/*! 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 2 A compressed cylinder of gas con... [FREE SOLUTION] | 魅影直播

魅影直播

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 2

A compressed cylinder of gas contains \(2.74 \times 10^{3} \mathrm{g}\) of \(\mathrm{N}_{2}\) gas at a pressure of \(3.75 \times 10^{7} \mathrm{Pa}\) and a temperature of \(18.7^{\circ} \mathrm{C} .\) What volume of gas has been released into the atmosphere if the final pressure in the cylinder is \(1.80 \times 10^{5}\) Pa? Assume ideal behavior and that the gas temperature is unchanged.

Short Answer

Expert verified
The volume of gas released into the atmosphere is approximately \(16.15 \ \mathrm{m}^3\).

Step by step solution

01

Understand the Ideal Gas Law

The Ideal Gas Law is represented by the equation: \[PV = nRT\] Where: - P is the pressure of the gas (in Pascals) - V is the volume of the gas (in cubic meters) - n is the amount of substance (in moles) - R is the universal gas constant (\(8.314 \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)) - T is the temperature, in Kelvin (K)
02

Calculate the number of moles

First, we need to find the number of moles (n) of the N鈧� gas. We can do this by using the molar mass of N鈧� gas, which is \(28.02 \ \mathrm{g} \cdot \mathrm{mol}^{-1}\). The given mass of the gas is \(2.74 \times 10^3 \mathrm{g}\), so we can calculate the number of moles as follows: \[n = \frac{2.74 \times 10^3 \mathrm{g}}{28.02 \ \mathrm{g} \cdot \mathrm{mol}^{-1}}\] \(n \approx 97.79 \ \mathrm{mol}\)
03

Convert the temperature to Kelvin

The temperature is given in Celsius, but we need to convert it to Kelvin for the Ideal Gas Law. The conversion is as follows: \[T_\mathrm{K} = T_\mathrm{C} + 273.15\] So, \[T_\mathrm{K} = 18.7^{\circ} \mathrm{C} + 273.15\] \(T_\mathrm{K} \approx 291.85 \ \mathrm{K}\)
04

Calculate the initial volume

Use the Ideal Gas Law to find the initial volume (V鈧�) of the gas in the compressed cylinder: \[V_1 = \frac{nRT}{P_1}\] Where: - \(P_1 = 3.75 \times 10^7 \ \mathrm{Pa}\) (initial pressure) - R and n have been calculated in previous steps - T is in Kelvin So, \[V_1 = \frac{(97.79 \ \mathrm{mol})(8.314 \ \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1})(291.85 \ \mathrm{K})}{3.75 \times 10^7 \ \mathrm{Pa}}\] \(V_1 \approx 0.082 \ \mathrm{m}^3\)
05

Calculate the final volume

Use the Ideal Gas Law again to find the final volume (V鈧�) of the gas in the cylinder after the gas has been released: \[V_2 = \frac{nRT}{P_2}\] Where: - \(P_2 = 1.80 \times 10^5 \ \mathrm{Pa}\) (final pressure) So, \[V_2 = \frac{(97.79 \ \mathrm{mol})(8.314 \ \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1})(291.85 \ \mathrm{K})}{1.80 \times 10^5 \ \mathrm{Pa}}\] \(V_2 \approx 16.23 \ \mathrm{m}^3\)
06

Calculate the volume of gas released

Finally, subtract the initial volume (V鈧�) from the final volume (V鈧�) to find the volume of the gas (螖V) that has been released into the atmosphere: \[\Delta V = V_2 - V_1\] So, \[\Delta V = 16.23 \ \mathrm{m}^3 - 0.082 \ \mathrm{m}^3\] \(\Delta V \approx 16.15 \ \mathrm{m}^3\) So, the volume of gas released into the atmosphere is approximately 16.15 cubic meters.

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 魅影直播!

Key Concepts

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

Moles Calculation
When dealing with gases in chemistry, calculating the number of moles is a fundamental step. Moles measure the amount of substance present, which is a crucial concept in understanding gas reactions and behaviors. In this exercise, we need to determine the moles of nitrogen gas (N鈧�) contained in a cylinder. The process is straightforward once we know the mass and molar mass of the gas.
The molar mass of nitrogen gas (N鈧�) is 28.02 g/mol. We're given a mass of 2.74 x 10鲁 g. To find the moles, use the formula:\[ n = \frac{\text{mass of gas}}{\text{molar mass}} \]Substituting the given values:\[ n = \frac{2.74 \times 10^3 \text{ g}}{28.02 \text{ g/mol}} \]This calculation provides approximately 97.79 moles of nitrogen gas.
  • Moles help in understanding the quantity of gas present and are central to using the Ideal Gas Law.
  • Remember to use molar mass appropriate for the gas in question.
Temperature Conversion to Kelvin
In the Ideal Gas Law, temperature must be in Kelvin. Kelvin is the SI unit of temperature and starts at absolute zero, where all molecular motion ceases. This conversion is a crucial detail when using the equation, \( PV = nRT \).
The given temperature in Celsius needs to be converted to Kelvin. The conversion formula is:\[ T_{\text{K}} = T_{\text{C}} + 273.15 \] For a temperature of 18.7掳C:\[ T_{\text{K}} = 18.7 + 273.15 = 291.85 \text{ K} \]This ensures the Ideal Gas Law can be applied correctly.
  • This conversion standardizes temperature measurements for gas calculations.
  • Absolute zero in Kelvin is 0 K, equivalent to -273.15掳C.
Volume Calculation
The calculation of volume changes in the gas within the cylinder is key in this exercise. Using the Ideal Gas Law, we can calculate the initial and final volumes of gas in the cylinder.

To find the initial volume \( V_1 \), we rearrange the Ideal Gas Law to solve for volume:\[ V_1 = \frac{nRT}{P_1} \]Given:
  • Amount of substance \( n = 97.79 \text{ mol} \)
  • Universal gas constant \( R = 8.314 \text{ J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \)
  • Temperature \( T = 291.85 \text{ K} \)
  • Initial pressure \( P_1 = 3.75 \times 10^7 \text{ Pa} \)
This results in an initial volume:\[ V_1 \approx 0.082 \text{ m}^3 \]For the final volume \( V_2 \), the formula used is the same, but with the final pressure \( P_2 \):\[ V_2 = \frac{nRT}{P_2} \]With \( P_2 = 1.80 \times 10^5 \text{ Pa} \):\[ V_2 \approx 16.23 \text{ m}^3 \]Finally, to find the volume of gas released, subtract the initial volume from the final volume:\[ \Delta V = V_2 - V_1 \approx 16.15 \text{ m}^3 \]
  • This calculation shows how volume changes under different pressure conditions.
  • Gas volume is directly related to pressure, temperature, and moles present.

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

Calculate the pressure exerted by Ar for a molar volume of \(1.31 \mathrm{L} \mathrm{mol}^{-1}\) at \(426 \mathrm{K}\) using the van der Waals equation of state, The van der Waals parameters \(a\) and \(b\) for Ar are 1.355 bar \(\mathrm{dm}^{6} \mathrm{mol}^{-2}\) and \(0.0320 \mathrm{dm}^{3} \mathrm{mol}^{-1},\) respectively. Is the attractive or repulsive portion of the potential dominant under these conditions?

Consider a \(31.0 \mathrm{L}\) sample of moist air at \(60 .^{\circ} \mathrm{C}\) and one atm in which the partial pressure of water vapor is 0.131 atm. Assume that dry air has the composition 78.0 mole percent \(\mathrm{N}_{2}, 21.0\) mole percent \(\mathrm{O}_{2},\) and 1.00 mole percent Ar. a. What are the mole percentages of each of the gases in the sample? b. The percent relative humidity is defined as \(\% \mathrm{RH}=\) \(P_{H_{2}} o / P_{H_{2} O}^{*}\) where \(P_{H_{2}, O}\) is the partial pressure of water in the sample and \(P_{H, O}^{*}=0.197\) atm is the equilibrium vapor pressure of water at \(60 .^{\circ} \mathrm{C}\). The gas is compressed at \(60 .^{\circ} \mathrm{C}\) until the relative humidity is \(100 . \% .\) What volume does the mixture contain now? c. What fraction of the water will be condensed if the total pressure of the mixture is isothermally increased to 81.0 atm?

Calculate the pressure exerted by benzene for a molar volume of \(2.00 \mathrm{L}\) at \(595 \mathrm{K}\) using the Redlich-Kwong equation of state: \\[ \begin{aligned} P &=\frac{R T}{V_{m}-b}-\frac{a}{\sqrt{T}} \frac{1}{V_{m}\left(V_{m}+b\right)} \\\ &=\frac{n R T}{V-n b}-\frac{n^{2} a}{\sqrt{T}} \frac{1}{V(V+n b)} \end{aligned} \\] The Redlich-Kwong parameters \(a\) and \(b\) for benzene are \(452.0 \mathrm{bar} \mathrm{dm}^{6} \mathrm{mol}^{-2} \mathrm{K}^{1 / 2}\) and \(0.08271 \mathrm{dm}^{3} \mathrm{mol}^{-1},\) respec- tively. Is the attractive or repulsive portion of the potential dominant under these conditions?

Approximately how many oxygen molecules arrive each second at the mitochondrion of an active person with a mass of \(84 \mathrm{kg} ?\) The following data are available: Oxygen consumption is about \(40 . \mathrm{mL}\) of \(\mathrm{O}_{2}\) per minute per kilogram of body weight, measured at \(T=300 . \mathrm{K}\) and \(P=1.00 \mathrm{atm} .\) In an adult there are about \(1.6 \times 10^{10}\) cells per kg body mass. Each cell contains about 800 . mitochondria.

Carbon monoxide competes with oxygen for binding sites on the transport protein hemoglobin. CO can be poisonous if inhaled in large quantities. A safe level of CO in air is \(50 .\) parts per million (ppm). When the CO level increases to \(800 .\) ppm, dizziness, nausea, and unconsciousness occur, followed by death. Assuming the partial pressure of oxygen in air at sea level is 0.20 at \(m\), what proportion of \(\mathrm{CO}\) to \(\mathrm{O}_{2}\) is fatal?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

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.