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

Ideal gas particles are assumed to be volumeless and to neither attract nor repel each other. Why are these assumptions crucial to the validity of Dalton's law of partial pressures?

Short Answer

Expert verified
The assumptions of volumeless particles and no intermolecular forces in ideal gases are crucial for the validity of Dalton's law of partial pressures, as they ensure the pressure exerted by each individual gas in the mixture is independent of the other gases. This independence allows the sum of the partial pressures to accurately represent the total pressure exerted by the gas mixture. If these assumptions were not met, the proportions of different gases within the mixture and their behavior would be affected, potentially leading to inaccuracies in calculating partial pressures according to Dalton's law.

Step by step solution

01

Understanding Dalton's law of partial pressures

Dalton's law of partial pressures states that in a mixture of non-reacting gases, the total pressure exerted by the mixture is equal to the sum of the partial pressures of individual gases. Mathematically, this can be written as: \[P_\text{total} = P_1 + P_2 + P_3 + ... + P_n\] where \(P_\text{total}\) is the total pressure exerted by the gas mixture, and \(P_i\) represents the partial pressure of the \(i\)-th gas in the mixture.
02

Understanding the ideal gas law and its assumptions

The ideal gas law is given by the equation: \[PV = nRT\] where \(P\) is the pressure, \(V\) is the volume, \(n\) is the amount of substance (in moles), \(R\) is the ideal gas constant, and \(T\) is the temperature (in Kelvin). The ideal gas law is based on several assumptions, including: 1. Gas particles have no volume (or their volume is negligible compared to the volume of the container). 2. Gas particles neither attract nor repel each other (i.e., no intermolecular forces). 3. Gas particles are in constant, random motion, and their collisions with the walls of the container are completely elastic (i.e., no energy is lost in collisions).
03

Examining the importance of ideal gas assumptions for Dalton's law

Now that we have laid down the basics, let's examine why the assumptions of volumeless particles and absence of intermolecular forces are crucial for the validity of Dalton's law. - Volumeless particles: If gas particles had a significant volume compared to the volume of the container, each individual gas's volume would contribute to the total volume of the mixture. This would ultimately change the proportions of different gases present, potentially leading to inaccuracies in calculating the partial pressures according to Dalton's law. - No intermolecular forces: If gas particles attracted or repelled each other, their behavior (such as the rate of collisions with the container walls) would be affected. This would, in turn, result in inaccuracies in the calculation of partial pressures, as the assumption of independence between each gas component in Dalton's law would no longer be valid.
04

Conclusion

In conclusion, the assumptions that ideal gas particles are volumeless and neither attract nor repel each other are crucial to the validity of Dalton's law of partial pressures because these assumptions ensure that the pressure exerted by each individual gas in the mixture is independent of the other gases. This independence allows the sum of the partial pressures to accurately represent the total pressure exerted by the gas mixture.

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!

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 large flask with a volume of \(936 \mathrm{mL}\) is evacuated and found to have a mass of \(134.66 \mathrm{g}\). It is then filled to a pressure of 0.967 atm at \(31^{\circ} \mathrm{C}\) with a gas of unknown molar mass and then reweighed to give a new mass of 135.87 g. What is the molar mass of this gas?

A tank contains a mixture of \(52.5 \mathrm{g}\) oxygen gas and \(65.1 \mathrm{g}\) carbon dioxide gas at \(27^{\circ} \mathrm{C}\). The total pressure in the tank is \(9.21\) atm. Calculate the partial pressures of each gas in the container.

A spherical glass container of unknown volume contains helium gas at \(25^{\circ} \mathrm{C}\) and \(1.960\) atm. When a portion of the helium is withdrawn and adjusted to 1.00 atm at \(25^{\circ} \mathrm{C},\) it is found to have a volume of \(1.75 \mathrm{cm}^{3} .\) The gas remaining in the first container shows a pressure of \(1.710 \) atm. Calculate the volume of the spherical container.

An organic compound contains \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{O} .\) Combustion of \(0.1023 \mathrm{g}\) of the compound in excess oxygen yielded \(0.2766 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0991 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) A sample of \(0.4831 \mathrm{g}\) of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129 ). At \(\mathrm{STP}, 27.6 \mathrm{mL}\) of dry \(\mathrm{N}_{2}\) was obtained. In a third experiment, the density of the compound as a gas was found to be \(4.02 \mathrm{g} / \mathrm{L}\) at \(127^{\circ} \mathrm{C}\) and \(256\) torr. What are the empirical and molecular formulas of the compound?

A compressed gas cylinder contains \(1.00 \times 10^{3} \mathrm{g}\) argon gas. The pressure inside the cylinder is \(2050 .\) psi (pounds per square inch) at a temperature of \(18^{\circ} \mathrm{C}\). How much gas remains in the cylinder if the pressure is decreased to \(650 .\) psi at a temperature of \(26^{\circ} \mathrm{C} ?\)

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