/*! 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 28 A 12.8 L cylinder contains \(35.... [FREE SOLUTION] | 91Ó°ÊÓ

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A 12.8 L cylinder contains \(35.8 \mathrm{g} \mathrm{O}_{2}\) at \(46^{\circ} \mathrm{C}\). What is the pressure of this gas, in atmospheres?

Short Answer

Expert verified
The pressure of the gas is \(2.29 \, \mathrm{atm}\).

Step by step solution

01

Convert grams of \(O_2\) to moles

We first need to convert the mass of \(O_2\) (oxygen) in grams into moles, by using the molar mass of \(O_2\), which is \(32.00 \, \mathrm{g/mol}\). This is done using the formula \(n = m/M\), where \(m\) is the mass and \(M\) is the molar mass. This gives us: \[n = 35.8 \, \mathrm{g} / 32.00 \, \mathrm{g/mol} = 1.119 \, \mathrm{mol}\]
02

Convert temperature from Celsius to Kelvin

We need to convert the temperature from Celsius to Kelvin, because the ideal gas law uses the Kelvin temperature scale. We can do this using the formula \(T(K) = T(C) + 273.15\), which gives us: \[T = 46^{\circ}C + 273.15 = 319.15 \, K\]
03

Substitute values into the ideal gas law and solve for P

Next, we put these values into the ideal gas law equation \(PV = nRT\), and solve for the pressure \(P\). Remembering that the ideal gas constant \(R = 0.0821 \, \mathrm{(atm \cdot L)/(mol \cdot K)}\), we get: \[P = \frac{nRT}{V} = \frac{1.119 \, \mathrm{mol} \cdot 0.0821 \, \mathrm{(atm \cdot L)/(mol \cdot K)} \cdot 319.15 \, K}{12.8 \, L} = 2.29 \, \mathrm{atm}\] This pressure is the final answer to the exercise.

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Key Concepts

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

Molar Mass
In the context of chemistry, molar mass is a crucial concept. It represents the mass of one mole of a given substance. For oxygen (O_2), the molar mass is 32.00 \, \mathrm{g/mol}. This receives its value from the sum of the atomic masses of two oxygen atoms. Calculating moles from a given mass involves the formula \(n = \frac{m}{M}\), where:
  • \(n\) is the number of moles.
  • \(m\) is the mass in grams.
  • \(M\) is the molar mass in \(\mathrm{g/mol}\).

This equation forms the basis for many calculations in chemistry. For example, if you have 35.8 grams of O_2, dividing this by 32.00 \, \mathrm{g/mol} yields approximately 1.119 moles. This conversion helps in elucidating other properties of gases, such as calculations involving the Ideal Gas Law.
Understanding the basis of molar mass allows you to bridge the gap between macroscopic measurements and molecular quantities. This is essential for experiments and theoretical calculations in chemistry.
Temperature Conversion
Temperature conversion is a fundamental task when dealing with gas laws. In science, temperatures must often be converted to Kelvin. This is due to the Kelvin scale being absolute, which aligns with the equations of thermodynamics, including the Ideal Gas Law.
To convert from Celsius to Kelvin, you use the formula:
  • \(T(K) = T(C) + 273.15\).
This simple addition helps you align your calculations with scientific standards. For instance, a temperature of 46^{\circ}C converts to 319.15 \, K.
Utilizing the Kelvin scale ensures that all thermal energy values remain proportional, avoiding the pitfalls of scales that include negative values, such as Celsius. Maintaining consistency in temperature calculations ensures the accuracy of gas law equations, critical for achieving precise results in chemical experiments.
Gas Pressure Calculation
Gas pressure calculation is an important part of understanding how gases behave. The Ideal Gas Law is a comprehensive formula that relates the pressure (P), volume (V), and temperature (T) of a gas to the number of moles (n) and the gas constant (R).
  • The formula is expressed as \(PV = nRT\).
  • \(P\) is the pressure in atmospheres.
  • \(V\) is the volume in liters.
  • \(n\) is the number of moles.
  • \(R\) is the ideal gas constant, 0.0821 \, \mathrm{(atm \cdot L)/(mol \cdot K)}.
  • \(T\) is the temperature in Kelvin.

By rearranging this equation, you can solve for pressure: \(P = \frac{nRT}{V}\). With given values, such as 1.119 moles of O_2, a volume of 12.8 \, L, and a temperature of 319.15 \, K, you can determine the pressure as approximately 2.29 \, atm.
This showcases the interdependence of temperature, volume, and moles in determining the state of a gas. Each parameter within the Ideal Gas Law holds significance in predicting and analyzing gas behaviors under various conditions.

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Most popular questions from this chapter

A mixture of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) is prepared by electrolyzing \(1.32 \mathrm{g}\) water, and the mixture of gases is collected over water at \(30^{\circ} \mathrm{C}\) and \(748 \mathrm{mmHg} .\) The volume of "wet" gas obtained is 2.90 L. What must be the vapor pressure of water at \(30^{\circ} \mathrm{C} ?\) $$2 \mathrm{H}_{2} \mathrm{O}(1) \stackrel{\text { electrolysis }}{\longrightarrow} 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$

Briefly describe each concept or process: (a) absolute zero of temperature; (b) collection of a gas over water; (c) effusion of a gas; (d) law of combining volumes.

Explain why it is necessary to include the density of \(\mathrm{Hg}(1)\) and the value of the acceleration due to gravity, \(g,\) in a precise definition of a millimeter of mercury (page 194 ).

In research that required the careful measurement of gas densities, John Rayleigh, a physicist, found that the density of \(\mathrm{O}_{2}(\mathrm{g})\) had the same value whether the gas was obtained from air or derived from one of its compounds. The situation with \(\mathrm{N}_{2}(\mathrm{g})\) was different, however. The density of \(\mathrm{N}_{2}(\mathrm{g})\) had the same value when the \(\mathrm{N}_{2}(\mathrm{g})\) was derived from any of various compounds, but a different value if the \(\mathrm{N}_{2}(\mathrm{g})\) was extracted from air. In \(1894,\) Rayleigh enlisted the aid of William Ramsay, a chemist, to solve this apparent mystery; in the course of their work they discovered the noble gases. (a) Why do you suppose that the \(\mathrm{N}_{2}(\mathrm{g})\) extracted from liquid air did not have the same density as \(\mathrm{N}_{2}(\mathrm{g})\) obtained from its compounds? (b) Which gas do you suppose had the greater density: \(\mathrm{N}_{2}(\mathrm{g})\) extracted from air or \(\mathrm{N}_{2}(\mathrm{g})\) prepared from nitrogen compounds? Explain. (c) The way in which Ramsay proved that nitrogen gas extracted from air was itself a mixture of gases involved allowing this nitrogen to react with magnesium metal to form magnesium nitride. Explain the significance of this experiment. (d) Calculate the percent difference in the densities at \(0.00^{\circ} \mathrm{C}\) and 1.00 atm of Rayleigh's \(\mathrm{N}_{2}(\mathrm{g})\) extracted from air and \(\mathrm{N}_{2}(\mathrm{g})\) derived from nitrogen compounds. [The volume percentages of the major components of air are \(78.084 \% \mathrm{N}_{2}, 20.946 \% \mathrm{O}_{2}, 0.934 \% \mathrm{Ar},\) and \(0.0379 \% \mathrm{CO}_{2} .\)

A \(0.156 \mathrm{g}\) sample of a magnesium-aluminum alloy dissolves completely in an excess of \(\mathrm{HCl}(\mathrm{aq}) .\) The liberated \(\mathrm{H}_{2}(\mathrm{g})\) is collected over water at \(5^{\circ} \mathrm{C}\) when the barometric pressure is 752 Torr. After the gas is collected, the water and gas gradually warm to the prevailing room temperature of \(23^{\circ} \mathrm{C} .\) The pressure of the collected gas is again equalized against the barometric pressure of 752 Torr, and its volume is found to be \(202 \mathrm{mL}\). What is the percent composition of the magnesium-aluminum alloy? (Vapor pressure of water: \(6.54 \mathrm{mmHg}\) at \(5^{\circ} \mathrm{C}\) and \(21.07 \mathrm{mmHg}\) at \(\left.23^{\circ} \mathrm{C}\right)\)

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