/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A 20.0 -L tank contains \(4.86 \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 20.0 -L tank contains \(4.86 \times 10^{-4} \mathrm{kg}\) of helium at \(18.0^{\circ} \mathrm{C} .\) The molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol} .\) (a) How many moles of helium are in the tank? (b) What is the pressure in the tank, in pascals and in atmospheres?

Short Answer

Expert verified
(a) 0.1215 moles of helium; (b) Pressure: 146.5 Pa or 1.45 x 10^-3 atm.

Step by step solution

01

Convert Mass to Moles

We start by converting the mass of helium in the tank into moles using the given molar mass. The formula for converting mass to moles is \( n = \frac{m}{M} \), where \( n \) is the number of moles, \( m \) is the mass (in grams), and \( M \) is the molar mass (in grams per mole). First, convert 4.86 x 10^{-4} kg to grams: \[ 4.86 \times 10^{-4} \text{ kg} \times 1000 \frac{\text{g}}{\text{kg}} = 0.486 \text{ g} \]. Now compute the moles: \[ n = \frac{0.486 \text{ g}}{4.00 \text{ g/mol}} = 0.1215 \text{ mol} \].
02

Convert Celsius to Kelvin

The temperature needs to be in Kelvin for all gas law calculations. Convert Celsius to Kelvin using the formula: \( K = ^\circ \text{C} + 273.15 \). So, \( 18.0 ^\circ\text{C} = 18.0 + 273.15 = 291.15 \text{ K} \).
03

Calculate Pressure in Pascals

Use the Ideal Gas Law to find pressure. The Ideal Gas Law is expressed as: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the ideal gas constant \( 8.314 \text{ J/(mol·K)} \), and \( T \) is temperature in Kelvin. Rearranging for pressure: \[ P = \frac{nRT}{V} \]. Substitute the known values: \[ P = \frac{0.1215 \text{ mol} \times 8.314 \text{ J/(mol·K)} \times 291.15 \text{ K}}{0.0200 \text{ m}^3} \approx 146.5 \text{ Pa} \].
04

Convert Pressure to Atmospheres

Now convert the pressure from pascals to atmospheres. Use the conversion: 1 atm = 101325 Pa. \[ P_{\text{atm}} = \frac{146.5 \text{ Pa}}{101325 \text{ Pa/atm}} \approx 1.45 \times 10^{-3} \text{ atm} \].

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moles Calculation
Calculating the number of moles is a fundamental step in understanding the behavior of gases using the Ideal Gas Law. Moles are a measure of the quantity of a substance. To find the moles of helium in the tank, we use its mass and molar mass. Remember, the molar mass of helium is given as 4.00 g/mol.

First, convert the mass from kilograms to grams since that matches the unit for molar mass better. For the given problem, the mass of helium is 0.486 grams. You then calculate the moles by dividing the mass by the molar mass using the formula: \[ n = \frac{m}{M} \] where \( n \) is the number of moles, \( m \) is the mass in grams, and \( M \) is the molar mass in grams per mole. This gives us 0.1215 moles of helium.

Understanding this conversion is critical as it bridges between the given mass and how many moles of gas we're working with in calculations like using the Ideal Gas Law.
Pressure Conversion
Pressure conversion is an essential skill as pressure may be given or needed in various units depending on the context or region. We often deal with pressure in both pascals (the SI unit) and atmospheres (atm), a convenient unit for many practical applications.

In this exercise, after calculating the pressure in pascals using the Ideal Gas Law, we converted it to atmospheres. The conversion factor used is that 1 atm equals 101325 pascals. To convert, divide the pressure in pascals by this conversion factor: \[ P_{\text{atm}} = \frac{P_{\text{Pa}}}{101325} \]

Given that the pressure in the tank was found to be 146.5 Pa, the conversion to atmospheres results in approximately \( 1.45 \times 10^{-3} \text{ atm} \). This small number reflects the low pressure when considering helium's very low mass and the large volume of the tank.
Temperature Conversion
When dealing with gases, temperature is crucial, and must always be in Kelvin for gas law calculations. Kelvin is the absolute temperature scale used in scientific calculations, where 0 K is absolute zero. To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature. This conversion is necessary because the Ideal Gas Law \( PV = nRT \) requires temperature in Kelvin.

In the original problem, the temperature of the helium gas is given as 18.0°C. Converting it to Kelvin involves adding 273.15, resulting in 291.15 K. This temperature conversion ensures that the values used in calculations align with the demands of the Ideal Gas Law, providing accurate results for pressure and volume relationships.
Helium Gas Properties
Helium is a noble gas, known for its non-reactive nature, colorlessness, and low density. In terms of its properties, helium's unique characteristics make it different from many other gases. It has a low molar mass of 4.00 g/mol, contributing to its low density and making it ideal for use in applications such as balloons and airships.

Because of its low boiling point and inertness, helium is used in cryogenics and as a protective gas for welding. When using it in gas law calculations, these properties mean that often pressures and densities are much lower than with heavier gases. The low atomic mass of helium also implies that a small mass results in a larger volume compared to heavier gases under similar conditions, demonstrating why it occupies a larger volume with lower mass in the Ideal Gas Law calculations. Understanding these properties helps in comprehending why helium behaves the way it does in different scientific and practical contexts.

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

Helium gas with a volume of \(2.60 \mathrm{L},\) under a pressure of 0.180 atm and at a temperature of \(41.0^{\circ} \mathrm{C},\) is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol} .\)

Deviations from the Ideal-Gas Equation. For carbon dioxide gas \(\left(\mathrm{CO}_{2}\right),\) the constants in the van der Waals equation are \(a=0.364 \mathrm{J} \cdot \mathrm{m}^{3} / \mathrm{mol}^{2}\) and \(b=4.27 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol} .\) (a) If 1.00 mol of \(\mathrm{CO}_{2}\) gas at 350 \(\mathrm{K}\) is confined to a volume of \(400 \mathrm{cm}^{3},\) find the pressure of the gas using the ideal-gas equation and the van der Waals equation. (b) Which equation gives a lower pressure? Why? What is the percentage difference of the van der Waals equation result from the ideal-gas equation result? (c) The gas is kept at the same temperature as it expands to a volume of 4000 \(\mathrm{cm}^{3} .\) Repeat the calculations of parts (a) and (b). (d) Explain how your calculations show that the van der Waals equation is equivalent to the ideal-gas equation if \(n / V\) is small.

A diver observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm \()\) to the surface (where the pressure is 1.00 atm). The temperature at the bottom is \(4.0^{\circ} \mathrm{C},\) and the temperature at the surface is \(23.0^{\circ} \mathrm{C}\) . (a) What is the ratio of the volume of the bubble as it reaches the surface to its volume at the bottom? (b) Would it be safe for the diver to hold his breath while ascending from the bottom of the lake to the surface? Why or why not?

You have several identical balloons. You experimentally determine that a balloon will break if its volume exceeds 0.900 L. The pressure of the gas inside the balloon equals air pressure 1.00 atm). (a) If the air inside the balloon is at a constant temperature of \(22.0^{\circ} \mathrm{C}\) and behaves as an ideal gas, what mass of air can you blow into one of the balloons before it bursts? (b) Repeat part (a) if the gas is helium rather than air.

It is possible to make crystalline solids that are only one layer of atoms thick. Such "two-dimensional" crystals can be created by depositing atoms on a very flat surface. (a) If the atoms in such a two-dimensional crystal can move only within the plane of the crystal, what will be its molar heat capacity near room temperature? Give your answer as a multiple of \(R\) and in \(\mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) ) At very low temperatures, will the molar heat capacity of a two-dimensional crystal be greater than, less than, or equal to the result you found in part (a)? Explain why.

See all solutions

Recommended explanations on Physics 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.