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A helium balloon occupies \(8.0 \mathrm{L}\) at \(20^{\circ} \mathrm{C}\) and 1.0 -atm pressure. The balloon rises to an altitude where the air pressure is 0.65 atm and the temperature is \(-10^{\circ} \mathrm{C}\). What's its volume when it reaches equilibrium at the new altitude?

Short Answer

Expert verified
After performing the calculation, the volume when the balloon reaches the new altitude is approximately \(\approx 10.5 \mathrm{L}\).

Step by step solution

01

Convert Temperatures to Kelvin

For gas law problems, temperatures must be in Kelvin. Convert the initial and final temperatures from Celsius to Kelvin by adding 273 to each temperature. Thus, \( T_1 = 20^{\circ} \mathrm{C} + 273 = 293 \mathrm{K} \) and \( T_2 = -10^{\circ} \mathrm{C} + 273 = 263 \mathrm{K} \). Always remember that temperature in any gas law should always be in Kelvin.
02

Identify Given Values and Required Value

From the problem, the initial pressure and volume are \( P_1 = 1.0 \mathrm{atm} \) and \( V_1 = 8.0 \mathrm{L} \) respectively, and the final pressure is \( P_2 = 0.65 \mathrm{atm} \). The volume at the new altitude \( V_2 \) is what we need to find.
03

Use the Gas Law to Find New Volume

Substitute all the known values into the gas law equation \( P_1V_1/T_1 = P_2V_2/T_2 \) and solve for \( V_2 \). After rearranging the equation to find \( V_2 \), it becomes \( V_2 = P_1V_1T_2 / (P_2T_1) \). Plugging in all the values we have \( V_2 = (1.0 \mathrm{atm} * 8.0 \mathrm{L} * 263 \mathrm{K})/(0.65 \mathrm{atm} * 293 \mathrm{K}) \).

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Key Concepts

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

Pressure and Volume Relationship
In gases, the relationship between pressure and volume is governed by Boyle's Law, which is part of the Ideal Gas Law. This relationship can be understood with a simple idea: when pressure increases, the volume decreases, and vice versa. This inverse relationship assumes that the temperature and the amount of gas stay constant.
To illustrate, imagine a balloon. Squeeze it, and it takes up less space—its volume decreases because the pressure applied increases. On the other hand, if you release your grip, the pressure decreases, allowing the balloon to expand and the volume to increase.
  • If you know the initial volume and pressure of a gas, and the final pressure, you can find the final volume using the equation: \[ P_1V_1 = P_2V_2 \]
  • Remember, the temperatures must remain the same throughout the process for Boyle's Law to apply directly.
In the balloon's case, as it rises, the air pressure decreases, which means the balloon can expand to a larger volume.
Temperature Conversion to Kelvin
When working with gases, it is crucial to express temperatures in Kelvin. This is because the Kelvin scale is an absolute scale that starts at absolute zero, the lowest possible temperature where all molecular motion stops. Hence, Kelvin is compatible with the mathematical equations used in gas laws.
To convert from Celsius to Kelvin, simply add 273 to the temperature in Celsius. This conversion is straightforward and vital for ensuring correct calculations.
  • For example, if you have a temperature of \(20^{\circ} \mathrm{C}\), convert it to Kelvin by calculating \(20 + 273 \), which equals \(293 \mathrm{K}\).
  • Similarly, \(-10^{\circ} \mathrm{C}\) would convert to \(263 \mathrm{K}\).
Using Kelvin helps maintain the integrity of the equations because it provides a true measure of the thermal energy present in the gas.
Altitude Effects on Gases
The behavior of gases changes with altitude due to variations in atmospheric pressure and temperature. At higher altitudes, the air pressure is lower because there is less air above that part of the atmosphere. This decrease in pressure allows gases to expand if the temperature conditions also change appropriately.
Temperature often decreases with altitude too, affecting gas volume according to the Ideal Gas Law, which combines Boyle's, Charles', and Avogadro's laws into a single equation: \[ PV = nRT \]where:
  • \(P\) is the pressure
  • \(V\) is the volume
  • \(n\) is the number of moles of gas
  • \(R\) is the ideal gas constant
  • \(T\) is the temperature in Kelvin
As a balloon ascends and the air becomes thinner, the temperature drop can counteract some of the expansion from lower pressures, depending on the exact conditions. Understanding these concepts allows us to predict changes in gas volume as altitude changes, a crucial idea if you're ever dealing with gas-filled objects like balloons, or high-altitude weather phenomena.

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Most popular questions from this chapter

Your professor asks you to order a tank of argon gas for a lab experiment. You obtain a "type \(\mathrm{C}^{\prime \prime}\) gas cylinder with interior volume 6.88 L. The supplier claims it contains 45 mol of argon. You measure its pressure to be 14 MPa at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) Did you get what you paid for?

If a 1 -megaton nuclear bomb were exploded deep in the Greenland ice cap, how much ice would it melt? Assume the ice is initially at about its freezing point, and consult Appendix C for the appropriate energy conversion.

You're on a team planning a mission to Venus to collect atmospheric samples for analysis. The design specs call for a \(1-\mathrm{L}\) sample container, while the scientists want at least 1 mol of gas. Venus's atmospheric pressure is 90 times that of Earth's, and its average temperature is \(730 \mathrm{K}\). Will the design work?

The Intergovernmental Panel on Climate Change estimates that Greenland is losing ice, as a result of global warming, at approximately 250 Pg/year. (a) Find the energy needed to melt 250 Pg of ice. (b) Greenland's ice melt results most immediately from an imbalance between incoming and outgoing energy-an imbalance created largely by the absorption of infrared radiation by human-produced greenhouse gases. Use your answer to part (a) to express Greenland's energy imbalance in watts per square meter of the Greenland ice sheet's surface area. That your result is larger than the global imbalance of somewhat less than \(1 \mathrm{W} / \mathrm{m}^{2}\) shows that the impact of global warming is greater in the Arctic.

A solar-heated house stores energy in 5.0 tons of Glauber salt ( \(\left.\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}\right),\) which melts at \(90^{\circ} \mathrm{F}\). The heat of fusion of Glauber salt is 104 Btu/lb and the specific heats of the solid and liquid are, respectively, \(0.46 \mathrm{Btu} / \mathrm{lb} \cdot^{\circ} \mathrm{F}\) and \(0.68 \mathrm{Btu} / \mathrm{b} \cdot^{\circ} \mathrm{F}\). After a week of sunny weather, the storage medium is all liquid at \(95^{\circ} \mathrm{F}\). Then comes a cloudy period during which the house loses heat at an average of \(20,000 \mathrm{Btu} / \mathrm{h}\). (a) How long is it before the temperature of the storage medium drops below \(60^{\circ} \mathrm{F} ?\) (b) How much of this time is spent at \(90^{\circ} \mathrm{F} ?\)

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