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A balloon containing \(5.0 \mathrm{dm}^{3}\) of gas at \(14^{\circ} \mathrm{C}\) and \(100.0\) kPa rises to an altitude of \(2000 . \mathrm{m}\), where the temperature is \(20^{\circ} \mathrm{C}\). The pressure of gas in the balloon is now \(79.0 \mathrm{kPa}\). What is the volume of gas in the balloon?

Short Answer

Expert verified
The volume of the gas in the balloon at 2000m is approximately 6.398 dm³.

Step by step solution

01

Understand the problem

We need to find the new volume of the gas in the balloon at higher altitude, given changes in temperature and pressure. We will use the Combined Gas Law for this purpose.
02

Write down the Combined Gas Law equation

The Combined Gas Law equation is given by: \[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \] where \(P\) represents pressure, \(V\) is volume, and \(T\) is the temperature in Kelvin.
03

Convert Celsius temperatures to Kelvin

Convert the given temperatures from degrees Celsius to Kelvin using the conversion formula: \[ T(K) = T(°C) + 273.15 \]For the initial condition: \[ T_1 = 14 + 273.15 = 287.15 \ K \]For the final condition: \[ T_2 = 20 + 273.15 = 293.15 \ K \]
04

Assign the initial and final values

Assign the initial conditions:- \(P_1 = 100.0 \ \text{kPa}\)- \(V_1 = 5.0 \ \text{dm}^3\)- \(T_1 = 287.15 \ \text{K}\)And the final conditions:- \(P_2 = 79.0 \ \text{kPa}\)- \(T_2 = 293.15 \ \text{K}\)
05

Rearrange the Combined Gas Law to solve for V2

Rearrange the equation to solve for the final volume \(V_2\):\[ V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1} \]
06

Insert the known values and solve for V2

Substitute the known values into the rearranged equation:\[V_2 = \frac{100.0 \times 5.0 \times 293.15}{79.0 \times 287.15}\]Calculate the result:\[ V_2 \approx 6.398 \ \text{dm}^3 \]
07

Interpret the result

The calculated volume \(V_2\) indicates the volume of the gas in the balloon at the altitude of 2000 meters.

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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 a set of relationships that describe how gases behave under various conditions of pressure, volume, and temperature. One of the key equations combining these variables is the Combined Gas Law. The Combined Gas Law is particularly useful when dealing with a gas under two different sets of conditions. The expression for the Combined Gas Law is:\[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \] In this equation, \(P\) represents the pressure, \(V\) is the volume, and \(T\) is the temperature measured in Kelvin. This law arises from a combination of Boyle's Law, Charles's Law, and Gay-Lussac's Law, which account for the interdependence of pressure, volume, and temperature.
  • Boyle's Law: It states that the pressure of a gas is inversely proportional to its volume at constant temperature.
  • Charles's Law: It states that the volume of a gas is directly proportional to its temperature at constant pressure.
  • Gay-Lussac's Law: It states that the pressure of a gas is directly proportional to its temperature at constant volume.
These relationships allow us to predict how a gas will respond under changing conditions of pressure, volume, and temperature and answer practical problems such as predicting the volume of a balloon at different altitudes.
Temperature Conversion
Temperature conversion is essential when working with gas laws, as gas laws require temperatures to be expressed in Kelvin. Celsius can be converted to Kelvin using the formula:\[ T(K) = T(°C) + 273.15 \]Kelvin is the SI unit of temperature and starts at absolute zero, the point at which molecular motion ceases. Because many scientific calculations, like those involving gas laws, rely on an absolute scale, Kelvin is preferred over Celsius.To illustrate:
  • Convert an initial temperature of \(14^\circ C\): \[ T_1 = 14 + 273.15 = 287.15 \ K \]
  • Convert a final temperature of \(20^\circ C\): \[ T_2 = 20 + 273.15 = 293.15 \ K \]
Converting temperatures from Celsius to Kelvin ensures that we accommodate the concept of absolute values in our calculations, thus preserving the mathematical integrity of gas law equations.
Gas Volume Calculation
Calculating the volume of a gas under varying conditions is a practical application of the Combined Gas Law. This involves solving for the new volume \(V_2\) given the changes in pressure and temperature.Starting with the Combined Gas Law:\[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \]To find \(V_2\), rearrange to:\[ V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1} \]
  • Step 1: Insert the initial conditions: \(P_1 = 100.0 \ kPa\), \(V_1 = 5.0 \ dm^3\), \(T_1 = 287.15 \ K\).
  • Step 2: Insert the final conditions: \(P_2 = 79.0 \ kPa\), \(T_2 = 293.15 \ K\).
  • Step 3: Solve for \(V_2\): \[ V_2 = \frac{100.0 \times 5.0 \times 293.15}{79.0 \times 287.15} \]
  • Result: \[ V_2 \approx 6.398 \ dm^3 \]
The calculation reveals the final volume of the balloon's gas at a new altitude and temperature. This example demonstrates how understanding pressure, volume, and temperature interrelations allows us to predict changes in a gas’s behavior.

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Most popular questions from this chapter

Two identical He-filled balloons, each with a volume of \(20 \mathrm{~L}\), are allowed to rise into the atmosphere. One rises to an altitude of \(3000 \mathrm{~m}\) while the other rises to \(6000 \mathrm{~m}\). a. Assuming that the balloons are at the same temperature, which balloon has the greater volume? b. What information would you need in order to calculate the volume of each of the balloons at their respective heights?

Butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is an easily liquefied gaseous fuel. Calculate the density of butane gas at \(0.897\) atm and \(24^{\circ} \mathrm{C}\). Give the answer in grams per liter

What is the ratio of rates of effusion of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) under the same conditions?

A rigid \(1.0\) - \(\mathrm{L}\) container at \(75^{\circ} \mathrm{C}\) is fitted with a gas pressure gauge. A \(1.0\) -mol sample of ideal gas is introduced into the container. a. What would the pressure gauge in the container be reading in \(\mathrm{mmHg}\) ? b. Describe the interactions in the container that are causing the pressure. c. Say the temperature in the container were increased to \(150^{\circ} \mathrm{C}\). Describe the effect this would have on the pressure, and, in terms of kinetic theory, explain why this change occurred.

A vessel containing \(39.5 \mathrm{~cm}^{3}\) of helium gas at \(25^{\circ} \mathrm{C}\) and \(106 \mathrm{kPa}\) was inverted and placed in cold ethanol. As the gas contracted, ethanol was forced into the vessel to maintain the same pressure of helium. If this required \(7.7 \mathrm{~cm}^{3}\) of ethanol, what was the final temperature of the helium?

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