A 5.0-L flask contains \(0.60 \mathrm{~g} \mathrm{O}_{2}\) at a temperature of \(22^{\circ} \mathrm{C}\). What is the pressure (in atm) inside the flask?

Short Answer

Expert verified
So: \(T(K) = 22 + 273.15 = 295.15~K\) #tag_title#Step 2: Calculate moles of gas#tag_content#Next, we need to determine the amount of gas in moles. We know that the molar mass of \(O_2\) is 32.00 g/mol, so we can calculate the moles of gas as follows: \(n = \dfrac{mass}{molar~mass}\) Thus: \(n = \dfrac{0.60~g}{32.00~g/mol} = 0.01875~mol\) #tag_title#Step 3: Apply the Ideal Gas Law#tag_content#Now, we can use the Ideal Gas Law formula (PV = nRT) to find the pressure inside the flask. We know the volume V is 5.0 L and the temperature T is 295.15 K. The Ideal Gas Constant R is 0.0821 L atm/mol K. Solving for P, we get: \(P = \dfrac{nRT}{V}\) Plugging in the values: \(P = \dfrac{0.01875 \times 0.0821 \times 295.15}{5}\) Finally, we calculate the pressure: \(P \approx 0.913~atm\)

Step by step solution

01

Convert temperature to Kelvin

First, we need to convert the temperature from Celsius to Kelvin. The formula for that is: \(T(K) = T(°C) + 273.15\)

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 \(15.0\) - \(\mathrm{L}\) tank is filled with \(\mathrm{H}_{2}\) to a pressure of \(2.00 \times 10^{2}\) atm. How many balloons (each \(2.00 \mathrm{~L}\) ) can be inflated to a pressure of \(1.00\) atm from the tank? Assume that there is no temperature change and that the tank cannot be emptied below \(1.00 \mathrm{~atm}\) pressure.

Calculate the pressure exerted by \(0.5000 \mathrm{~mol} \mathrm{~N}_{2}\) in a 1.0000-L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results.

You have two containers each with 1 mol of xenon gas at \(15^{\circ} \mathrm{C}\). Container A has a volume of \(3.0 \mathrm{~L}\), and container \(\mathrm{B}\) has a volume of \(1.0 \mathrm{~L}\). Explain how the following quantities compare between the two containers. a. the average kinetic energy of the \(\mathrm{Xe}\) atoms b. the force with which the Xe atoms collide with the container walls c. the root mean square velocity of the Xe atoms d, the collision frequency of the Xe atoms (with other atoms) e. the pressure of the Xe sample

Do all the molecules in a 1 -mol sample of \(\mathrm{CH}_{4}(g)\) have the same kinetic energy at \(273 \mathrm{~K}\) ? Do all molecules in a \(1-\mathrm{mol}\) sample of \(\mathrm{N}_{2}(g)\) have the same velocity at \(546 \mathrm{~K}\) ? Explain.

In the "Méthode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ Fermentation of \(750 . \mathrm{mL}\) grape juice (density \(=1.0 \mathrm{~g} / \mathrm{cm}^{3}\) ) is allowed to take place in a bottle with a total volume of \(825 \mathrm{~mL}\) until \(12 \%\) by volume is ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\). Assuming that the \(\mathrm{CO}_{2}\) is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C}\) ? (The density of ethanol is \(0.79 \mathrm{~g} / \mathrm{cm}^{3}\).)

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