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Chapter 6: Problem 37

A \(100.0\) -mL bubble of hot gases at \(225^{\circ} \mathrm{C}\) and \(1.80 \mathrm{~atm}\) is emitted from an active volcano. What is the new volume, in milliliters, of the bubble outside the volcano where the temperature is \(-25^{\circ} \mathrm{C}\) and the pressure is \(0.80 \mathrm{~atm}\), if the amount of gas remains the same?

Short Answer

Expert verified
The new volume is approximately 112.1 mL.

Step by step solution

01

Identify the known variables

First, identify the given values: Initial volume: 100 mL Initial temperature: \(225^{\text{o}} \text{C} + 273.15 = 498.15 \text{K}\) Initial pressure: 1.80 atm Final temperature: \(-25^{\text{o}} \text{C} + 273.15 = 248.15 \text{K}\) Final pressure: 0.80 atm
02

Use the Combined Gas Law

The Combined Gas Law relates pressure, volume, and temperature: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). We need to solve for the final volume \(V_2\).
03

Rearrange the equation to solve for \(V_2\)

Rearrange the formula to solve for \(V_2\): \(V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}\)
04

Substitute the values into the equation

Substitute the known values into the equation: \(V_2 = \frac{1.80 \text{ atm} \times 100 \text{ mL} \times 248.15 \text{ K}}{498.15 \text{ K} \times 0.80 \text{ atm}}\)
05

Calculate the final volume

Perform the calculation: \(V_2 = \frac{1.80 \times 100 \times 248.15}{498.15 \times 0.80}\) \( = \frac{44667}{398.52}\) \(\text{鈮� 112.1 mL}\)

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Key Concepts

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

Gas Laws
The gas laws are fundamental principles used to describe how gases behave. They relate the pressure, volume, and temperature of a gas. One key law is Boyle's Law, which states that pressure and volume are inversely related at a constant temperature. Another is Charles's Law, which says that volume and temperature are directly related at constant pressure. The Combined Gas Law integrates these relationships into a single equation, \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\), allowing us to calculate changes in gas properties under varying conditions.

Understanding these laws simplifies the study of gases by providing a clear methodology to predict their behavior. Always remember to use Kelvin for temperature calculations to avoid incorrect results.
Pressure-Volume-Temperature Relationship
The relationship between pressure, volume, and temperature is crucial in understanding gas behavior. The Combined Gas Law is \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). This equation helps us understand how changing one or more of these properties affects the others. For example:
  • Increasing the temperature of a gas, while keeping pressure constant, will increase its volume.
  • Increasing the pressure, while keeping temperature constant, will decrease volume.
  • Cooling a gas decreases its volume if pressure is constant.

In the original exercise, we used this relationship to find the final volume of a gas bubble emitted from a volcano. By manipulating the equation, we can solve for the unknown property when given the others. This is especially useful in real-world applications like weather balloons and engineering calculations.
Volcano Gases
Volcanoes emit various gases, often under high pressure and temperature. Common gases include water vapor, carbon dioxide, and sulfur dioxide. These gases' behavior can be predicted and analyzed using gas laws. When a volcano emits gases, they often undergo significant changes in pressure and temperature as they escape into the atmosphere.

For instance, in the exercise, the gas bubble originally inside the volcano at high pressure (1.80 atm) and temperature (225掳C) expanded as it reached lower pressure (0.80 atm) and temperature (-25掳C) conditions outside. Calculating the final volume of such a bubble can help scientists predict eruption impacts and forecast the dispersion of volcanic gases.

Understanding these principles is not only essential for academic purposes but also crucial for practical applications in geology and environmental science.

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Most popular questions from this chapter

Two flasks of equal volume and at the same temperature contain different gases. One flask contains \(1.00 \mathrm{~g}\) of \(\mathrm{Ne}\), and the other flask contains \(1.00 \mathrm{~g}\) of He. Which of the following statements are correct? Explain your answers. a. Both flasks contain the same number of atoms. b. The pressures in the flasks are the same. c. The flask that contains helium has a higher pressure than the flask that contains neon. d. The densities of the gases are the same.

A weather balloon has a volume of \(750 \mathrm{~L}\) when filled with helium at \(8{ }^{\circ} \mathrm{C}\) at a pressure of 380 torr. What is the new volume of the balloon, where the pressure is \(0.20 \mathrm{~atm}\) and the temperature is \(-45^{\circ} \mathrm{C}\) ?

Solve for the new pressure, in atm, for each of the following, if \(n\) and \(V\) are constant: a. A gas with an initial pressure of \(1.20\) atm at \(75^{\circ} \mathrm{C}\) is cooled to \(-32^{\circ} \mathrm{C}\). b. A sample of \(\mathrm{N}_{2}\) with an initial pressure of \(780 . \mathrm{mmHg}\) at \(-75^{\circ} \mathrm{C}\) is heated to \(28^{\circ} \mathrm{C}\).

Indicate whether the final volume in each of the following is the same, larger, or smaller than the initial volume, if pressure and amount of gas do not change: a. A volume of \(505 \mathrm{~mL}\) of air on a cold winter day at \(-15^{\circ} \mathrm{C}\) is breathed into the lungs, where body temperature is \(37^{\circ} \mathrm{C}\). b. The heater used to heat the air in a hot-air balloon is turned off. c. A balloon filled with helium at the amusement park is left in a car on a hot day.

An accident to the head can affect the ability of a person to ventilate (breathe in and out). a. What would happen to the partial pressures of oxygen and carbon dioxide in the blood if a person cannot properly ventilate? b. When a person who cannot breathe properly is placed on a ventilator, an air mixture is delivered at pressures that are alternately above the air pressure in the person's lung, and then below. How will this move oxygen gas into the lungs, and carbon dioxide out?
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