Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 59

The air in a hot-air balloon at 744 torr is heated from \(17^{\circ} \mathrm{C}\) to \(60.0^{\circ} \mathrm{C}\). Assuming that the amount (mol) of air and the pressure remain constant, what is the density of the air at each temperature? (The average molar mass of air is \(28.8 \mathrm{~g} / \mathrm{mol} .)\)

Short Answer

Expert verified
At 17°C, the density is approximately 1.18 g/L. At 60°C, the density is approximately 1.03 g/L.

Step by step solution

01

Convert Temperatures to Kelvin

First, convert the given temperatures from Celsius to Kelvin using the formula: \[ T(K) = T(^{\text{C}}) + 273.15 \] At 17^{\text{C}}, the temperature in Kelvin is: \[ T_1 = 17 + 273.15 = 290.15 \text{ K} \] At 60^{\text{C}}, the temperature in Kelvin is: \[ T_2 = 60 + 273.15 = 333.15 \text{ K} \]
02

Apply the Ideal Gas Law (Initial Condition)

Use the ideal gas law to find the initial density of air at 17^{\text{C}}. The ideal gas law is: \[ PV = nRT \] Rearrange to find density (\( \rho \)): \[ \rho = \frac{MP}{RT} \] where \( M \) is the molar mass, \( P \) is the pressure, \( R \) is the gas constant (0.0821 L·atm/(mol·K)), and \( T \) is the temperature.Given: Pressure (P) = 744 torr (convert to atm: 744 torr / 760 torr/atm = 0.979 atm) Molar Mass (M) = 28.8 g/mol Temperature (T_1) = 290.15 K\[ \rho_1 = \frac{28.8 \times 0.979}{0.0821 \times 290.15} \approx 1.18 \ g/L \]
03

Apply the Ideal Gas Law (Final Condition)

Now find the density of air at 60^{\text{C}}. Using the same formula: \[ \rho = \frac{MP}{RT} \] Given: Pressure (P) = 0.979 atm Molar Mass (M) = 28.8 g/mol Temperature (T_2) = 333.15 K\[ \rho_2 = \frac{28.8 \times 0.979}{0.0821 \times 333.15} \approx 1.03 \ g/L \]
04

Final Answer

The densities of air are: At 17^{\text{C}}, the density is approximately \( 1.18 \ g/L \). At 60^{\text{C}}, the density is approximately \( 1.03 \ g/L \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Density
Density is a measure of how much mass is contained within a given volume. It’s usually expressed in grams per liter (g/L) or kilograms per cubic meter (kg/m³).
For a gas, density can be affected by pressure, temperature, and the type of gas itself. In general, increased temperature decreases density - this is why hot air balloons rise when the air inside them is heated!

In calculation, we use the formula: \(\rho = \frac{M \times P}{R \times T}\),
where \(M\) is molar mass, \(P\) is pressure, \(R\) is the gas constant, and \(T\) is temperature.
Ideal Gas Law
The ideal gas law is a fundamental equation that describes the relationship between pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas. The formula is: \[ PV = nRT \]
Here:
* \(P\) is the pressure
* \(V\) is the volume
* \(n\) is the number of moles
* \(R\) is the universal gas constant (\(0.0821 \textrm{L} \textrm{​atm} /\textrm{mol} \textrm{​ K}\)
* \(T\) is the temperature in Kelvin

We can rearrange this equation to find density by using the fact that \(n\) (moles) is equal to the mass (m) divided by the molar mass (M): \[ \rho = \frac{MP}{RT} \]
This is exactly what we did in the exercise to find the density at two different temperatures.
Temperature Conversion
In many scientific calculations, the temperature needs to be in Kelvin (K), not Celsius (°C). The conversion formula from Celsius to Kelvin is simple:
\(T(\textrm{K}) = T(\textrm{°C}) + 273.15\)

For example:
* 17°C = 17 + 273.15 = 290.15 K
* 60°C = 60 + 273.15 = 333.15 K
This conversion is crucial because the Kelvin scale is based on absolute zero, the theoretical point where particles have minimum thermal motion.
Pressure Units
Pressure is the force exerted per unit area. It can be expressed in different units, including atmospheres (atm), torr, and pascals (Pa).

In the problem, the pressure is given in torr, but we need it in atmospheres for the ideal gas law calculation. The conversion factor is:
\textrm{1 atm} = \textrm{760 torr} \.

Therefore:
744 torr can be converted to atmospheres with:
\(\frac{744 \text{ torr}}{760 \text{ torr/atm}} \approx 0.979 \text{ atm} \)
This helps standardize measurements, making calculations easier.
Molar Mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol).

For air, the average molar mass is given as 28.8 g/mol.
This value represents the weighted average of the molar masses of the different gases that make up air, such as nitrogen, oxygen, and argon.

The molar mass is used in the ideal gas law to calculate other properties like density.
We can apply it as:
\( \rho = \frac{M \times P} {R \times T}\),
where Molar Mass (M) is essential for determining the mass in this context.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 355 -mL container holds 0.146 g of Ne and an unknown amount of Ar at \(35^{\circ} \mathrm{C}\) and a total pressure of \(626 \mathrm{mmHg}\). Calculate the number of moles of Ar present.

A balloon filled with \(1.92 \mathrm{~g}\) of helium has a volume of \(12.5 \mathrm{~L}\) What is the balloon's volume after \(0.850 \mathrm{~g}\) of helium has leaked out through a small hole (assume constant pressure and temperature)?

In preparation for a combustion demonstration, a professor fills a balloon with equal molar amounts of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2},\) but the demonstration has to be postponed until the next day. During the night, both gases leak through pores in the balloon. If \(35 \%\) of the \(\mathrm{H}_{2}\) leaks, what is the \(\mathrm{O}_{2} / \mathrm{H}_{2}\) ratio in the balloon the next day?

The Hawaian volcano Kilauea emits an average of \(1.5 \times 10^{3} \mathrm{~m}^{3}\) of gas each day, when corrected to \(298 \mathrm{~K}\) and \(1.00 \mathrm{~atm} .\) The mixture contains gases that contribute to global warming and acid rain, and some are toxic. An atmospheric chemist analyzes a sample and finds the following mole fractions: \(0.4896 \mathrm{CO}_{2}, 0.0146 \mathrm{CO}, 0.3710 \mathrm{H}_{2} \mathrm{O}\) \(0.1185 \mathrm{SO}_{2}, 0.0003 \mathrm{~S}_{2}, 0.0047 \mathrm{H}_{2}, 0.0008 \mathrm{HCl},\) and \(0.0003 \mathrm{H}_{2} \mathrm{~S} .\) How many metric tons (t) of each gas are emitted per year ( \(1 \mathrm{t}=1000 \mathrm{~kg}\) )?

What is the effect of the following on the volume of \(1 \mathrm{~mol}\) of an ideal gas? (a) Temperature decreases from \(800 \mathrm{~K}\) to \(400 \mathrm{~K}\) (at constant \(P\) ). (b) Temperature increases from \(250^{\circ} \mathrm{C}\) to \(500^{\circ} \mathrm{C}\) (at constant \(P\) ). (c) Pressure increases from 2 atm to 6 atm (at constant \(T\) ).
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free