A sample of N 2 O 3 (g) has a pressure of 0.017 atm. The temperature (in K) is doubled and the N 2 O 3 undergoes complete decomposition to NO 2 (g) and NO(g). Find the total pressure of the mixture of gases assuming constant volume and no additional temperature change.

Hey everyone we have a nine liter gas canister and it contains a mixture of neon and Xenon at 28 degrees Celsius. We have the partial pressure of the two gasses which is 0.45 atmosphere 0.25 atmosphere. And we have a 7.5 millimeter sample of methanol with a density of 0.792 g per meal leader. That's introduced into the canister and were asked what is the total pressure in the canister. We need to first find the most of methanol. Then we can plug this into the ideal gas law to find the partial pressure of methanol. Have 7.5 mL. We have the density of methanol which is 0.792 grams per one male leader. They got the molar mass in one wall of methanol which is 12.11 g. Last four that was 1.8 g plus 15 0.999 grams. Would you give us 32.42 grams. When we get 0.185 malls of methanol. So now we can plug this into the ideal gas law which is P. V. Equals N. R. T. We can rearrange this is all for the pressure which is N. R. T. B. So we're looking for the pressure and we have 0. malls. The value of art is 0.8206 Leaders Times atmosphere. Not about malls Calvin. And the temperature is 28 degrees Celsius. We need to convert this to Calvin. So we're gonna add 273 15. Get 301.15 kelvin. Sophie plug this in. We're gonna get 0.185 Smalls time 0.8 to our sex. Later. Some atmosphere about moles times, Calvin times 301.15 kelvin provided by nine liters. The pressure of 0.58 atmosphere. Now we can find the total pressure. Using the partial pressure of methanol. Recall that the total pressure because the pressure of gas one plus the pressure of gas too. Plus the pressure of gas three for the total pressure. Can I have 0.4 fine atmosphere 0.25 atmosphere 0.508 atmosphere over the total pressure we get 1. atmosphere. Thanks for watching my video and I hope it's up for

Ch.6 - Gases

All textbooks

Tro 6th Edition

Ch.6 - Gases

Problem 136

Chapter 6, Problem 136

A sample of N₂O₃(g) has a pressure of 0.017 atm. The temperature (in K) is doubled and the N₂O₃ undergoes complete decomposition to NO₂(g) and NO(g). Find the total pressure of the mixture of gases assuming constant volume and no additional temperature change.

Verified step by step guidance

Identify the initial conditions: The initial pressure of N2O3 is 0.017 atm.

Write the balanced chemical equation for the decomposition: N2O3(g) → NO2(g) + NO(g).

Determine the mole ratio from the balanced equation: 1 mole of N2O3 produces 1 mole of NO2 and 1 mole of NO.

Apply the ideal gas law concept: Since the volume is constant and temperature is doubled, the pressure of each gas will also double.

Calculate the total pressure: Add the pressures of NO2 and NO, considering the initial pressure of N2O3 and the effect of temperature doubling.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas through the equation PV = nRT. This law is essential for understanding how changes in temperature and pressure affect gas behavior. In this scenario, the initial pressure of N2O3 and the effects of temperature doubling on gas pressure are analyzed using this principle.

Stoichiometry of Gas Reactions

Stoichiometry involves the calculation of reactants and products in chemical reactions. For the decomposition of N2O3 into NO2 and NO, the stoichiometric coefficients indicate the molar relationships between the gases produced. Understanding these relationships is crucial for determining the total pressure of the gas mixture after the reaction.

Dalton's Law of Partial Pressures

Dalton's Law states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas. This concept is vital for calculating the total pressure after the decomposition of N2O3, as it allows for the addition of the partial pressures of NO2 and NO produced from the reaction, assuming no volume change.