Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 60

\(\mathrm{N}_{2} \mathrm{O}\) is a gas commonly used to help sedate patients in medicine and dentistry due to its mild anesthetic and analgesic properties; it is also nonflammable. If a cylinder of \(\mathrm{N}_{2} \mathrm{O}\) is at 10.5 atm and has a volume of 5.00 \(\mathrm{L}\) at \(298 \mathrm{K},\) how many moles of \(\mathrm{N}_{2} \mathrm{O}\) gas are present? The gas from the cylinder is emptied into a large balloon at 745 torr. What is the volume of the balloon at 298 \(\mathrm{K}\) ?

Short Answer

Expert verified
The number of moles of N₂O gas present in the cylinder is approximately 2.13 moles. After the gas is emptied into a large balloon at 745 torr and 298 K, the volume of the balloon is approximately 53.2 L.

Step by step solution

01

Identify the given information for the first problem

The cylinder contains N₂O gas at a pressure of 10.5 atm, a volume of 5.00 L, and a temperature of 298 K. We will use these values along with the ideal gas law constant R=0.0821 L.atm/mol.K to find the number of moles (n) of N₂O gas present.
02

Apply the Ideal Gas Law formula for the first problem

The ideal gas law formula is PV=nRT. We know the values for P, V and T, we will solve for the number of moles (n). \( n = \frac{P * V}{R * T} \)
03

Calculate the number of moles of N₂O gas for the first problem

Substitute the known values into the formula: \( n = \frac{(10.5\: atm) * (5.00\: L)}{(0.0821\: L.atm/mol.K) * (298\: K)} \) After the calculation, we get: \( n \approx 2.13\: moles \) So, there are approximately 2.13 moles of N₂O gas present in the cylinder. Now, we will proceed to the second part of the problem.
04

Identify the given information for the second problem

The gas is emptied into a balloon at 745 torr and has a pressure of 745 torr which is approximately equal to 0.979 atm (1 atm = 760 torr). Temperature remains at 298 K, and we have already found the number of moles of N₂O gas, which is 2.13 moles.
05

Apply the Ideal Gas Law formula for the second problem

We will use the ideal gas law formula again, this time to find the balloon's volume (V). \( V = \frac{n * R * T}{P} \)
06

Calculate the volume of the balloon for the second problem

Substitute the known values into the formula: \( V = \frac{(2.13\: moles) * (0.0821\: L.atm/mol.K) * (298\: K)}{0.979\: atm} \) After the calculation, we get: \( V \approx 53.2\: L \) The volume of the balloon at 298 K when filled with the N₂O gas is approximately 53.2 L.

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.

Start your free trial

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 glass vessel contains 28 g of nitrogen gas. Assuming ideal behavior, which of the processes listed below would double the pressure exerted on the walls of the vessel? a. Adding 28 g of oxygen gas b. Raising the temperature of the container from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\) c. Adding enough mercury to fill one-half the container d. Adding 32 g of oxygen gas e. Raising the temperature of the container from \(30 .^{\circ} \mathrm{C}\) to \(60 .^{\circ} \mathrm{C}\)

You have a helium balloon at 1.00 atm and \(25^{\circ} \mathrm{C} .\) You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is 79.0\(\%\) nitrogen and 21.0\(\%\) oxygen by volume. The “lift” of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C} ?\) Explain. b. Calculate the temperature of the air required for the hot-air balloon to provide the same lift as the helium balloon at 1.00 atm and $25^{\circ} \mathrm{C}\( . Assume atmospheric conditions are 1.00 atm and \)25^{\circ} \mathrm{C} .$

If 27.1 g of \(\mathrm{Ar}(g)\) occupies a volume of 4.21 \(\mathrm{L}\) , what volume will 1.29 moles of \(\mathrm{Ne}(g)\) occupy at the same temperature and pressure?

Freon- 12\(\left(\mathrm{CF}_{2} \mathrm{Cl}_{2}\right)\) is commonly used as the refrigerant in central home air conditioners. The system is initially charged to a pressure of 4.8 atm. Express this pressure in each of the fol lowing units \((1 \mathrm{atm}=14.7 \mathrm{psi}).\) a. mm Hg b. torr c. Pa d. psi

A hot-air balloon is filled with air to a volume of \(4.00 \times\) $10^{3} \mathrm{m}^{3}\( at 745 torr and \)21^{\circ} \mathrm{C}$ . The air in the balloon is then heated to \(62^{\circ} \mathrm{C},\) causing the balloon to expand to a volume of \(4.20 \times 10^{3} \mathrm{m}^{3} .\) What is the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon? (Hint: Openings in the balloon allow air to flow in and out. Thus the pressure in the balloon is always the same as that of the atmosphere.)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation
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