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 fixed quantity of gas at \(21^{\circ} \mathrm{C}\) exhibits a pressure of 752 torr and occupies a volume of 5.12 L. \((\mathbf{a})\) Calculate the volume the gas will occupy if the pressure is increased to 1.88 atm while the temperature is held constant. \((\mathbf{b})\) Calculate the volume the gas will occupy if the temperature is increased to\(175^{\circ} \mathrm{C}\) while the pressure is held constant.

Short Answer

Expert verified
(a) Using Boyle's Law, the volume of the gas will be \(2.71 \, \text{L}\) when the pressure is increased to 1.88 atm while holding the temperature constant. (b) Using Charles's Law, the volume of the gas will be \(7.84 \, \text{L}\) when the temperature is increased to \(175^{\circ} \mathrm{C}\) while holding the pressure constant.

Step by step solution

01

(a) Boyle's Law

Boyle's Law states that at constant temperature, the volume of a gas is inversely proportional to its pressure: \[ P_1V_1 = P_2V_2 \] Given: Initial Pressure, \(P_1 = 752 \, \text{torr}\) Initial Volume, \(V_1 = 5.12 \, \text{L}\) Final Pressure, \(P_2 = 1.88 \, \text{atm}\) First, we need to convert the pressure units to be consistent. Let's convert both pressures to the same unit (atm): \[1 \, \text{atm} = 760 \, \text{torr}\] \[P_1 = \frac{752 \, \text{torr}}{760 \, \text{torr/atm}} = 0.989 \, \text{atm}\] Now, we will use Boyle's Law formula to find the final volume \(V_2\): \[0.989 \, \text{atm} \cdot 5.12 \, \text{L} = 1.88 \, \text{atm} \cdot V_2\]
02

(a) Solution

Dividing both sides by 1.88 atm, we get: \[V_2 = \frac{0.989 \, \text{atm} \cdot 5.12 \, \text{L}}{1.88 \, \text{atm}}\] \[V_2 = 2.71 \, \text{L}\] So, the volume of the gas will be \(2.71 \, \text{L}\) when the pressure is increased to 1.88 atm while holding the temperature constant.
03

(b) Charles's Law

Charles's Law states that at constant pressure, the volume of a gas is directly proportional to its temperature: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] Given: Initial Temperature, \(T_1 = 21^{\circ} \mathrm{C} + 273.15 = 294.15 \, \text{K}\) Final Temperature, \(T_2 = 175^{\circ} \mathrm{C} + 273.15 = 448.15 \, \text{K}\) Initial Volume, \(V_1 = 5.12 \, \text{L}\) We will use Charles's Law formula to find the final volume \(V_2\): \[ \frac{5.12 \, \text{L}}{294.15 \, \text{K}} = \frac{V_2}{448.15 \, \text{K}} \]
04

(b) Solution

Multiplying both sides by 448.15 K, we get: \[V_2 = \frac{5.12 \, \text{L}}{294.15 \, \text{K}} \cdot 448.15 \, \text{K}\] \[V_2 = 7.84 \, \text{L}\] So, the volume of the gas will be \(7.84 \, \text{L}\) when the temperature is increased to \(175^{\circ} \mathrm{C}\) while holding the pressure constant.

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.

Boyle's Law
Boyle's Law is a fundamental concept in gas laws chemistry that describes the pressure-volume relationship for a fixed amount of gas at constant temperature. This law is mathematically expressed as
\( P_1V_1 = P_2V_2 \), where
- \( P_1 \) and \( P_2 \) are the initial and final pressures, and
- \( V_1 \) and \( V_2 \) are the initial and final volumes of the gas.
To visualize Boyle's Law, imagine squeezing a balloon. As you squeeze (increase pressure), the volume of the balloon decreases; when you release it, the balloon expands (volume increases).

Application in Exercises

When solving problems involving Boyle's Law, it's important to make sure the pressure units are consistent. Once you've established uniform units, you can algebraically solve for the unknown variable. This law is especially useful in conditions where the temperature does not change such as in isothermal processes.
Charles's Law
Charles's Law focuses on the temperature-volume relationship of gases at constant pressure. It suggests that as the temperature of a gas increases, so does its volume and vice versa. This relationship is linear and can be represented as
\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where
- \( V_1 \) and \( V_2 \) correspond to initial and final volumes, and
- \( T_1 \) and \( T_2 \) are the initial and final temperatures measured in Kelvins (K).
Imagine heating a closed and flexible container filled with gas: as it warms, the gas expands, demonstrating Charles's Law in action.

Practical Uses

This principle is regularly used to predict the behavior of gases when they are heated or cooled. In homework problems, always convert temperatures to Kelvin and solve for the variable in question. Charles's Law is at play in daily life, from balloons contracting in cold air to thermal expansion in automobile engines.
Pressure-Volume Relationship
The pressure-volume relationship in gases, also derived from Boyle's Law, indicates an inverse proportionality between the pressure and volume of a gas when temperature remains unchanged.
As pressure on a contained gas increases, its volume decreases, provided the temperature is kept constant, and conversely, if the pressure decreases, the volume will increase. A classic example is a syringe: as you push the plunger, the pressure inside increases while the volume of the gas decreases.

Significance in Solving Problems

This relationship is paramount in problem-solving where the temperature is controlled. Understanding the inverse nature of pressure and volume can help students better predict and calculate responses of gases under different pressures. It's vital to stress consistency in units of pressure when applying the formulas.
Temperature-Volume Relationship
The temperature-volume relationship is a direct relationship highlighted by Charles's Law. This relationship shows that a gas's volume will increase with a rise in temperature when the pressure is kept constant.
The law applies to ideal gases and is reliant on absolute temperatures (Kelvin scale). When explaining this relationship to students, it's important to stress that as atoms and molecules in a gas move more quickly with increased temperature, they need more space, hence the volume increases.

Temperature Conversion – Key to Solutions

One should not forget that solving textbook exercises requires converting Celsius to Kelvin, as only absolute temperatures reflect the true energy of the particles in the gas. The Kelvin temperature can be found by adding 273.15 to the Celsius temperature. As volume changes proportionally to temperature, this law helps explain phenomena like inflating balloons or the danger of leaving aerosol cans in high heat.

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

Both Jacques Charles and Joseph Louis Guy-Lussac were avid balloonists. In his original flight in \(1783,\) Jacques Charles used a balloon that contained approximately \(31,150\) L of \(\mathrm{H}_{2}\) . He generated the \(\mathrm{H}_{2}\) using the reaction between iron and hy- drochloric acid: $$\mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ How many kilograms of iron were needed to produce this volume of \(\mathrm{H}_{2}\) if the temperature was \(22^{\circ} \mathrm{C}\) ?

As discussed in the "Chemistry Put to Work" box in Section 10.8 , enriched uranium can be produced by effusion of gaseous \(\mathrm{UF}_{6}\) across a porous membrane. Suppose a process were developed to allow effusion of gaseous uranium atoms, U(g). Calculate the ratio of effusion rates for \(^{235} \mathrm{U}\) and \(^{238} \mathrm{U},\) and compare it to the ratio for \(\mathrm{UF}_{6}\) given in the essay.

(a) The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20 \(\mathrm{g} / \mathrm{mL}\) . The density of mercury is 13.6 \(\mathrm{g} / \mathrm{mL} .\) What do you predict for the height of a barometer column based on 1 -iodododecane, when the atmospheric pressure is 749 torr? (b) What is the pressure, in atmospheres, on the body of a diver if he is 21 ft below the surface of the water when the atmospheric pressure is 742 torr?

(a) What conditions are represented by the abbreviation STP? (b) What is the molar volume of an ideal gas at STP? (c) Room temperature is often assumed to be \(25^{\circ} \mathrm{C}\) . Calculate the molar volume of an ideal gas at \(25^{\circ} \mathrm{C}\) and 1 atm pressure. (d) If you measure pressure in bars instead of atmospheres, calculate the corresponding value of \(R\) in L-bar/mol-K.

A rigid vessel containing a \(3 : 1\) mol ratio of carbon dioxide and water vapor is held at \(200^{\circ} \mathrm{C}\) where it has a total pressure of 2.00 atm. If the vessel is cooled to \(10^{\circ} \mathrm{C}\) so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.

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