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Chapter 13: Problem 31

For each of the following sets of volume/temperature data, calculate the missing quantity after the change is made. Assume that the pressure and the amount of gas remain the same. a. \(V=2.03 \mathrm{~L}\) at \(24 \mathrm{C} ; V=3.01 \mathrm{~L}\) at \(? \mathrm{C}\) b. \(V=127 \mathrm{~mL}\) at \(273 \mathrm{~K} ; V=? \mathrm{~mL}\) at \(373 \mathrm{~K}\) c. \(V=49.7 \mathrm{~mL}\) at \(34 \mathrm{C} ; V=?\) at \(350 \mathrm{~K}\)

Short Answer

Expert verified
a. \(T_2 \approx 166.33 \mathrm{C}\) b. \(V_2 \approx 173.13 \mathrm{~mL}\) c. \(V_2 \approx 56.54 \mathrm{~mL}\)

Step by step solution

01

Problem a: Finding the missing temperature

Given: \(V_1 = 2.03 \mathrm{~L}\) at \(T_1 = 24 \mathrm{C}\), \(V_2 = 3.01 \mathrm{~L}\) at \(T_2 = ?\mathrm{C}\) Step 1: Convert the temperature from Celsius to Kelvin (Using \(T(K) = T(掳C) + 273.15\)): \(T_1 = 24 + 273.15 = 297.15 \mathrm{K}\) Step 2: Use the Charles' law formula: \(V_1 / T_1 = V_2 / T_2\): \({2.03 \mathrm{~L}} / {297.15 \mathrm{K}} = {3.01 \mathrm{~L}} / {T_2}\) Step 3: Solve for \(T_2\): \(T_2 = 3.01 \mathrm{~L} * 297.15 \mathrm{K} / 2.03 \mathrm{~L} \approx 439.48 \mathrm{K}\) Step 4: Convert the final temperature from Kelvin back to Celsius: \(T(掳C) = T(K) - 273.15\), \(T_2 = 439.48 \mathrm{K}- 273.15 \approx{166.33 \mathrm{C}}\)
02

Problem b: Finding the missing volume

Given: \(V_1 = 127 \mathrm{~mL}\) at \(T_1 = 273 \mathrm{~K}\), \(V_2 = ? \mathrm{~mL}\) at \(T_2 = 373 \mathrm{~K}\) Step 1: Use the Charles' law formula \(V_1 / T_1 = V_2 / T_2\): \({127 \mathrm{~mL}}/{273 \mathrm{~K}} = {V_2}/{373 \mathrm{~K}}\) Step 2: Solve for \(V_2\): \(V_2 = {127 \mathrm{~mL}}*{373 \mathrm{~K}}/{273 \mathrm{~K}} \approx 173.13 \mathrm{~mL}\)
03

Problem c: Finding the missing volume

Given: \(V_1 = 49.7 \mathrm{~mL}\) at \(T_1 = 34 \mathrm{C}\), \(V_2 = ? \mathrm{~mL}\) at \(T_2 = 350 \mathrm{~K}\) Step 1: Convert the initial temperature from Celsius to Kelvin (Using \(T(K) = T(掳C) + 273.15\)): \(T_1 = 34 + 273.15 = 307.15 \mathrm{K}\) Step 2: Use the Charles' law formula \(V_1 / T_1 = V_2 / T_2\): \({49.7 \mathrm{~mL}}/{307.15 \mathrm{~K}} = {V_2}/{350 \mathrm{~K}}\) Step 3: Solve for \(V_2\): \(V_2 = {49.7 \mathrm{~mL}}*{350 \mathrm{~K}}/{307.15 \mathrm{~K}} \approx 56.54 \mathrm{~mL}\)

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Key Concepts

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

Understanding Gas Laws
Gas laws are fundamental principles that describe the behavior of gases and how they respond to changes in temperature, pressure, and volume. Among these laws, Charles' Law is particularly important as it specifically addresses the temperature-volume relationship of a gas when pressure is held constant.

According to Charles' Law, the volume of a gas expands when heated and contracts when cooled, assuming the amount of gas and pressure remain unaltered. This law applies to ideal gases, which are theoretical gases that perfectly follow the gas laws. Real gases approximate this behavior under many conditions, allowing us to use Charles' Law to make useful predictions in both scientific and everyday applications.

When solving Charles' Law problems, it's vital to ensure that temperatures are measured in absolute units, which means using the Kelvin scale. This is because the Kelvin scale starts at absolute zero, the point where the particles of an ideal gas would theoretically stop moving entirely.
Exploring the Temperature-Volume Relationship
The temperature-volume relationship is a core aspect of Charles' Law, which states that the volume of a gas is directly proportional to its temperature when pressure and the amount of gas are held constant. This means that if the temperature of a gas increases, its volume will increase proportionally, and if the temperature decreases, the volume will decrease accordingly.

To understand this concept, picture a balloon partially filled with air on a cold day. As the balloon is exposed to warmth, it will expand. This phenomenon occurs because the gas molecules move faster and push harder against the balloon's walls as temperature rises, causing the increase in volume. Conversely, when exposed to cold, the balloon shrinks due to the decreased movement of gas molecules.

This fundamental principle can be observed in numerous real-life situations, such as why a car tire can become overinflated on a hot day or why a hot air balloon rises when the air inside it is heated.
Applying Charles' Law Formula
Charles' Law formula is a mathematical representation of the relationship between the volume and temperature of a gas. The formula is expressed as \( V_1 / T_1 = V_2 / T_2 \), where \( V \) represents the volume and \( T \) represents the temperature of the gas. Subscripts 1 and 2 distinguish between the initial and final states of the gas, respectively.

In using this formula, remember that all temperatures must be in Kelvin. To convert Celsius to Kelvin, add 273.15 to the Celsius temperature. This step is crucial because any calculations with temperatures in Celsius would yield incorrect results.

When solving for the missing quantity, whether it be volume or temperature, simply cross-multiply and solve for the unknown. For example, if you know the initial volume and temperature (\(\V_1, T_1\)) and the final temperature (\(\T_2\)), you can find the final volume (\(\V_2\)) by rearranging the formula to \(\V_2 = V_1 \times T_2 / T_1\). It's essential to handle the units carefully and ensure that all quantities are consistent to avoid errors in calculations.

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Most popular questions from this chapter

At what temperature would \(4.25 \mathrm{~g}\) of oxygen gas, \(\mathrm{O}_{2}\), exert a pressure of \(784 \mathrm{~mm} \mathrm{Hg}\) in a 2.51 -L container?

144\. A sample of oxygen gas has a volume of \(125 \mathrm{~L}\) at 25 ' \(\mathrm{C}\) and a pressure of 0.987 atm. Calculate the volume of this oxygen sample at STP.

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.

If a gaseous mixture is made of \(2.41 \mathrm{~g}\) of \(\mathrm{He}\) and \(2.79 \mathrm{~g}\) of \(\mathrm{Ne}\) in an evacuated 1.04-L container at 25 ' \(C\), what will be the partial pressure of each gas and the total pressure in the container?

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe ): $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) $$ What volume of pure \(\mathrm{O}_{2}(g),\) collected at \(27 \mathrm{C}\) and 764 torr, would be generated by decomposition of \(125 \mathrm{~g}\) of a \(50.0 \%\) by mass hydrogen peroxide solution?
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