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If 3.25 moles of argon gas occupies a volume of \(100 .\) L at a particular temperature and pressure, what volume does 14.15 moles of argon occupy under the same conditions?

Short Answer

Expert verified
Under the same temperature and pressure conditions, 14.15 moles of argon gas would occupy a volume of approximately \(435.38\) L.

Step by step solution

01

Understand the problem and the relationship between moles and volume

We need to find the volume occupied by 14.15 moles of argon, given that 3.25 moles of argon occupy 100 L under the same temperature and pressure conditions. Since the temperature and pressure are constant, we can use the proportionality between moles and volume to solve for the unknown volume.
02

Set up the proportion

We can set up a proportion to compare the volume and moles of argon as follows: \(\frac{V_1}{n_1} = \frac{V_2}{n_2}\) where \(V_1\) is the volume occupied by the first amount of moles (100 L), \(n_1\) is the first amount of moles (3.25 moles), \(V_2\) is the volume occupied by the second amount of moles (what we are trying to find), and \(n_2\) is the second amount of moles (14.15 moles).
03

Solve for the unknown volume

Input the given values and solve for \(V_2\): \(\frac{100 \text{ L}}{3.25 \text{ moles}} = \frac{V_2}{14.15 \text{ moles}}\) Cross-multiply: \(100 \text{ L} \times 14.15 \text{ moles} = 3.25 \text{ moles} \times V_2\) Divide both sides by 3.25 moles: \(V_2 = \frac{100 \text{ L} \times 14.15 \text{ moles}}{3.25 \text{ moles}}\) Calculate the result: \(V_2 \approx 435.38 \text{ L}\)
04

State the final answer

Under the same temperature and pressure conditions, 14.15 moles of argon gas would occupy a volume of approximately 435.38 L.

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Key Concepts

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

Molar Volume of Gas
When studying gases, it's fundamental to understand that gases occupy space, and how much space they occupy is influenced by the amount of gas present. This concept is captured by the term 'molar volume of gas', which is the volume that one mole of a gas occupies at a defined temperature and pressure. For many gases at standard temperature and pressure (STP), which is 0 °C and 1 atm, the molar volume is approximately 22.4 liters.

It’s imperative to keep in mind that the molar volume can vary with changing conditions. For instance, in the given exercise, the volume corresponding to 3.25 moles of argon is 100 liters, which indicates a unique set of temperature and pressure conditions specific to the problem.
Avogadro's Law
Avogadro's Law states that equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of molecules. In simpler terms, volume is directly proportional to the number of moles of gas, as long as the temperature and pressure are constant. This law is a crucial underpinning of the proportional relationship used in the exercise.

According to this law, if you have a certain volume of gas with a known number of moles, like the 100 L of argon equal to 3.25 moles in our exercise, and maintain the same temperature and pressure, doubling the moles would double the volume. It’s the reason we can use a simple ratio to find unknown volumes or moles in stoichiometric calculations for gases.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the reactants and products in a chemical reaction. In the context of gases, stoichiometry is used to relate the molar amounts of gases involved in a reaction to their respective volumes.

Using the molar volume of a gas, one can convert between the number of moles and the volume the gas occupies. It plays a crucial role in calculations, such as the one found in the exercise, where stoichiometric principles allow us to determine the volume of gas that corresponds to a particular number of moles. This concept advocates the utility of a balanced equation and knowledge of the molar volume to predict the outcome of a chemical reaction involving gases.
Gas Laws

Understanding the Behavioral Patterns of Gases

Gas laws are a set of rules describing the behavior of gases and how they respond to changes in temperature, volume, and pressure. The most fundamental gas laws include Boyle's Law, Charles's Law, Gay-Lussac's Law, and the Combined Gas Law. Each of these explains a specific aspect of gas behavior:
  • Boyle's Law states that pressure and volume are inversely proportional at constant temperature.
  • Charles's Law shows that volume and temperature are directly proportional at constant pressure.
  • Gay-Lussac's Law describes how pressure and temperature are directly proportional at constant volume.
  • The Combined Gas Law combines all three of these relationships into one equation.
In our exercise, since the temperature and pressure remain constant while dealing with volumes and moles, we're actually applying a simplified version of the gas laws, sometimes referred to as Avogadro's Law or the direct proportionality of moles to volume when temperature and pressure are held constant.

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Most popular questions from this chapter

Although we generally think of combustion reactions as involving oxygen gas, other rapid oxidation reactions are also referred to as combustions. For example, if magnesium metal is placed into chlorine gas, a rapid oxidation takes place, and magnesium chloride is produced. $$ \operatorname{Mg}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{MgCl}_{2}(s) $$ What volume of chlorine gas, measured at STP, is required to react completely with 1.02 g of magnesium?

At what temperature will a 1.0 -g sample of neon gas exert a pressure of \(500 .\) torr in a \(5.0-\mathrm{L}\) container?

A widely used weather instrument called a barometer can be built from a long, thin tube of glass that is sealed at one end. The tube is completely filled with mercury and then inverted into a small pool of mercury. The level of the mercury inside the tube drops initially but then stabilizes at some height. A measure of the height of the column of mercury once it stabilizes is a measure of pressure in \(\mathrm{mm} \mathrm{Hg}\) (or torr). Which of the following is the best explanation of how this barometer works? a. Air pressure outside the tube (pressure of the atmosphere) counterbalances the weight of the mercury inside the tube. b. Air pressure inside the tube causes the mercury to move in the tube until the air pressure inside and outside the tube are equal. c. Air pressure outside the tube causes the mercury to move in the tube until the air pressure inside and outside the tube are equal. d. The vacuum that is formed at the top of the tube of mercury (once the mercury level in the tube has dropped some) holds up the mercury. e. I have no idea how a barometer works.

Given each of the following sets of values for an ideal gas, calculate the unknown quantity. a. \(P=782 \mathrm{~mm} \mathrm{Hg} ; V=? ; n=0.210 \mathrm{~mol} ; T=27 \quad \mathrm{C}\) b. \(P=? \mathrm{~mm} \mathrm{Hg} ; V=644 \mathrm{~mL} ; n=0.0921 \mathrm{~mol} ; T=303 \mathrm{~K}\) c. \(P=745 \mathrm{~mm} \mathrm{Hg} ; V=11.2 \mathrm{~L} ; n=0.401 \mathrm{~mol} ; T=? \mathrm{~K}\)

For each of the following sets of volume/temperature data, calculate the missing quantity. Assume that the pressure and the mass of gas remain constant. a. \(V=25.0 \mathrm{~L}\) at \(0 \quad \mathrm{C} ; V=50.0 \mathrm{~L}\) at \(? \mathrm{C}\) b. \(V=247 \mathrm{~mL}\) at 25 ' \(\mathrm{C} ; V=255 \mathrm{~mL}\) at \(? \mathrm{C}\) c. \(V=1.00 \mathrm{~mL}\) at \(2272 \mathrm{C} ; V=?\) at \(25 \mathrm{C}\)

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