/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 A hot-air balloon is filled with... [FREE SOLUTION] | 91Ó°ÊÓ

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A hot-air balloon is filled with air to a volume of \(4.00 \times\) \(10^{3} \mathrm{m}^{3}\) at \(745\) torr and \(21^{\circ} \mathrm{C}\). The air in the balloon is then heated to \(62^{\circ} \mathrm{C},\) causing the balloon to expand to a volume of \(4.20 \times 10^{3} \mathrm{m}^{3} .\) What is the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon? (Hint: Openings in the balloon allow air to flow in and out. Thus the pressure in the balloon is always the same as that of the atmosphere.)

Short Answer

Expert verified
The ratio of the number of moles of air in the heated balloon to the original number of moles is approximately 0.949.

Step by step solution

01

Identify the given information and convert to SI units

We are given the initial volume (V1), initial temperature (T1), final volume (V2), final temperature (T2), and atmospheric pressure (P), which will let us use the ideal gas law formula nRT = PV. First, we need to convert the pressure to SI units and the temperatures to kelvin. Initial volume: V1 = 4.00 x 10³ m³ Final volume: V2 = 4.20 x 10³ m³ Pressure: P = 745 torr Initial temperature: T1 = 21°C Final temperature: T2 = 62°C Remember that 1 atm = 760 torr, and we also need to convert the pressure to pascals (Pa) for which 1 atm = 101325 Pa. Also, remember that to convert from Celsius to Kelvin, we need to add 273.15 to the Celsius value.
02

Convert the pressure and temperature

Let's convert the pressure and temperature values to their SI units: P (in atm) = 745 torr × (1 atm / 760 torr) ≈ 0.98 atm P (in Pa) = 0.98 atm × 101325 Pa/atm ≈ 99278 Pa T1 (in K) = 21°C + 273.15 = 294.15 K T2 (in K) = 62°C + 273.15 = 335.15 K Now we have P = 99278 Pa, T1 = 294.15 K, and T2 = 335.15 K.
03

Use the ideal gas law to relate number of moles and volume

Since the pressure is constant inside and outside the balloon, we can relate the initial and final number of moles (n1 and n2) and the volumes (V1 and V2) using the ideal gas law formula. The universal gas constant (R) is given by R = 8.314 J/(mol K). For the initial state, n1 * R * T1 = P * V1, and for the final state, n2 * R * T2 = P * V2. Divide the first equation by the second equation: (n1 * R * T1) / (n2 * R * T2) = (P * V1) / (P * V2) Because the pressure is constant, we can cancel out R, P, and T: n1 / n2 = (V1 / T1) / (V2 / T2) Now we can find the ratio of the number of moles of air in the heated balloon to the original number of moles.
04

Calculate the ratio of the number of moles

Plug in the given values for volumes and temperatures into the formula we got in step 3 and calculate the ratio. n1 / n2 = (V1 / T1) / (V2 / T2) = (4.00 x 10³ m³ / 294.15 K) / (4.20 x 10³ m³ / 335.15 K) n1 / n2 ≈ 0.949 The ratio of the number of moles of air in the heated balloon to the original number of moles is approximately 0.949.

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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 to solving problems related to the behavior of gases under different physical conditions. One of the foundational principles is the ideal gas law, which describes how the pressure (P), volume (V), temperature (T), and number of moles (n) of a gas are related under ideal conditions. The ideal gas law is expressed mathematically as \( PV = nRT \), where R is the universal gas constant.

This law proves to be invaluable when determining the state of a gas when any of the other state variables are altered, as seen in the exercise with the hot-air balloon. As long as the gas behaves ideally, which means it follows the kinetic molecular theory, the ideal gas law allows us to predict how changes in volume, temperature, and the number of moles (if the balloon is open) will affect pressure, or vice versa.
Molar Volume
Molar volume is another key concept, particularly when dealing with gases. It refers to the volume occupied by one mole of a substance. In the context of gases, the molar volume at standard temperature and pressure (STP, which is 273.15 K and 1 atm) is \(22.4 L/mol\) for an ideal gas.

In our balloon scenario, understanding molar volume helps explain the relationship between the number of moles of air in the balloon and the balloon's volume at different temperatures. Since the ideal gas law assumes the gas molecules do not interact and take up no volume themselves, variations in real gas behavior may occur at high pressures or low temperatures. However, for our hypothetical balloon and the conditions given, we can assume ideal behavior for a straightforward calculation.
Temperature Conversion
When working with the ideal gas law, it is crucial to convert temperature to an absolute scale. Kelvin (K) is the SI unit for temperature used in gas law equations. To convert Celsius to Kelvin, add 273.15 to the Celsius value.

In our exercise, we are given temperatures in Celsius: T1 at \(21^\circ C\) and T2 at \(62^\circ C\). To properly apply the ideal gas law, we convert these to Kelvin by adding 273.15, giving us T1 as \(294.15 K\) and T2 as \(335.15 K\). This step is fundamental, as temperature directly affects the kinetic energy of the gas particles and, by extension, the pressure and volume.
Pressure Units Conversion
In dealing with gases, pressure measurements come in various units, including atmospheres (atm), torr, and pascals (Pa). It is essential to convert pressures to the same units when conducting calculations. Since the SI unit for pressure is pascals, we often need to convert atm or torr to Pa.

For our exercise, pressure was originally given in torr. To convert it to atm, we used the relationship \(1 atm = 760 torr\). Furthermore, to get the pressure in pascals, we used \(1 atm = 101325 Pa\). These conversions allowed us to use \(99278 Pa\) as the standard pressure in our application of the ideal gas law, ensuring consistency and accuracy in our calculations.

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Most popular questions from this chapter

You have a sealed, flexible balloon filled with argon gas. The atmospheric pressure is \(1.00\) atm and the temperature is \(25^{\circ} \mathrm{C}\). Assume that air has a mole fraction of nitrogen of \(0.790\), the rest being oxygen. a. Explain why the balloon would float when heated. Make sure to discuss which factors change and which remain constant, and why this matters. b. Above what temperature would you heat the balloon so that it would float?

An organic compound contains \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{O} .\) Combustion of \(0.1023 \mathrm{g}\) of the compound in excess oxygen yielded \(0.2766 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0991 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) A sample of \(0.4831 \mathrm{g}\) of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129 ). At \(\mathrm{STP}, 27.6 \mathrm{mL}\) of dry \(\mathrm{N}_{2}\) was obtained. In a third experiment, the density of the compound as a gas was found to be \(4.02 \mathrm{g} / \mathrm{L}\) at \(127^{\circ} \mathrm{C}\) and \(256\) torr. What are the empirical and molecular formulas of the compound?

The total volume of hydrogen gas needed to fill the Hindenburg was \(2.0 \times 10^{8} \mathrm{L}\) at 1.0 atm and \(25^{\circ} \mathrm{C}\). Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(l)\) is \(-286 \mathrm{kJ} / \mathrm{mol},\) how much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted to form water?

Consider an equimolar mixture (equal number of moles) of two diatomic gases \(\left(A_{2} \text { and } B_{2}\right)\) in a container fitted with a piston. The gases react to form one product (which is also a gas) with the formula \(A_{x} B_{y}\). The density of the sample after the reaction is complete (and the temperature returns to its original state) is \(1.50\) times greater than the density of the reactant mixture. a. Specify the formula of the product, and explain if more than one answer is possible based on the given data. b. Can you determine the molecular formula of the product with the information given or only the empirical formula?

Which statement best explains why a hot-air balloon rises when the air in the balloon is heated? a. According to Charles's law, the temperature of a gas is directly related to its volume. Thus the volume of the balloon increases, making the density smaller. This lifts the balloon. b. Hot air rises inside the balloon, and this lifts the balloon. c. The temperature of a gas is directly related to its pressure. The pressure therefore increases, and this lifts the balloon. d. Some of the gas escapes from the bottom of the balloon, thus decreasing the mass of gas in the balloon. This decreases the density of the gas in the balloon, which lifts the balloon. e. Temperature is related to the root mean square velocity of the gas molecules. Thus the molecules are moving faster, hitting the balloon more, and thus lifting the balloon. Justify your choice, and for the choices you did not pick, explain what is wrong with them.

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