Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 68

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 $26^{\circ} \mathrm{C} ?$

Short Answer

Expert verified
There are approximately 359.14 grams of Argon gas remaining in the cylinder when the pressure is decreased to 650 psi at a temperature of 26°C.

Step by step solution

01

1. Convert mass of Argon to moles

To calculate the moles of Argon, we can use the equation: moles = mass / molar mass. The molar mass of Argon is 39.95 g/mol. Therefore: moles = \( \frac{1.00 \times 10^{3} \mathrm{g}}{39.95 \mathrm{g/mol}} \) moles = 25.03 mol
02

2. Convert psi to atm

We need to convert pressure from psi to atm, because the ideal gas law uses pressure in atm. The conversion factor is 1 atm = 14.696 psi. Therefore: Initial pressure = \( \frac{2050 \mathrm{psi}}{14.696 \mathrm{psi/atm}} \) = 139.58 atm Final pressure = \( \frac{650 \mathrm{psi}}{14.696 \mathrm{psi/atm}} \) = 44.24 atm
03

3. Convert Celsius to Kelvin

We need to convert temperature from degrees Celsius to Kelvin, because the ideal gas law uses temperature in Kelvin. To convert, add 273.15 to the Celsius temperature. Therefore: Initial Temperature = 18°C + 273.15 = 291.15 K Final Temperature = 26°C + 273.15 = 299.15 K
04

4. Calculate the volume of the cylinder

Using the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 \( \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}} \)), and T is temperature, we can calculate the volume of the cylinder: V = \( \frac{nRT}{P} \) = \( \frac{(25.03 \mathrm{mol})(0.0821 \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}})(291.15 \mathrm{K})}{139.58 \mathrm{atm}} \) V = 4.86 L
05

5. Calculate the moles of Argon remaining

We can use the ideal gas law again but with the final pressure and temperature values to determine the moles of Argon remaining in the cylinder: n_final = \( \frac{PV}{RT} \) = \( \frac{(44.24 \mathrm{atm})(4.86 \mathrm{L})}{(0.0821 \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}})(299.15 \mathrm{K})} \) n_final = 8.99 mol
06

6. Convert moles of Argon back to grams

Finally, we can convert the moles of Argon remaining back to grams using the molar mass of Argon (39.95 g/mol): Mass remaining = (8.99 mol) (39.95 g/mol) = 359.14 g There are approximately 359.14 grams of Argon gas remaining in the cylinder.

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

Which noble gas has the smallest density at STP? Explain

A gas sample containing 1.50 moles at \(25^{\circ} \mathrm{C}\) exerts a pressure of 400. torr. Some gas is added to the same container and the temperature is increased to 50\(\cdot^{\circ} \mathrm{C}\). If the pressure increases to 800. torr, how many moles of gas were added to the container? Assume a constant-volume container.

Equal moles of sulfur dioxide gas and oxygen gas are mixed in a flexible reaction vessel and then sparked to initiate the formation of gaseous sulfur trioxide. Assuming that the reaction goes to completion, what is the ratio of the final volume of the gas mixture to the initial volume of the gas mixture if both volumes are measured at the same temperature and pressure?

Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 \(\mathrm{K}\). Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 \(\mathrm{K}\). This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon

Methanol (CH_l3 \(\mathrm{OH}\) ) can be produced by the following reaction: $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$$ Hydrogen at STP flows into a reactor at a rate of 16.0 L/min. Carbon monoxide at STP flows into the reactor at a rate of 25.0 L/min. If 5.30 g methanol is produced per minute, what is the percent yield of the reaction?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

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