A hot-air balloon is filled with air to a volume of \(4.00 \times 10^{3} \mathrm{~m}^{3}\) at 745 torr and \(21^{\circ} \mathrm{C}\). The air in the balloon is then heated to \(62^{\circ} \mathrm{C}\), causing the balloon to expand to a volume of \(4.20 \times 10^{3} \mathrm{~m}^{3}\). What is the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon? (Hint: Openings in the balloon allow air to flow in and out. Thus the pressure in the balloon is always the same as that of the atmosphere.)

Short Answer

Expert verified
The ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon is approximately 1.197.

Step by step solution

01

Convert temperatures to Kelvin

To use the ideal gas law, we must first convert the given temperatures from Celsius to Kelvin: Original temperature: \(21^{\circ} \mathrm{C} + 273.15 = 294.15\mathrm{K}\) Heated temperature: \(62^{\circ} \mathrm{C} + 273.15 = 335.15\mathrm{K}\)
02

Calculate moles for original volume/temperature

Using the ideal gas law, we will set up the following equation: \(PV = nRT\) We will use the given values for the original volume and pressure to find the original number of moles (n1), where R is the gas constant 0.0821 L⋅atm/mol⋅K: \((745\,\text{torr})(4.00 \times 10^{3}\,\text{m}^{3}) = n1 (0.0821\,\mathrm{L\cdot atm/mol\cdot K})(294.15\,\mathrm{K})\) Since we know 1 atm = 760 torr and 1 m³ = 1000 L, we need to convert the values to be consistent with the units of the gas constant R: \((\frac{745\,\text{torr}}{760\,\text{torr/atm}})(\frac{4.00 \times 10^{3}\,\text{m}^{3}}{0.001\,\text{m}^{3}/\text{L}}) = n1 (0.0821\,\mathrm{L\cdot atm/mol\cdot K})(294.15\,\mathrm{K})\)
03

Calculate moles for heated volume/temperature

Similarly, we will find the number of moles for the heated volume and temperature (n2): \((745\,\text{torr})(4.20 \times 10^{3}\,\text{m}^{3}) = n2 (0.0821\,\mathrm{L\cdot atm/mol\cdot K})(335.15\,\mathrm{K})\) As before, we convert the values to be consistent with the units of the gas constant R: \((\frac{745\,\text{torr}}{760\,\text{torr/atm}})(\frac{4.20 \times 10^{3}\,\text{m}^{3}}{0.001\,\text{m}^{3}/\text{L}}) = n2 (0.0821\,\mathrm{L\cdot atm/mol\cdot K})(335.15\,\mathrm{K})\)
04

Determine the ratio

To find the ratio of the number of moles after heating (n2) to the original number of moles (n1), we divide the expression for n2 by the expression for n1: \(Ratio=\frac{n2}{n1}=\frac{(\frac{745}{760})(\frac{4.20\times10^{3}}{0.001})}{(\frac{745}{760})(\frac{4.00\times10^{3}}{0.001})}\cdot\frac{335.15}{294.15}\) Notice that some terms cancel out in the equation: \(Ratio=\frac{4.20\times10^{3}}{4.00\times10^{3}}\cdot\frac{335.15}{294.15}\)
05

Calculate the final ratio

Now, we can simplify the expression to get the final ratio: \(Ratio=\frac{1.05}{1}\cdot\frac{335.15}{294.15}\) \(Ratio=1.05\cdot1.13968\) \(Ratio \approx 1.197\) Thus, the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon is approximately 1.197.

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Key Concepts

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

Gas Constant
The gas constant, denoted as R, is a fundamental aspect of the ideal gas law and serves as a bridge to connect pressure, volume, temperature, and moles of an ideal gas. Its value is determined by the units used for pressure, volume, and temperature. In most physics and chemistry problems, the value of the gas constant is provided to students as 0.0821 L⋅atm/mol⋅K.

This constant is key when using the ideal gas law formula, which is given as:\[ PV = nRT \]Here, P represents the pressure of the gas, V is the volume, n is the number of moles of gas, and T is the temperature in Kelvin. The gas constant R ensures that each of these variables are in equilibrium, so to speak, allowing us to solve for any one of them if we have the other three. Understanding the units and how they relate is crucial, especially when converting between different systems of measurement—like torr to atmospheres or liters to cubic meters—as seen in our balloon example.
Moles of Gas
Moles provide a way to quantify the amount of a substance. In the context of gases, the mole allows us to relate the microscopic world of atoms and molecules to the macroscopic, measurable quantities of gas that we observe. One mole of any substance contains Avogadro's number of particles (approximately \(6.022 \times 10^{23}\) particles), symbolizing a bridge between the microscopic and macroscopic scales.

Using the ideal gas law, we can calculate the number of moles (n) of a gas present in a system when given the other variables (P, V, and T), as demonstrated in our exercise. This is done with the rearranged equation:\[ n = \frac{PV}{RT} \]It is important to note that when dealing with moles, the temperature must be in Kelvin, and we should ensure that our units for pressure and volume align with those of the gas constant. By calculating the moles of gas, we can also find out how the amount of gas changes with variables like temperature and volume, thus understanding the system's behavior under different conditions.
Pressure Temperature Volume Relationship
The relationship between pressure, temperature, and volume is at the core of the behavior of gases. As per the ideal gas law, these quantities are interdependent; changing one can affect the others. The law is usually stated as:\[ PV = nRT \]This equation tells us that for a fixed amount of gas (constant n) and a constant R, if the temperature (in Kelvin) increases, the volume must increase if the pressure remains the same. Conversely, if the volume decreases, the pressure must increase if the temperature is held constant. In the hot-air balloon example, the balloon expands as the gas inside is heated. The increase in volume is directly proportional to the increase in temperature since the pressure is consistent with the atmospheric pressure.

The pressure temperature volume relationship is essential in various applications, including predicting weather patterns, designing engines, and even making culinary delights like the perfect loaf of bread. It reminds us of the dynamic quality of gases and the elegant predictability of their behavior when understood through the lens of the ideal gas law.

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Most popular questions from this chapter

Some very effective rocket fuels are composed of lightweight liquids. The fuel composed of dimethylhydrazine \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2} \mathrm{H}_{2}\right]\) mixed with dinitrogen tetroxide was used to power the Lunar Lander in its missions to the moon. The two components react according to the following equation: \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2} \mathrm{H}_{2}(l)+2 \mathrm{~N}_{2} \mathrm{O}_{4}(l) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)\) If \(150 \mathrm{~g}\) dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at \(27^{\circ} \mathrm{C}\) in an evacuated 250-L tank, what is the partial pressure of nitrogen gas produced and what is the total pressure in the tank assuming the reaction has \(100 \%\) yield?

A container is filled with an ideal gas to a pressure of \(40.0\) atm at \(0^{\circ} \mathrm{C}\). a. What will be the pressure in the container if it is heated to \(45^{\circ} \mathrm{C}\) ? b. At what temperature would the pressure be \(1.50 \times 10^{2}\) atm? c. At what temperature would the pressure be \(25.0 \mathrm{~atm} ?\)

Consider the reaction between \(50.0 \mathrm{~mL}\) liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}\) (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ), and \(22.8 \mathrm{~L} \mathrm{O}_{2}\) at \(27^{\circ} \mathrm{C}\) and a pressure of \(2.00 \mathrm{~atm}\). The products of the reaction are \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(g)\). Calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion.

A person accidentally swallows a drop of liquid oxygen, \(\mathrm{O}_{2}(l)\), which has a density of \(1.149 \mathrm{~g} / \mathrm{mL}\). Assuming the drop has a volume of \(0.050 \mathrm{~mL}\), what volume of gas will be produced in the person's stomach at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and a pressure of \(1.0 \mathrm{~atm}\) ?

Do all the molecules in a 1 -mol sample of \(\mathrm{CH}_{4}(g)\) have the same kinetic energy at \(273 \mathrm{~K}\) ? Do all molecules in a \(1-\mathrm{mol}\) sample of \(\mathrm{N}_{2}(g)\) have the same velocity at \(546 \mathrm{~K}\) ? Explain.

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