/*! 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 30 A balloon filled with \(1.92 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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)?

Short Answer

Expert verified
The balloon's volume is approximately 6.963 L.

Step by step solution

01

Understand the Ideal Gas Law

The Ideal Gas Law is given by \[ PV = nRT \] where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature. Since pressure and temperature are constant, the volume is directly proportional to the number of moles (\(V \propto n\)).
02

Calculate Initial Moles of Helium

The molar mass of helium (He) is approximately \(4.00 \text{ g/mol}\). Use this to calculate the initial moles of helium: \[ n_i = \frac{1.92 \text{ g}}{4.00 \text{ g/mol}} = 0.48 \text{ mol} \]
03

Calculate Remaining Moles of Helium

Determine how much helium remains after \(0.850 \text{ g}\) has leaked out: \[ \text{Remaining mass of helium} = 1.07 \text{ g} \text{ (1.92 g - 0.850 g)} \] Now, calculate the moles of helium remaining: \[ n_f = \frac{1.07 \text{ g}}{4.00 \text{ g/mol}} = 0.2675 \text{ mol} \]
04

Use Proportional Relationship to Find Final Volume

Since volume is proportional to moles, use the initial and final moles to find the final volume: \[ V_f = V_i \times \frac{n_f}{n_i} \] Substitute the values: \[ V_f = 12.5 \text{ L} \times \frac{0.2675 \text{ mol}}{0.480 \text{ mol}} = 6.963 \text{ L} \]

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.

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.

moles of gas
The concept of moles is fundamental in chemistry. A mole is a standard unit in chemistry that measures the amount of substance. When dealing with gases, we use the molar mass to convert grams of a substance to moles.
In the exercise, we started with 1.92 grams of helium. The molar mass of helium is approximately 4.00 grams per mole. Using this, we convert the initial mass of helium to moles by dividing the mass by the molar mass: \( n_i = \frac{1.92 \text{ g}}{4.00 \text{ g/mol}} = 0.48 \text{ mol} \).
After some helium leaks out, we need to figure out how many grams are left. We perform another division to find the remaining moles: \( n_f = \frac{1.07 \text{ g}}{4.00 \text{ g/mol}} = 0.2675 \text{ mol} \).
Understanding moles helps us to relate the amount of gas (mass) to how it behaves in terms of volume and other properties.
proportional relationship
In this problem, we are using the Ideal Gas Law to understand how the volume of a gas changes when the amount of gas changes, assuming constant temperature and pressure. The Ideal Gas Law is given by \( PV = nRT \).
Pressure \( P \) and temperature \( T \) are held constant, which simplifies the relationship to \( V \rightpropto n \), meaning volume is directly proportional to the number of moles.
This means if the number of moles of gas decreases, the volume of the gas also decreases in proportion. We directly use this proportionality: \( V_f = V_i \times \frac {n_f}{n_i} \). Here, \( V_i \) is the initial volume, \( n_f \) is the final number of moles, and \( n_i \) is the initial number of moles.
By plugging numbers, we discovered how much the volume changes: \( V_f = 12.5 \text{ L} \times \frac{0.2675 \text{ mol}}{0.48 \text{ mol}} = 6.963 \text{ L} \).
This shows how knowing the relationship between volume and moles can help determine changes due to leaks.
constant pressure and temperature
Under constant pressure and temperature, the gas laws tell us that the volume of the gas is directly proportional to the number of moles of gas present. This fundamental understanding comes from the Ideal Gas Law, \( PV = nRT \), where pressure and temperature are constants.
In simpler terms, when the amount (moles) of gas decreases, the volume also decreases if the pressure and temperature don’t change. This directly impacts our calculations and shows why keeping these conditions stable is crucial.
In the given problem, the constant conditions mean our calculations stay valid as we account for the leaked helium. The direct proportionality between volume and moles eases the computation, reinforcing why learning these constants and their impacts is essential for solving gas-related problems.
Through such problems, students grasp how theoretical constructs like the Ideal Gas Law apply practically, bolstered by the constants that dictate gas behavior.

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

Sulfur dioxide emissions from coal-burning power plants are removed by flue- gas desulfurization. The flue gas passes through a scrubber, and a slurry of wet calcium carbonate reacts with it to form carbon dioxide and calcium sulfite. The calcium sulfite then reacts with oxygen to form calcium sulfate, which is sold as gypsum. (a) If the sulfur dioxide concentration is 1000 times higher than its mole fraction in clean, dry air \(\left(2 \times 10^{-10}\right),\) how much calcium sulfate \((\mathrm{kg})\) can be made from scrubbing \(4 \mathrm{GL}\) of flue gas ( \(\left.1 \mathrm{GL}=1 \times 10^{9} \mathrm{~L}\right)\) ? A state-of-the-art scrubber removes at least \(95 \%\) of the sulfur dioxide. (b) If the mole fraction of oxygen in air is \(0.209,\) what volume \((\mathrm{L})\) of air at \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) is needed to react with all the calcium sulfite?

An equimolar mixture of Ne and Xe is accidentally placed in a container that has a tiny leak. After a short while, a very small proportion of the mixture has escaped. What is the mole fraction of Ne in the effusing gas?

A sample of chlorine gas is confined in a \(5.0-\mathrm{L}\) container at 328 torr and \(37^{\circ} \mathrm{C} .\) How many moles of gas are in the sample?

An environmental chemist sampling industrial exhaust gases from a coal-burning plant collects a \(\mathrm{CO}_{2}-\mathrm{SO}_{2}-\mathrm{H}_{2} \mathrm{O}\) mixture in a \(21-\mathrm{L}\) steel tank until the pressure reaches \(850 .\) torr at \(45^{\circ} \mathrm{C}\) (a) How many moles of gas are collected? (b) If the \(\mathrm{SO}_{2}\) concentration in the mixture is \(7.95 \times 10^{3}\) parts per million by volume (ppmv), what is its partial pressure? [Hint: \(\mathrm{ppmv}=\left(\right.\) volume of component/volume of mixture) \(\left.\times 10^{6} .\right]\)

Why can we use a gas mixture, such as air, to study the general behavior of an ideal gas under ordinary conditions?

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.