A helium balloon has a volume of \(12.4 \mathrm{~L}\) when the pressure is \(0.885 \mathrm{~atm}\) and the temperature is \(22^{\circ} \mathrm{C}\). The balloon is cooled at a constant pressure until the temperature is \(-18^{\circ} \mathrm{C}\). What is the volume of the balloon at this stage?

Short Answer

Expert verified
The volume of the balloon at -18°C is approximately 10.7 L.

Step by step solution

01

Identify the Variables

The initial volume (\( V_1 \text{ in L} \) or liters) is 12.4 L, the initial temperature (\( T_1 \text{ in K} \) or kelvin) is 22°C, which is equal to 295 K (\( T_{\text{kelvin}} = T_{\text{celsius}} + 273 \)), the initial pressure (\( P_1 \text{ in atm} \) or atmospheres) is 0.885 atm, and the final temperature (\( T_2 \text{ in K} \) or kelvins) is -18°C, which is equal to 255 K. The pressure remains constant, which will be relevant for the law we will use. We want to find out the final volume (\( V_2 \) in liters).
02

Apply Charles's Law

Since the pressure is held constant while the temperature changes, we can use Charles's Law which states that volume is directly proportional to temperature in kelvins. This gives us the equation \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\).
03

Convert Celsius Temperatures to Kelvin

First, convert both Celsius temperatures to Kelvin. The initial temperature in kelvin is \(\frac{22 + 273}{1} = 295 \text{K}\) and the final temperature in kelvin is \(\frac{-18 + 273}{1} = 255 \text{K}\).
04

Rearrange Charles's Law to Solve for Final Volume

We can rearrange Charles's Law to solve for the final volume (\(V_2\)): \(\frac{V_1}{T_1} = \frac{V_2}{T_2} \) gives \(\frac{T_2}{T_1} \times V_1 = V_2 \) by multiplying both sides by \(\frac{T_2}{T_1} \) and then by \(\frac{V_1}{1} \) to isolate \(\frac{V_2}{1}\).
05

Plug in the Known Values

Now, plug in the values into the rearranged equation: \(\frac{255 \text{K}}{295 \text{K}} \times 12.4 \text{L} = V_2\).
06

Calculate the Final Volume

Calculating the final volume, we have: \(\frac{255}{295} \times 12.4 \text{L} \approx 10.7 \text{L}\). Hence, the final volume is approximately 10.7 liters.

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Key Concepts

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

Gas Laws
Understanding gas laws is essential for explaining how gases behave under various conditions of temperature, pressure, and volume. These laws describe the relationships between these three variables. One of the fundamental gas laws is Charles's Law, which we've applied in solving our exercise about a helium balloon's volume change at different temperatures.

Charles's Law specifically states that for a given mass of gas at constant pressure, the volume is directly proportional to its temperature in kelvins. This means that if you increase the temperature, the volume will also increase, and vice versa, as long as the pressure doesn't change. Other important gas laws include Boyle's Law, which describes the pressure-volume relationship, and Avogadro's Law, which relates volume and the amount of substance when pressure and temperature remain constant.

By combining these laws, scientists can predict how gases will react to different conditions, which is essential in fields like chemistry, physics, meteorology, and engineering.
Volume-Temperature Relationship
The volume-temperature relationship of gases is elegantly represented by Charles's Law, which reveals a direct and proportional relationship between these two properties. In simpler terms, as the temperature of a gas increases, the volume it occupies will also increase, assuming the pressure and amount of gas are kept constant.

This behavior is based on the kinetic molecular theory, which states that particles in a gas move more quickly as they are heated, thus requiring more space and leading to an increase in volume. In our exercise with the helium balloon, when the temperature decreases, the kinetic energy of the helium gas particles also decreases. This reduction in energy causes a decrease in the gas volume, which is why the balloon shrinks when cooled.

When solving problems involving the volume-temperature relationship, it is essential to use the Kelvin scale, as it begins at absolute zero, the point where particles theoretically have no kinetic energy. This allows the proportional relationship of Charles's Law to hold true mathematically.
Kelvin Temperature Scale
The Kelvin temperature scale is pivotal when working with gas laws. It is an absolute temperature scale, starting at absolute zero where all molecular motion ceases. Unlike the Celsius or Fahrenheit scales, the Kelvin scale directly relates to the energy levels of particles, with each unit on the scale representing a true fraction of thermal energy.

This scale is essential when applying Charles's Law which relates volume and temperature of gases. As we've seen in the balloon exercise, the first step to finding our solution was to convert the temperatures from Celsius to Kelvin. This helps avoid negative temperature values that would make the proportional relationship nonsensical and also ensures the accuracy of our calculations since Charles's Law formula requires temperatures to be in kelvins.

The conversion is straightforward: to convert Celsius to Kelvin, add 273.15 to your Celsius temperature. So, for our calculations, when the balloon's temperature was given as -18°C, we added 273 to find the Kelvin equivalent, ensuring we could plug this value into our Charles's Law equation for a correct volume at the new, lower temperature.

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Most popular questions from this chapter

(a) Suppose that \(4.00 \mathrm{~L}\) of methane at a pressure of 800 . Torr is transferred to a vessel of volume \(2.40 \mathrm{~L}\). What is the final pressure of methane if the change occurs at constant temperature? (b) A fluorinated organic gas in a cylinder is compressed from an initial volume of \(936 \mathrm{~mL}\) at \(158 \mathrm{~Pa}\) to \(468 \mathrm{~mL}\) at the same temperature. What is the final pressure?

(a) The van der Waals parameters for helium are \(a=\) \(3.41 \times 10^{-2} \mathrm{~L}^{2} \cdot\) atm \(\cdot \mathrm{mol}{ }^{-2}\) and \(b=2.38 \times 10^{-2} \mathrm{~L}^{-\mathrm{mol}^{-1}}\). Calculate the apparent volume (in \(\mathrm{pm}^{3}\) ) and radius (in pm) of a helium atom as determined from the van der Waals parameters. (b) Estimate the volume of a helium atom on the basis of its atomic radius. (c) How do these quantities compare? Should they be same? Discuss.

A flask of volume \(5.00 \mathrm{~L}\) is evacuated and \(43.78 \mathrm{~g}\) of solid dinitrogen tetroxide, \(\mathrm{N}_{2} \mathrm{O}_{4}\), is introduced at \(-196^{\circ} \mathrm{C}\). The sample is then warmed to \(25^{\circ} \mathrm{C}\), during which time the \(\mathrm{N}_{2} \mathrm{O}_{4}\) vaporizes and some of it dissociates to form brown \(\mathrm{NO}_{2}\) gas. The pressure slowly increases until it stabilizes at \(2.96\) atm. (a) Write a balanced equation for the reaction. (b) If the gas in the flask at \(25^{\circ} \mathrm{C}\) were all \(\mathrm{N}_{2} \mathrm{O}_{4}\), what would the pressure be? (c) If all the gas in the flask converted into \(\mathrm{NO}_{2}\), what would the pressure be? (d) What are the mole fractions of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) once the pressure stabilizes at \(2.96 \mathrm{~atm}\) ?

The pressure recorded for an argon gas cylinder is \(44.0 \mathrm{lb} \cdot\) inch \(^{-2}\). Convert this pressure into (a) kPa; (b) Torr; (c) bar; (d) atm.

The density of a gaseous compound of phosphorus is \(0.943\) \(\mathrm{g} \cdot \mathrm{L}^{-1}\) at \(420 . \mathrm{K}\) when its pressure is 727 Torr. (a) What is the molar mass of the compound? (b) If the compound remains gaseous, what would be its density at \(1.00 \mathrm{~atm}\) and \(298 \mathrm{~K}\) ?

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