Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A bubble of air is underwater at temperature \(15^{\circ} \mathrm{C}\) and pressure \(1.5\) bar. If the bubble rises to the surface where the temperature is \(25^{\circ} \mathrm{C}\) and the pressure is \(1.0\) bar, what will happen to the volume of the bubble? (a) Volume will become greater by a factor of \(1.6\), (b) Volume will become greater by a factor of \(1.1 .\) (c) Volume will become smaller by a factor of \(0.70\) (d) Volume will become greater by a factor of \(2.5\).

Short Answer

Expert verified
The volume of the bubble will become greater by a factor of approximately 1.6.

Step by step solution

01

Identify the Relevant Gas Law

To determine the change in volume of the bubble, we need to use the combined gas law, which is stated as \(P_1V_1/T_1 = P_2V_2/T_2\), where \(P\) is the pressure, \(V\) is the volume, and \(T\) is the absolute temperature (in Kelvin).
02

Convert Temperatures to Kelvin

First, convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature. The initial temperature is \(T_1 = 15 + 273.15 = 288.15 \text{K}\) and the final temperature is \(T_2 = 25 + 273.15 = 298.15 \text{K}\).
03

Apply the Combined Gas Law

With the known values, we input them into the combined gas law: \[(1.5 \text{bar})(V_1)/(288.15 \text{K}) = (1.0 \text{bar})(V_2)/(298.15 \text{K})\]. To find the factor by which the volume increases, we need to isolate \(V_2/V_1\).
04

Calculate the Volume Ratio

Rearrange the equation to solve for \(V_2/V_1\): \[V_2/V_1 = (1.5/1.0) \times (298.15/288.15)\]. By performing the calculations, we find the factor to be approximately 1.6.
05

Choose the Correct Answer

Since the volume ratio is approximately 1.6, this corresponds to the bubble's volume becoming greater by a factor of 1.6. Therefore, the correct choice is option (a).

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 Vaia!

Key Concepts

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

Gas Laws
Understanding gas laws is fundamental in predicting how gases will react under various conditions of temperature, volume, and pressure. The gas laws are a set of rules that describe the relationship between these three variables. Each combination of variables has its gas law, but the combined gas law brings together Charles's Law, Boyle's Law, and Gay-Lussac's Law into one comprehensive equation.

This unified equation is key to solving problems where more than one property changes, as often happens in the real world. For example, if we consider a bubble rising from underwater to the surface, not only does the pressure change due to the ascent, but there may also be a change in temperature from the depths to the warmer surface waters. Here, the combined gas law is a powerful tool that helps us understand and predict how the bubble will behave.
Volume and Temperature Relationship
The volume and temperature relationship in gases is described by Charles's Law, which states that, at a constant pressure, the volume of a gas is directly proportional to its absolute temperature. In simpler terms, as the temperature of a gas increases, so does its volume, provided that the pressure remains unchanged.

For example, when heating a balloon, the air inside expands and the balloon grows larger. Conversely, cooling the balloon makes the air contract and the balloon shrinks. It's vital to use the Kelvin scale for temperature here because it starts at absolute zero, meaning there are no negative temperatures, allowing for a consistent proportional relationship.

Real-Life Example

In the case of the underwater bubble, as it rises and moves into warmer water at the surface, the temperature increases. According to Charles's Law, we would expect the volume of the gas to increase accordingly if the pressure were held constant.
Pressure and Volume Relationship
Boyle's Law eloquently states the pressure-volume relationship for gases, which is inversely proportional. This means that if the temperature is constant, an increase in the volume of the gas leads to a decrease in its pressure, and vice versa.

Imagine squeezing a closed, flexible container with a fixed quantity of gas inside; as you decrease the volume, the pressure inside the container rises. Likewise, if you expand the container, the internal pressure drops. This concept is imperative for understanding how gases will behave when confined in containers of varying sizes.

Examining Our Bubble

In our exercise, the bubble's pressure decreases as it rises to the surface. According to Boyle's Law, we would expect its volume to increase since the external water pressure exerted on the bubble is reduced. However, to precisely calculate this effect, we must consider both the temperature change and the change in pressure, which is where the combined gas law plays its pivotal role.

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

I is surface molecule whereas II is interior molecule. Choose the correct one. (a) I results net attraction into the liquid. (b) II are attracted in all directions. (c) Both (a) and (b). (d) Neither (a) nor (b).

What is the relationship between thermal energy and intermolecular interaction energy of a substance in three states in terms of \(X\) and \(Y ?\) Gas \(\longrightarrow\) Liquid \(\longrightarrow\) Solid (a) \(X\)-Thermal energy, \(Y\)-Intermolecular interactions (b) \(X\)-Thermal energy, \(Y\)-Thermal energy (c) \(X\)-Intermolecular interactions, \(Y\)-Thermal energy (d) \(X\)-Intermolecular interactions, \(Y\)-Intermolecular interactions

Compressibility factor of a gas is given by the equation \(Z=\frac{P V}{n R T} .\) On this basis, mark the correct statement. (a) When \(Z>1\), real gases get compressed easily. (b) When \(Z=1\) real gases get compressed easily. (c) When \(Z>1\), real gases are difficult to compress. (d) When \(Z=1\), real gases are difficult to compress

The relations between various variables of gaseous substances are given along with their formulae. Mark the incorrect relationship. (a) Density and molar mass : \(M=\frac{d R T}{P}\) (b) Universal gas constant, \(P, V, T: R=\frac{P V}{n T}\) (c) Volume and pressure: \(V_{2}=\frac{P_{2} V_{1}}{P_{1}}\) (d) Volume and temperature: \(V_{2}=\frac{V_{1} T_{2}}{T_{1}}\)

If the ratio of masses of \(\mathrm{SO}_{3}\) and \(\mathrm{O}_{2}\) gases confined in a vessel is \(1: 1\), then the ratio of their partial pressures would be (a) \(5: 2\) (b) \(2: 5\) (c) \(2: 1\) (d) \(1: 2\)

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.

Sign-up for free