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

Draw a qualitative graph to show how the first property varies with the second in each of the following (assume 1 mole of an ideal gas and \(T\) in kelvin). a. \(P V\) versus \(V\) with constant \(T\) b. \(P\) versus \(T\) with constant \(V\) c. \(T\) versus \(V\) with constant \(P\) d. \(P\) versus \(V\) with constant \(T\) e. \(P\) versus \(1 / V\) with constant \(T\) f. \(P V / T\) versus \(P\)

Short Answer

Expert verified
a. The graph of PV versus V with constant T will be a straight line with a positive slope. b. The graph of P versus T with constant V will be a straight line with a positive slope. c. The graph of T versus V with constant P will be a straight line with a positive slope. d. The graph of P versus V with constant T will be a downward curve (hyperbola). e. The graph of P versus 1/V with constant T will be a straight line with a positive slope. f. The graph of PV/T versus P will be a horizontal straight line with no slope.

Step by step solution

01

a. Graph of PV versus V with constant T

In this scenario, temperature is constant, and we need to plot the relation between PV and V. From the ideal gas law, we know that PV = nRT. Since temperature is constant, we relate PV with V as follows: PV = Constant We can conclude that the graph will be a straight line with a positive slope.
02

b. Graph of P versus T with constant V

In this scenario, we need to plot the relationship between pressure and temperature while keeping the volume constant. By rearranging the ideal gas law equation, we get: P = (nR/V)T When volume is constant, the relation between P and T becomes directly proportional. Thus, the graph will be a straight line with a positive slope.
03

c. Graph of T versus V with constant P

In this scenario, we will plot the relationship between temperature and volume while keeping the pressure constant. Rearranging the ideal gas law equation, we get: T = (PV)/(nR) When pressure is constant, the relation between V and T becomes directly proportional. The graph will be a straight line with a positive slope.
04

d. Graph of P versus V with constant T

In this scenario, we need to plot the relationship between pressure and volume while keeping the temperature constant. By rearranging the ideal gas law equation, we get: P = (nRT)/V When temperature is constant, the relation between P and V becomes inversely proportional. Thus, the graph will be a downward curve known as a hyperbola.
05

e. Graph of P versus 1/V with constant T

In this scenario, we need to plot the relationship between pressure and reciprocal of the volume, keeping the temperature constant. From the ideal gas law, we know that P = (nRT)/V. Rewriting this equation in terms of 1/V, we get: P = (nR)T(V^-1) In this case, pressure and the reciprocal of the volume are directly related. Therefore, the graph will be a straight line with a positive slope.
06

f. Graph of PV/T versus P

In this scenario, we are asked to plot the relationship between (PV/T) and pressure. From the ideal gas law, we know that PV = nRT. If we divide both sides by T, we get: PV/T = nR Since we are considering one mole of an ideal gas, n = 1. Therefore: PV/T = R Since R is a constant, the graph of PV/T versus P will be a horizontal straight line with no slope.

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.

Pressure-Volume Relationship
The relationship between pressure and volume is elegantly described by Boyle's Law. When the temperature is held constant, the product of pressure (P) and volume (V) for a gas remains constant. This means that as the volume of the gas decreases, the pressure increases, and vice versa. The relationship can be represented by the equation\[ PV = ext{constant} \]This is an example of an inverse relationship. If you were to draw a graph of pressure versus volume while keeping the temperature constant, you would see a downward curve known as a hyperbola. This representation helps in understanding that pressure and volume change in opposite directions under constant temperature conditions. Additionally, if we plot pressure against the reciprocal of volume (1/V), the graph becomes a straight line.To sum up:
  • Increased volume results in decreased pressure (and vice versa).
  • The graph of pressure versus volume is a hyperbola.
  • The graph of pressure versus 1/V is a straight line.
Temperature-Volume Relationship
Charles's Law explores the relationship between temperature and volume, providing insight into how gases expand when heated. According to this law, the volume of a gas is directly proportional to its temperature when pressure is held constant. Hence, when you increase the temperature, the gas molecules have more energy and thus, occupy more space. This can be mathematically expressed as:\[ V \propto T \quad \text{or} \quad \frac{V}{T} = ext{constant} \]When graphing volume against temperature at a constant pressure, the graph will be a straight line with a positive slope. In such a graph, as the temperature increases, the volume does too, illustrating their direct proportionality.Key points:
  • Higher temperature increases volume.
  • The graph is a straight line with a positive slope.
  • Direct proportionality between temperature and volume.
Pressure-Temperature Relationship
The relationship between pressure and temperature is explained by Gay-Lussac's Law. When the volume is held constant, the pressure of a gas is directly proportional to its absolute temperature. So, if you increase the temperature of a gas, its pressure also increases provided the volume does not change. This relationship is given by:\[ P \propto T \quad \text{or} \quad \frac{P}{T} = ext{constant} \]When plotting pressure against temperature with a constant volume, you will observe a linear graph with a positive slope. This shows that as the temperature of the gas increases, so does the pressure, showcasing their direct relationship.Summary:
  • Pressure increases with an increase in temperature.
  • The graph is a straight line with a positive slope.
  • A direct relationship between pressure and temperature.
Qualitative Graphs
Qualitative graphs are an excellent tool for visualizing relationships in gas laws. They help establish clear visual interpretations of theoretical concepts like the interdependence of gas properties. - **Pressure vs. Volume graphs**: typically showcase a hyperbola when temperature is constant, highlighting an inverse relationship. - **Temperature vs. Volume graphs**: display a straight line when pressure is constant, illustrating a direct proportionality. - **Pressure vs. Temperature graphs**: also show a straight line when volume is constant, again displaying direct proportionality. By utilizing these graphs, students and scientists alike can predict how alterations in one property affect others. These visual tools simplify complex ideas into more digestible forms. In essence:
  • Help in visualizing complex relationships.
  • Show whether properties are directly or inversely related.
  • Assist in conceptual understanding of the gas behaviors.
Gas Behavior
Understanding gas behavior is crucial in comprehending various natural and industrial processes. Gas laws encapsulate the fundamental aspects influencing how gases behave under different conditions.
  • Boyle’s Law explains that at constant temperature, the pressure and volume of a gas are inversely related.
  • Charles’s Law states that at constant pressure, the volume and temperature of a gas are directly proportional.
  • Gay-Lussac's Law shows that at constant volume, pressure and temperature are directly proportional.
These laws collectively form the Ideal Gas Law expressed as:\[ PV = nRT \]Here, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. This equation helps predict gas behavior under various scenarios, offering a baseline understanding essential in fields like chemistry and engineering.To recap:
  • Describes fundamental gas behaviors.
  • Helps predict how gases will react to changes in conditions.
  • Essential for scientific and engineering applications.

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 2.00-L sample of \(\mathrm{O}_{2}(g)\) was collected over water at a total pressure of 785 torr and \(25^{\circ} \mathrm{C}\). When the \(\mathrm{O}_{2}(g)\) was dried (water vapor removed), the gas had a volume of \(1.94 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\) and 785 torr. Calculate the vapor pressure of water at \(25^{\circ} \mathrm{C}\).

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of 200. L/min at \(1.50 \mathrm{~atm}\) and ambient temperature. Air is added to the chamber at \(1.00 \mathrm{~atm}\) and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(g)\), three times as much oxygen as is necessary is reacted. Assuming air is 21 mole percent \(\mathrm{O}_{2}\) and \(79 \mathrm{~mole}\) percent \(\mathrm{N}_{2}\), calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that \(95.0 \%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\). The remainder was present as carbon in \(\mathrm{CO}\). Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O} .\) Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.

Consider the following samples of gases at the same temperature. Arrange each of these samples in order from lowest to highest: a. pressure b. average kinetic energy c. density d. root mean square velocity Note: Some samples of gases may have equal values for these attributes. Assume the larger containers have a volume twice the volume of the smaller containers, and assume the mass of an argon atom is twice the mass of a neon atom.

An organic compound containing only C, H, and N yields the following data. i. Complete combustion of \(35.0 \mathrm{mg}\) of the compound produced \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\). ii. A \(65.2\) -mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129), giving \(35.6 \mathrm{~mL}\) of dry \(\mathrm{N}_{2}\) at 740 . torr and \(25^{\circ} \mathrm{C}\). iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{~mL} / \mathrm{min}\). The effusion rate of argon gas, under identical conditions, is \(26.4 \mathrm{~mL} / \mathrm{min}\). What is the molecular formula of the compound?

A 2.50-L container is filled with \(175 \mathrm{~g}\) argon. a. If the pressure is \(10.0 \mathrm{~atm}\), what is the temperature? b. If the temperature is \(225 \mathrm{~K}\), what is the pressure?

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