Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 49

A hydrogen balloon at Earth's surface has a volume of \(5.0 \mathrm{m}^{3}\) on a day when the temperature is \(27^{\circ} \mathrm{C}\) and the pressure is \(1.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2} .\) The balloon rises and expands as the pressure drops. What would the volume of the same number of moles of hydrogen be at an altitude of \(40 \mathrm{km}\) where the pressure is \(0.33 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}\) and the temperature is \(-13^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The volume of the hydrogen balloon at 40 km altitude is approximately 14,995.75 m³.

Step by step solution

01

Convert temperature to Kelvin

Before using the combined gas law formula, we need to convert the temperatures to Kelvin. To do so, we add 273.15 to the Celsius temperature. Initial Temperature (T1) = \(27^{\circ} \mathrm{C} + 273.15 = 300.15 \mathrm{K}\) Final Temperature (T2) = \(-13^{\circ} \mathrm{C} + 273.15 = 260.15 \mathrm{K}\)
02

Plug in the values into the combined gas law formula

We will now input the given values and the temperatures in Kelvin into the combined gas law formula: (P1 * V1) / T1 = (P2 * V2) / T2 where P1 = \(1.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) V1 = \(5.0 \mathrm{m}^{3}\) T1 = \(300.15 \mathrm{K}\) P2 = \(0.33 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}\) T2 = \(260.15 \mathrm{K}\)
03

Solve for V2

Now, we will solve for V2 (the final volume at 40 km altitude). Rearrange the combined gas law formula to solve for V2: V2 = (P1 * V1 * T2) / (P2 * T1) Plug in the values: V2 = (\((1.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2})\) * \((5.0 \mathrm{m}^{3})\) * \((260.15 \mathrm{K})\)) / (\((0.33 \times 10^{3} \mathrm{N} / \mathrm{m}^{2})\) * \((300.15 \mathrm{K})\)) Now, calculate V2: V2 ≈ \(14,995.75 \mathrm{m}^{3}\) So, the volume of the hydrogen balloon at 40 km altitude is approximately \(14,995.75 \mathrm{m}^{3}\).

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

A flat square of side \(s_{0}\) at temperature \(T_{0}\) expands by \(\Delta s\) in both length and width when the temperature increases by \(\Delta T\). The original area is \(s_{0}^{2}=A_{0}\) and the final area is $\left(s_{0}+\Delta s\right)^{2}=A .\( Show that if \)\Delta s \ll s_{0}$$$\frac{\Delta A}{A_{0}}=2 \alpha \Delta T$$(Although we derive this relation for a square plate, it applies to a flat area of any shape.)
In plants, water diffuses out through small openings known as stomatal pores. If \(D=2.4 \times 10^{-5} \mathrm{m}^{2} / \mathrm{s}\) for water vapor in air, and the length of the pores is \(2.5 \times 10^{-5} \mathrm{m},\) how long does it take for a water molecule to diffuse out through the pore?
An emphysema patient is breathing pure \(\mathrm{O}_{2}\) through a face mask. The cylinder of \(\mathrm{O}_{2}\) contains \(0.60 \mathrm{ft}^{3}\) of \(\mathrm{O}_{2}\) gas at a pressure of \(2200 \mathrm{lb} / \mathrm{in}^{2} .\) (a) What volume would the oxygen occupy at atmospheric pressure (and the same temperature)? (b) If the patient takes in \(8 \mathrm{L} / \mathrm{min}\) of \(\mathrm{O}_{2}\) at atmospheric pressure, how long will the cylinder last?
(a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees? (b) At what temperature (if any) does the numerical value of kelvins equal the numerical value of Fahrenheit degrees?
A cylinder in a car engine takes \(V_{i}=4.50 \times 10^{-2} \mathrm{m}^{3}\) of air into the chamber at \(30^{\circ} \mathrm{C}\) and at atmospheric pressure. The piston then compresses the air to one-ninth of the original volume \(\left(0.111 \mathrm{V}_{\mathrm{i}}\right)\) and to 20.0 times the original pressure \(\left(20.0 P_{\mathrm{i}}\right) .\) What is the new temperature of the air?
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Turning Points in Physics

Read Explanation

Physics of Motion

Read Explanation

Geometrical and Physical Optics

Read Explanation

Physical Quantities And Units

Read Explanation

Quantum Physics

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