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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 is approximately 1.97 L.

Step by step solution

01

Convert temperatures to Kelvin

First, convert the Celsius temperatures into Kelvin: Initial Temperature (T1) = \(23^{\circ} \mathrm{C} = 23 + 273.15 = 296.15\) K Final Temperature (T2) = \(-31^{\circ} \mathrm{C} = -31 + 273.15 = 242.15\) K
02

Convert pressure units to atmospheres

Given the pressure in torr, we need to convert it into atmospheres: Final Pressure (P2) = \(220\) torr \(= \frac{220}{760} = 0.289\) atm
03

Apply the Combined Gas Law

The combined gas law is given as: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) We're given P1, V1, T1, P2, and T2 and need to find V2. Rearrange the formula for V2: \(V_2 = \frac{P_1V_1T_2}{P_2T_1}\)
04

Calculate the final volume (V2)

Substitute the given values into the equation: \(V_2 = \frac{(1.00 \mathrm{~atm})(1.00 \mathrm{~L})(242.15 \mathrm{~K})}{(0.289 \mathrm{~atm})(296.15 \mathrm{~K})}\) \(V_2 \approx 2.97\) L
05

Calculate the change in volume

Now that we have the final volume, we can calculate the change in volume: Change in volume = V2 - V1 Change in volume = \(2.97\) L - \(1.00\) L Change in volume \(\approx 1.97\) L The change in volume of the balloon as it ascends is approximately 1.97 L.

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Key Concepts

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

Gas Laws
Gas laws are essential for understanding how gases behave under different conditions of temperature, pressure, and volume. One of these laws, which is highly applicable to our problem, is the Combined Gas Law. This law integrates three separate gas laws—Boyle's Law, Charles's Law, and Gay-Lussac's Law—into a single equation that helps us predict how a change in one of these variables affects the others in a confined space.

The Combined Gas Law is expressed as:
  • Boyle's Law: Pressure and volume are inversely proportional at constant temperature.
  • Charles's Law: Volume and temperature are directly proportional at constant pressure.
  • Gay-Lussac's Law: Pressure and temperature are directly proportional at constant volume.
The formula used to represent the Combined Gas Law is \[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\]. This equation allows us to calculate unknown values when some of the gas properties change while assuming a fixed amount of gas.
Temperature Conversion
Temperature conversion is crucial when using gas laws, as these laws require temperatures to be in an absolute scale like Kelvin, rather than Celsius or Fahrenheit. Kelvin is the SI unit of temperature, starting from absolute zero, making it ideal for scientific calculations.

To convert temperatures from Celsius to Kelvin, use the formula:
\[\text{Kelvin} = \text{Celsius} + 273.15\].

This formula ensures accurate temperature-related calculations in the gas laws. In the given exercise, temperatures of \(23^{\circ}\mathrm{C}\) and \(-31^{\circ}\mathrm{C}\) are converted to Kelvin as \(296.15\, \text{K}\) and \(242.15\, \text{K}\) respectively. Kelvin conversion aligns the calculations to the conditions where gas laws can be reliably applied.
Pressure Conversion
Pressure conversion is another important step in solving problems involving gas laws because different units of pressure are commonly used. The standard unit of pressure in the context of gas laws is the atmosphere (atm), but measurements may be given in other units like torr or pascal.

In this exercise, the pressure was given in torr, which we needed to convert into atmospheres. The conversion factor between torr and atmosphere is:
\[1\, \text{atm} = 760\, \text{torr}\].

This means that you can convert the pressure from torr to atm by dividing by 760. Thus, the final pressure of 220 torr converted to atmospheres is calculated as follows:
\[\text{atmospheres} = \frac{220}{760} \approx 0.289\, \text{atm}\].Proper conversion of pressure units is essential for correctly applying the Combined Gas Law.
Volume Calculation
Volume calculation in gas law problems determines how gases expand or contract when subjected to changes in temperature and pressure. In the context of our exercise, we used the Combined Gas Law to compute the change in volume of a balloon filled with helium as it ascends.

The reformulated equation for finding the final volume \(V_2\) is:
\[V_2 = \frac{P_1V_1T_2}{P_2T_1}\].

Here, our given values were \(P_1 = 1.00\, \text{atm}\), \(V_1 = 1.00\, \text{L}\), \(T_1 = 296.15\, \text{K}\), \(P_2 = 0.289\, \text{atm}\), and \(T_2 = 242.15\, \text{K}\).
The calculation provides a final volume of approximately 2.97 L. Finally, finding the volume change is straightforward:
Change in volume = \(V_2 - V_1 = 2.97\, \text{L} - 1.00\, \text{L} \approx 1.97\, \text{L}\).This measures how much the balloon's volume has increased due to reduced pressure and temperature.

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Most popular questions from this chapter

Complete the following table for an ideal gas. $$ \begin{array}{l|lcc|} & P(\mathrm{~atm}) & V(\mathrm{~L}) & n(\mathrm{~mol}) & T \\ \hline \text { a. } & 5.00 & & 2.00 & 155^{\circ} \mathrm{C} \\ \hline \text { b. } & 0.300 & 2.00 & & 155 \mathrm{~K} \\ \hline \text { c. } & 4.47 & 25.0 & 2.01 & \\ \hline \text { d. } & & 2.25 & 10.5 & 75^{\circ} \mathrm{C} \\ \hline \end{array} $$

Consider the following chemical equation. $$ 2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ If \(25.0 \mathrm{~mL} \mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) $$ What volume of pure \(\mathrm{O}_{2}(\mathrm{~g})\), collected at \(27^{\circ} \mathrm{C}\) and 746 torr, would be generated by decomposition of \(125 \mathrm{~g}\) of a \(50.0 \%\) by mass hydrogen peroxide solution? Ignore any water vapor that may be present.

In the "Méthode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ Fermentation of \(750 . \mathrm{mL}\) grape juice (density \(=1.0 \mathrm{~g} / \mathrm{cm}^{3}\) ) is allowed to take place in a bottle with a total volume of \(825 \mathrm{~mL}\) until \(12 \%\) by volume is ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\). Assuming that the \(\mathrm{CO}_{2}\) is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C}\) ? (The density of ethanol is \(0.79 \mathrm{~g} / \mathrm{cm}^{3} .\).)

We state that the ideal gas law tends to hold best at low pressures and high temperatures. Show how the van der Waals equation simplifies to the ideal gas law under these conditions.

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