/*! 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 37 (a) Calculate the number of mole... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Calculate the number of molecules in a deep breath of air whose volume is \(2.25 \mathrm{~L}\) at body temperature, \(37^{\circ} \mathrm{C},\) and a pressure of \(97.99 \mathrm{kPa} .(\mathbf{b})\) The adult blue whale has a lung capacity of \(5.0 \times 10^{3} \mathrm{~L}\). Calculate the mass of air (assume an average molar mass of \(28.98 \mathrm{~g} / \mathrm{mol}\) ) contained in an adult blue whale's lungs at \(0.0^{\circ} \mathrm{C}\) and \(101.33 \mathrm{kPa}\), assuming the air behaves ideally.

Short Answer

Expert verified
(a) About \(5.16 \times 10^{22}\) molecules; (b) About 6.49 kg of air.

Step by step solution

01

Understand Ideal Gas Law

To solve these types of problems, we need to use the Ideal Gas Law, which is given by the equation: \[ PV = nRT \] where \( P \) is pressure in pascals (Pa), \( V \) is volume in cubic meters (m³), \( n \) is the number of moles, \( R \) is the universal gas constant (8.314 J/mol·K), and \( T \) is temperature in Kelvin (K).
02

Convert Units for Part (a)

For part (a), convert the volume from liters to cubic meters: \( V = 2.25 \text{ L} = 2.25 \times 10^{-3} \text{ m}^3 \).Convert the temperature from Celsius to Kelvin: \( T = 37 + 273.15 = 310.15 \text{ K} \).Convert the pressure from kPa to Pa: \( P = 97.99 \text{ kPa} = 97990 \text{ Pa} \).
03

Calculate the Number of Moles for Part (a)

Rearrange the Ideal Gas Law to solve for \( n \): \[ n = \frac{PV}{RT} \]Substitute the values: \( n = \frac{97990 \times 2.25 \times 10^{-3}}{8.314 \times 310.15} \). This gives \( n \approx 0.0857 \text{ moles} \).
04

Calculate the Number of Molecules for Part (a)

Use Avogadro's number to find the number of molecules: \[ N = n \times 6.022 \times 10^{23} \]Substitute the value of \( n \) from Step 3: \( N = 0.0857 \times 6.022 \times 10^{23} \), which results in \( N \approx 5.16 \times 10^{22} \) molecules.
05

Convert Units for Part (b)

For part (b), convert the lung capacity from liters to cubic meters: \( V = 5.0 \times 10^3 \text{ L} = 5.0 \text{ m}^3 \).Convert the temperature from Celsius to Kelvin: \( T = 0 + 273.15 = 273.15 \text{ K} \).Convert the pressure from kPa to Pa: \( P = 101.33 \text{ kPa} = 101330 \text{ Pa} \).
06

Calculate the Number of Moles for Part (b)

Rearrange the Ideal Gas Law to solve for \( n \): \[ n = \frac{PV}{RT} \]Substitute the values: \( n = \frac{101330 \times 5.0}{8.314 \times 273.15} \). This gives \( n \approx 223.97 \text{ moles} \).
07

Calculate the Mass of Air for Part (b)

The mass can be calculated using the number of moles and the molar mass: \[ \text{mass} = n \times \text{molar mass} \]Substitute the values: \( \text{mass} = 223.97 \times 28.98 \), which results in \( \text{mass} \approx 6488.23 \text{ g} = 6.49 \text{ kg} \).

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mole calculations
Mole calculations are a fundamental part of chemistry used to determine the number of particles in a given quantity of substance. In our exercise, moles (\(n\)) were calculated using the Ideal Gas Law formula: \[n = \frac{PV}{RT}\].
This formula highlights the relationship between pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and moles (\(n\)) of a gas.
Written out, to find moles, we:
  • Multiply the pressure (\(P\)) of the gas by its volume (\(V\)).
  • Divide by the product of the universal gas constant (\(R = 8.314\) J/mol·K) and the temperature (\(T\).) in Kelvin.
It's important to ensure all units are consistent with the equation—pressure in pascals, volume in cubic meters, and temperature in Kelvin.
For example, if you calculated that there are 0.0857 moles of gas in a deep breath of air, it describes a portion of air containing that number of mole quantities of particles.
Avogadro's number
Avogadro's number is crucial for converting between moles and molecules or atoms. It's defined as \(6.022 \times 10^{23}\), the number of atoms, ions, or molecules in one mole of a substance.
This number allows chemists to convert between the macroscopic scale that can be measured directly (grams, liters) and the microscopic scale of atoms and molecules.
So, to get the total number of molecules from the moles calculated, the equation is straightforward:
  • Using \(N = n \times 6.022 \times 10^{23}\), where \(n\) is the number of moles, you'll accurately find the microscopic representation in numbers of particles.
For part (a) of your exercise, after finding 0.0857 moles of air, you would multiply by Avogadro's number to find there are approximately \(5.16 \times 10^{22}\) molecules.
Unit conversion in chemistry
Unit conversion plays a critical role in chemistry for maintaining consistency and ensuring correct calculations.
In many chemical processes and calculations, converting units is essential as different measurements may be initially given in non-standard units. Here’s a breakdown of common conversions:
  • Volume: Convert from liters to cubic meters by using \(1 \text{L} = 0.001 \text{m}^3\).
  • Pressure: Convert from kilopascals (\(\text{kPa}\)) to pascals (\(\text{Pa}\)) by multiplying by 1,000 (1 \(\text{kPa}\) = 1,000 \(\text{Pa}\)).
  • Temperature: Convert from Celsius to Kelvin by adding 273.15 to the Celsius temperature.
These conversions ensure that when applying the Ideal Gas Law, everything fits together neatly, ensuring the integrity of calculations.
Gas laws applications
The Ideal Gas Law is a powerful tool for solving many practical and theoretical gas-related questions. It describes how gases behave under various conditions of temperature, pressure, and volume and is expressed as \(PV = nRT\).
In real-world applications, understanding gas laws allows predictions and calculations for various situations, such as:
  • Calculating the amount of gas in a container, like lung capacity for living organisms.
  • Understanding changes in conditions, like how temperature changes affect pressure at constant volume.
  • Predicting outcomes in industrial processes, automotive engines, or weather patterns.
In part (b) of the exercise, using the Ideal Gas Law allowed calculating not just the number of moles but also the corresponding mass of air, enabling comprehensive analysis of gas content both in everyday situations and complex systems.

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

Consider a lake that is about \(40 \mathrm{~m}\) deep. A gas bubble with a diameter of 1.0 mm originates at the bottom of a lake where the pressure is \(405.3 \mathrm{kPa}\). Calculate its volume when the bubble reaches the surface of the lake where the pressure is 98 kPa, assuming that the temperature does not change.

(a) If the pressure exerted by ozone, \(\mathrm{O}_{3}\), in the stratosphere is \(304 \mathrm{~Pa}\) and the temperature is \(250 \mathrm{~K}\), how many ozone molecules are in a liter? (b) Carbon dioxide makes up approximately \(0.04 \%\) of Earth's atmosphere. If you collect a \(2.0-\mathrm{L}\) sample from the atmosphere at sea level ( \(101.33 \mathrm{kPa}\) ) on a warm day \(\left(27^{\circ} \mathrm{C}\right),\) how many \(\mathrm{CO}_{2}\) molecules are in your sample?

Consider the sample of gas depicted here. What would the drawing look like if the volume and temperature remained constant while you removed enough of the gas to decrease the pressure by a factor of \(2 ?\) [Section 10.3\(]\) (a) It would contain the same number of molecules. (b) It would contain half as many molecules. (c) It would contain twice as many molecules. (d) There is insufficient data to say.

A sample of \(5.00 \mathrm{~mL}\) of diethylether \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5},\right.\) density \(=0.7134 \mathrm{~g} / \mathrm{mL}\) ) is introduced into a 6.00 -L vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), whose partial pressures are \(P_{\mathrm{N}_{2}}=21.08 \mathrm{kPa}\) and \(P_{\mathrm{O}_{2}}=76.1 \mathrm{kPa}\). The temperature is held at \(35.0^{\circ} \mathrm{C}\), and the diethylether totally evaporates. (a) Calculate the partial pressure of the diethylether. (b) Calculate the total pressure in the container.

A plasma-screen TV contains thousands of tiny cells tilled with a mixture of Xe, Ne, and He gases that emits light of specific wavelengths when a voltage is applied. A particular plasma cell, \(0.900 \mathrm{~mm} \times 0.300 \mathrm{~mm} \times 10.0 \mathrm{~mm}\), contains \(4 \%\) Xe in a 1:1 Ne:He mixture at a total pressure of \(66.66 \mathrm{kPa}\). Calculate the number of Xe, Ne, and He atoms in the cell and state the assumptions you need to make in your calculation.

See all solutions

Recommended explanations on Chemistry Textbooks

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.