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The gas in an aerosol can is at a pressure of \(3.0 \mathrm{atm}\) at \(23^{\circ} \mathrm{C} .\) What will the pressure of the gas in the can be if the temperature is raised to \(400^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The final pressure of the gas will be approximately 6.83 atm.

Step by step solution

01

Understand the Equation

We're dealing with a situation where pressure and temperature change, so we'll use the ideal gas law transform: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). This is derived from the combined gas law, assuming volume and moles of gas are constant.
02

Convert Temperatures to Kelvin

Convert both temperatures from Celsius to Kelvin using the formula \( T(K) = T(°C) + 273.15 \). For the initial temperature, \( T_1 = 23^{\circ} C = 23 + 273.15 = 296.15 \ K \). For the final temperature, \( T_2 = 400^{\circ} C = 400 + 273.15 = 673.15 \ K \).
03

Set Up the Equation

Using the initial and final pressures and temperatures, set up the equation: \( \frac{3.0 \ atm}{296.15 \ K} = \frac{P_2}{673.15 \ K} \). We will solve this equation for \( P_2 \).
04

Solve for Final Pressure

To find \( P_2 \), rearrange the equation to: \( P_2 = \frac{3.0 \ atm \times 673.15 \ K}{296.15 \ K} \).
05

Calculate the Final Pressure

Perform the calculation: \( P_2 = \frac{3.0 \ atm \times 673.15}{296.15} \approx 6.83 \ atm \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure and Temperature Relationship
The relationship between pressure and temperature of a gas is a fundamental aspect of the ideal gas law. These two quantities are directly proportional when the volume and the amount of gas remain constant. This means if the temperature of a gas increases, its pressure will also increase, assuming that the gas does not expand in volume or escape. When pressure and temperature are linked through the equation \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), it implies that the ratio of the initial pressure over the initial temperature is equal to the ratio of the final pressure over the final temperature. This equation helps predict how pressure will change as temperature varies. Importantly, the temperature must be in Kelvin for this equation to hold true, as Kelvin is the absolute scale used in gas laws.
In simple terms, if you heat up a gas, the particles inside start moving more vigorously. Since the particles are moving faster, they collide with the walls of their container more often and with greater force, leading to increased pressure. The opposite happens when the temperature decreases. Understanding this relationship is crucial in real-life applications such as safety protocols for storage of pressurized containers.
Kelvin Temperature Conversion
Conversion to Kelvin from Celsius is vital when dealing with ideal gas law equations because Kelvin is the SI unit for temperature in these calculations. The Kelvin scale starts at absolute zero, making it a true representation of thermal energy. To convert a temperature from Celsius to Kelvin, we add 273.15 to the Celsius temperature. For instance, if you have a temperature of \(23^{\circ} C\), it converts to Kelvin as follows:
\[ T(K) = 23 + 273.15 = 296.15 \, K \]
Similarly, a temperature of \(400^{\circ} C\) would convert to:
\[ T(K) = 400 + 273.15 = 673.15 \, K \]
This adjustment is necessary because all gas law calculations rely on absolute temperature, which avoids issues that arise with negative Celsius temperatures and ensures the equations work properly. Always remember to perform this conversion before plugging temperatures into any gas-related formulas.
Combined Gas Law
The combined gas law is a powerful tool in understanding the behavior of gaseous substances. It merges three fundamental gas laws: Boyle's Law, Charles's Law, and Gay-Lussac's Law. Each of these laws describes how temperature, volume, and pressure relate to each other in specific ways when two of these variables are kept constant. The combined gas law is expressed as \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).

In cases where the volume \(V\) is constant, it simplifies to \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), which we used in our exercise. This simplified form shows that when a gas is contained in a fixed volume, its pressure is directly related to its temperature.
  • Boyle's Law describes the relationship between pressure and volume at constant temperature.
  • Charles's Law relates volume and temperature at constant pressure.
  • Gay-Lussac's Law associates pressure and temperature at constant volume, which particularly pertains to the exercise of interest.
The combined gas law is vital for solving complex problems involving closed systems, where knowing how these variables interact can predict outcomes and suggest solutions under different conditions.

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Most popular questions from this chapter

A sample of \(\mathrm{SO}_{2}\) gas has a volume of \(5.2 \mathrm{L}\). It is heated at constant pressure from \(30 .\) to \(90 .^{\circ} \mathrm{C}\). What is its new volume?

Can water and dimethyl sulfoxide, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{S}=\mathrm{O}, \mathrm{mo}\)lecules form hydrogen bonds between them?

Isooctane, which has a chemical formula \(\mathrm{C}_{8} \mathrm{H}_{18}\), is the component of gasoline from which the term octane rating derives. (a) Write the balanced chemical equation for the combustion of isooctane. (b) The density of isooctane is \(0.792 \mathrm{g} / \mathrm{mL}\). How many \(\mathrm{kg}\) of \(\mathrm{CO}_{2}\) are produced each year by the annual U.S. gasoline consumption of \(4.6 \times 10^{10} \mathrm{L} ?\) (c) What is the volume in liters of this \(\mathrm{CO}_{2}\) at \(\mathrm{STP}\) ? (d) The chemical formula for isooctane can be represented by \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\). Draw a Lewis structure of isooctane. (e) Another molecule with the same molecular formula is octane, which can be represented by: When comparing isooctane and octane, one structure is observed to have a boiling point of \(99^{\circ} \mathrm{C}\), while another is known to have a boiling point of \(125^{\circ} \mathrm{C} .\) Which substance, isooctane or octane, is expected to have the higher boiling point? (f) Determine whether isooctane or octane is expected to have the greater vapor pressure.

A sample of the inhalation anesthetic gas Halothane, \(\mathrm{C}_{2} \mathrm{HBrClF}_{3},\) in a \(500-\mathrm{mL}\) cylinder has a pressure of 2.3 atm at \(0^{\circ} \mathrm{C}\). What will be the pressure of the gas if its temperature is warmed to \(37^{\circ} \mathrm{C}\) (body temperature)?

Answer true or false. (a) Avogadro's law states that equal volumes of gas at the same temperature and pressure contain equal numbers of molecules. (b) At STP, one mole of uranium hexafluoride (UF \(\mathrm{MW} 352 \text { amu }),\) the gas used in uranium enrichment programs, occupies a volume of 352 L. (c) If two gas samples have the same temperature, volume, and pressure, then both contain the same number of molecules. (d) The value of Avogadro's number is \(6.02 \times 10^{23} \mathrm{g} / \mathrm{mol}\) (e) Avogadro's number is valid only for gases at STP. (f) The ideal gas law is \(P V=n R T\). (g) When using the ideal gas law for calculations, temperature must be in degrees Celsius. (h) If one mole of ethane \(\left(\mathrm{CH}_{3} \mathrm{CH}_{3}\right)\) gas occupies \(20.0 \mathrm{L}\) at 1.00 atm, the temperature of the gas is \(244 \mathrm{K}\) (i) One mole of helium (MW 4.0 amu) gas at STP occupies twice the volume of one mole of hydrogen \((\mathrm{MW} 2.0 \mathrm{amu})\).

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