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Chapter 5: Problem 61

A sealed balloon is filled with \(1.00 \mathrm{~L}\) helium at \(23^{\circ} \mathrm{C}\) and \(1.00\) atm. The balloon rises to a point in the atmosphere where the pressure is 220 . torr and the temperature is \(-31^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from \(1.00\) atm to a pressure of 220 . torr?

Short Answer

Expert verified
The change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 torr is 2.29 L.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert both the given temperatures from Celsius to Kelvin by adding 273.15 K to each temperature: Temperature T1: \(23^\circ C + 273.15 K = 296.15 K\) Temperature T2: \(-31^\circ C + 273.15 K = 242.15 K\)
02

Convert pressure to atm

Next, we need to convert the given pressure (P2) from torr to atm. We know that 1 atm = 760 torr, so we can use the conversion factor: Pressure P2: \(\frac{220 \, torr}{760 \, torr/atm} = 0.2895 \, atm\)
03

Apply the combined gas law

Now, we can apply the combined gas law equation to find the final volume (V2) of the balloon: \(\frac{1.00 \, L}{296.15 \, K} \times \frac{242.15 \, K}{0.2895 \, atm} = V2\) Solve for V2: \(V2 = 3.29 \, L\)
04

Find the change in volume

Now that we have the initial volume of the balloon (V1) and its final volume (V2), we can find the change in volume: Change in volume = \(V2 - V1\) Change in volume = \(3.29 \, L - 1.00 \, L\) Change in volume = \(2.29\, L\) The change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 torr is 2.29 L.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a mathematical relationship that describes the behavior of ideal gases. It is expressed with the formula \( PV = nRT \), where:
  • \( P \) is the pressure of the gas.
  • \( V \) is the volume.
  • \( n \) is the number of moles.
  • \( R \) is the ideal gas constant \((0.0821 \, L \cdot atm/mol \cdot K)\).
  • \( T \) is the temperature in Kelvin.
By understanding and employing this law, we can predict how a gas will behave under varying conditions such as temperature, pressure, or volume changes. The Ideal Gas Law helps us comprehend the relationship among these variables. However, in this particular problem, we use a derived form called the Combined Gas Law, which is useful when the amount of gas does not change. Remember to always convert temperatures to Kelvin when using these equations for accurate results.
Temperature Conversion
Temperature conversion plays a crucial role in gas law calculations. The Kelvin scale is used in gas law equations because it provides an absolute reference point. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
  • For example, if you have a temperature of \( 23^{\circ} C \), the conversion to Kelvin would be:
  • \( 23 + 273.15 = 296.15 \, K \)
Similarly, if the temperature is \(-31^{\circ} C\), you would add 273.15:
  • \(-31 + 273.15 = 242.15 \, K \)
Using Kelvin is important as it ensures the proportionality in gas law equations. Always double-check your temperature units before proceeding with any calculation involving gases.
Pressure Conversion
Pressure conversion helps us work seamlessly with gas laws by using consistent units. Often, pressure measurements may need to be converted. In this exercise, pressure was given in torr, but our calculation requires it in atm.
  • The conversion factor used is \( 1 \, atm = 760 \, torr \).
  • Given pressure in torr: \( 220 \, torr \).
  • Convert this to atm: \( \frac{220}{760} = 0.2895 \, atm \).
Accurate pressure conversion is vital to ensure that all gas law variables are compatible, which prevents calculation errors. Always remember to match your pressure unit to the standard required by the equation you are using.
Volume Change Calculation
Understanding volume change is essential to solving gas problems like the one we've tackled. By applying the Combined Gas Law, which is useful when the amount of gas remains constant, we can find how a gas's volume changes with pressure and temperature shifts.
  • The Combined Gas Law formula is \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \).
  • Using this law, we rearrange to solve for the final volume \( V_2 \).
  • Substituting known values: \( \frac{1.00 \, L}{296.15 \, K} \times \frac{242.15 \, K}{0.2895 \, atm} = V_2 \).
  • This gives \( V_2 = 3.29 \, L \).
Finally, to find the change in volume, subtract the initial volume \( V_1 \) from \( V_2 \), resulting in a 2.29 L increase. This change shows how the gas expanded as it ascended, due to lower pressure and temperature.

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Most popular questions from this chapter

Helium is collected over water at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) total pressure. What total volume of gas must be collected to obtain \(0.586 \mathrm{~g}\) helium? (At \(25^{\circ} \mathrm{C}\) the vapor pressure of water is \(23.8\) torr.)

You have a sealed, flexible balloon filled with argon gas. The atmospheric pressure is \(1.00 \mathrm{~atm}\) and the temperature is \(25^{\circ} \mathrm{C}\). Assume that air has a mole fraction of nitrogen of \(0.790\), the rest being oxygen. a. Explain why the balloon would float when heated. Make sure to discuss which factors change and which remain constant, and why this matters. Be complete. b. Above what temperature would you heat the balloon so that it would float?

An \(11.2-\mathrm{L}\) sample of gas is determined to contain \(0.50 \mathrm{~mol} \mathrm{~N}_{2}\). At the same temperature and pressure, how many moles of gas would there be in a 20.-L sample?

Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc. $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose \(240 . \mathrm{mL}\) of hydrogen gas is collected at \(30 .{ }^{\circ} \mathrm{C}\) and has a total pressure of \(1.032\) atm by this process. What is the partial pressure of hydrogen gas in the sample? How many grams of zinc must have reacted to produce this quantity of hydrogen? (The vapor pressure of water is 32 torr at \(30^{\circ} \mathrm{C}\).)

The partial pressure of \(\mathrm{CH}_{4}(g)\) is \(0.175\) atm and that of \(\mathrm{O}_{2}(g)\) is \(0.250\) atm in a mixture of the two gases. a. What is the mole fraction of each gas in the mixture? b. If the mixture occupies a volume of \(10.5 \mathrm{~L}\) at \(65^{\circ} \mathrm{C}\), calculate the total number of moles of gas in the mixture. c. Calculate the number of grams of each gas in the mixture.
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