Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 91

As a Boeing 747 gains altitude, the passenger cabin is pressurized. However, the cabin is not pressurized fully to atmospheric $\left(1.01 \times 10^{5} \mathrm{Pa}\right),$ as it would be at sea level, but rather pressurized to \(7.62 \times 10^{4} \mathrm{Pa}\). Suppose a 747 takes off from sea level when the temperature in the airplane is \(25.0^{\circ} \mathrm{C}\) and the pressure is \(1.01 \times 10^{5} \mathrm{Pa} .\) (a) If the cabin temperature remains at \(25.0^{\circ} \mathrm{C},\) what is the percentage change in the number of moles of air in the cabin? (b) If instead, the number of moles of air in the cabin does not change, what would the temperature be?

Short Answer

Expert verified
Also, what is the final temperature if the number of moles of air does not change? Answer: Using the provided steps, the percentage change in the number of moles of air is calculated as \(1 - \frac{P_2}{P_1} = 1 - \frac{7.62 × 10^4 \mathrm{Pa}}{1.01 × 10^5 \mathrm{Pa}} = 0.2465\) or \(24.65\%\). To calculate the final temperature if the number of moles does not change, we rearrange the combined gas law as \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) and solve for \(T_2\).

Step by step solution

01

(Step 1: Calculate the initial pressure and volume of the air inside the cabin)

Since we are given the pressure and temperature of the air inside the cabin, we can use the ideal gas law to calculate the initial volume (assuming an ideal gas): Here, we'll let \(P_1 = 1.01 \times 10^5 \mathrm{Pa},\ T_1 = 25.0^{\circ} \mathrm{C} = 298.15\ \mathrm{K}\).
02

(Step 2: Calculate the final volume of the air inside the cabin given the new pressure at altitude)

Now, we're given that the final pressure in the cabin at altitude is \(7.62 \times 10^4 \mathrm{Pa}\), which we'll label \(P_2\). Since the initial and final conditions of pressure and temperature are given, we can find the final volume of air in the cabin. Because the temperature remains the same, we can apply Boyle's Law, which states the following : \({P_1V_1=P_2V_2}\)
03

(Step 3: Calculate the percentage change in the number of moles of air)

By comparing the initial and final volume, we can calculate the percentage change in the number of moles of air, \((\Delta n)/n\). We can rewrite the ideal gas law as follows: \(\frac{P_1V_1}{RT_1} = n_1\) and \(\frac{P_2V_2}{RT_2} = n_2\) Since temperature remains constant, we can write the relationship between the initial and final number of moles as: \(\frac{\Delta n}{n} = 1 - \frac{n_2}{n_1} = 1 - \frac{P_2}{P_1}\). This can be used to compute the percentage change.
04

(Step 4: Calculate the final temperature if the number of moles of air does not change)

For this part, we assume that the number of moles of air does not change (\(n_1 = n_2\)). So, the ideal gas laws for initial and final states are: \(P_1V_1 = nRT_1\) and \(P_2V_2 = nRT_2\) As temperature changes, we can't apply Boyle's law here. Instead, we have to apply the combined gas law: \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\) Since the number of moles is constant, in this case, we can rearrange the equation and solve for \(T_2\).

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!

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

The mass of \(1 \mathrm{mol}\) of \(^{13} \mathrm{C}\) (carbon- 13 ) is $13.003 \mathrm{g}$. (a) What is the mass in \(u\) of one \(^{13} \mathrm{C}\) atom? (b) What is the mass in kilograms of one \(^{13} \mathrm{C}\) atom?
An iron cannonball of radius \(0.08 \mathrm{m}\) has a cavity of radius $0.05 \mathrm{m}$ that is to be filled with gunpowder. If the measurements were made at a temperature of \(22^{\circ} \mathrm{C}\) how much extra volume of gunpowder, if any, will be required to fill 500 cannonballs when the temperature is \(30^{\circ} \mathrm{C} ?\)
The temperature at which liquid nitrogen boils (at atmospheric pressure) is \(77 \mathrm{K}\). Express this temperature in (a) \(^{\circ} \mathrm{C}\) and (b) \(^{\circ} \mathrm{F}\).
Agnes Pockels \((1862-1935)\) was able to determine Avogadro's number using only a few household chemicals, in particular oleic acid, whose formula is \(\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_{2}\) (a) What is the molar mass of this acid? (b) The mass of one drop of oleic acid is \(2.3 \times 10^{-5} \mathrm{g}\) and the volume is $2.6 \times 10^{-5} \mathrm{cm}^{3} .$ How many moles of oleic acid are there in one drop? (c) Now all Pockels needed was to find the number of molecules of oleic acid. Luckily, when oleic acid is spread out on water, it lines up in a layer one molecule thick. If the base of the molecule of oleic acid is a square of side \(d\), the height of the molecule is known to be \(7 d .\) Pockels spread out one drop of oleic acid on some water, and measured the area to be \(70.0 \mathrm{cm}^{2}\) Using the volume and the area of oleic acid, what is \(d ?\) (d) If we assume that this film is one molecule thick, how many molecules of oleic acid are there in the drop? (e) What value does this give you for Avogadro's number?
Air at room temperature and atmospheric pressure has a mass density of $1.2 \mathrm{kg} / \mathrm{m}^{3} .$ The average molecular mass of air is 29.0 u. How many molecules are in \(1.0 \mathrm{cm}^{3}\) of air?
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Medical Physics

Read Explanation

Electromagnetism

Read Explanation

Translational Dynamics

Read Explanation

Turning Points in Physics

Read Explanation

Torque and Rotational Motion

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