of \(0.5 \ell\). If someone releases all the contents to fill an empty \(4.6 \ell\) box at standard pressure, what is the temperature of the contents inside the box?

Short answer

The temperature of the contents inside the box is approximately \(4.9^{\circ}\mathrm{C}\).

Step-by-step solution

Write down the ideal gas law for initial and final states

The ideal gas law is given by the formula \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature. Since the number of moles of gas remains constant throughout this process, we can write two equations, one for the initial state and one for the final state: Initial state: \(P_1V_1 = nRT_1\) Final state: \(P_2V_2 = nRT_2\)

Solve for nRT in both initial and final equations

We need to relate the initial and final states, so we need to solve for the common term in both equations, which is \(nRT\). First, let's convert the given temperatures from Celsius to Kelvin: Initial temperature: \(T_1 = 25^{\circ}\mathrm{C} + 273.15 = 298.15\,\mathrm{K}\) We want to solve for the final temperature, \(T_2\), inside the box at standard pressure: Initial equation: \(nRT_1 = P_1V_1\) Final equation: \(nRT_2 = P_2V_2\)

Divide the final equation by the initial equation to eliminate n and R

To eliminate the common terms and relate the initial and final states, we will divide the final equation by the initial equation: \(\frac{P_2V_2}{P_1V_1} = \frac{nRT_2}{nRT_1}\) Since \(n\) and \(R\) cancel out, we are left with: \(\frac{P_2V_2}{P_1V_1} = \frac{T_2}{T_1}\)

Plug in the given values and solve for T2

Now, we can plug in the given values (converting units to ensure consistency): \(\frac{P_2V_2}{P_1V_1} = \frac{T_2}{T_1}\) \(\frac{(1\,\mathrm{atm})(4.6\,\mathrm{L})}{(10\,\mathrm{atm})(0.5\,\mathrm{L})} = \frac{T_2}{298.15\,\mathrm{K}}\) Now, solve for \(T_2\): \(T_2 = 298.15\,\mathrm{K} \times \frac{(1\,\mathrm{atm})(4.6\,\mathrm{L})}{(10\,\mathrm{atm})(0.5\,\mathrm{L})}\) \(T_2 = 298.15\,\mathrm{K} \times \frac{4.6}{5} \) \(T_2 = 278.051\,\mathrm{K}\)

Convert the final temperature to Celsius

Finally, convert the temperature back to Celsius: Final temperature: \(T_2 = 278.051\,\mathrm{K} - 273.15 = 4.901^{\circ}\mathrm{C}\) The temperature of the contents inside the box is approximately \(4.9^{\circ}\mathrm{C}\).

Key concepts

Understanding the Ideal Gas Law

The ideal gas law is a crucial equation in understanding how gases behave under varying conditions of pressure, volume, and temperature. It is expressed as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant (8.314 J/mol·K), and \( T \) is the temperature in Kelvin. This law consolidates the combined gas law and Avogadro's hypothesis by including the number of moles in the equation, providing a comprehensive description of the state of an ideal gas.

It is particularly helpful in solving problems where the state of a gas changes from one condition to another. By assuming the gas behaves ideally, a condition which closely holds for many gases under a range of temperatures and pressures, the equation becomes a powerful tool to predict the final state of a gas, given that we keep one of the terms constant or know its change.

Gas Temperature Conversion

When working with gas laws, temperature must always be expressed in Kelvin to ensure the accuracy of calculations. The Kelvin scale is an absolute temperature scale, starting at absolute zero, where no thermal energy remains in a substance. To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. Similarly, converting back from Kelvin to Celsius requires subtracting 273.15 from the Kelvin temperature.

Understanding this conversion is essential since temperature changes can significantly impact pressure and volume in gas law calculations. Moreover, the use of Kelvin ensures that we do not encounter negative temperatures in our calculations, which could disrupt the mathematical relationships within the ideal gas law.

Standard Pressure and Volume

In the context of gas laws, 'standard' refers to certain agreed-upon conditions for measurements to make comparisons meaningful. Standard temperature and pressure (STP) are typically set at 0 degrees Celsius (273.15 K) and 1 atmosphere (atm) of pressure. A mole of an ideal gas at STP occupies 22.4 liters.

This standard is used as a reference point in many chemical calculations as it simplifies many aspects of the quantitative analysis of gases. When we say a gas is at standard pressure, it means that it is at 1 atm. Similarly, standard volume refers to the volume one mole of gas occupies at standard conditions. Remembering these benchmark figures comes in handy for problem-solving since they serve as useful comparison points or 'starting blocks' for calculations.

Applying Charles's Law

Charles's Law, one of the specific gas laws, describes how gases tend to expand when heated. The law states that at constant pressure, the volume of a given mass of an ideal gas is directly proportional to its temperature in Kelvin. Mathematically, this is expressed as \( V_1/T_1 = V_2/T_2 \) when pressure remains constant.

In real-life applications, Charles's law can help predict the behavior of gases in a variety of scenarios, such as balloon inflation, engine efficiency, or even the expansion of air in weather phenomena. Understanding this principle allows one to see the relationship between temperature and volume and how they are interdependent in maintaining the pressure of a gas sample.