/*! 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 16 Pressure of a mixture of \(4 \ma... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 16

Pressure of a mixture of \(4 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(2 \mathrm{~g}\) of \(\mathrm{H}_{2}\) confined in a bulb of \(1.0 \mathrm{~L}\) capacity at \(0^{\circ} \mathrm{C}\) is (a) \(25.18 \mathrm{~atm}\) (b) \(31.205 \mathrm{~atm}\) (c) \(40.215 \mathrm{~atm}\) (d) \(15.210 \mathrm{~atm}\)

Short Answer

Expert verified
The pressure of the mixture is 25.18 atm, so the answer is (a).

Step by step solution

01

Calculate the Number of Moles of O2

First, calculate the moles of \( \mathrm{O}_2 \). The molar mass of \( \mathrm{O}_2 \) is 32 g/mol. Therefore, the number of moles of \( \mathrm{O}_2 \) can be calculated as:\[\text{{moles of }} \mathrm{O}_2 = \frac{\text{{mass of }} \mathrm{O}_2}{\text{{molar mass of }} \mathrm{O}_2} = \frac{4\, \text{g}}{32\, \text{g/mol}} = 0.125\, \text{mol}\]
02

Calculate the Number of Moles of H2

Next, calculate the moles of \( \mathrm{H}_2 \). The molar mass of \( \mathrm{H}_2 \) is 2 g/mol. Calculate the number of moles as follows:\[\text{{moles of }} \mathrm{H}_2 = \frac{\text{{mass of }} \mathrm{H}_2}{\text{{molar mass of }} \mathrm{H}_2} = \frac{2\, \text{g}}{2\, \text{g/mol}} = 1\, \text{mol}\]
03

Calculate the Total Number of Moles

Add the number of moles of \( \mathrm{O}_2 \) and \( \mathrm{H}_2 \) to find the total number of moles:\[\text{Total moles} = 0.125 + 1 = 1.125\, \text{mol}\]
04

Use the Ideal Gas Law to Find Pressure

Use the ideal gas law to calculate the pressure. The ideal gas law is \( PV = nRT \).- \(P\) is the pressure in atmospheres, which we need to find.- \(V\) is the volume, given as 1.0 L.- \(n\) is the total moles, which is 1.125 mol.- \(R\) is the gas constant, \(0.0821 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}}\).- \(T\) is the temperature in Kelvin. At \(0^{\circ}\text{C}\), \(T = 273.15\, \text{K}\).Rearranging the ideal gas law gives:\[P = \frac{nRT}{V} = \frac{1.125 \cdot 0.0821 \cdot 273.15}{1.0} = 25.18\, \text{atm}\]
05

Match the Answer

Compare the calculated pressure with the provided options. The calculated pressure is \(25.18\, \text{atm}\), which matches option (a).

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Key Concepts

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

Moles Calculation
To find the number of moles in a given substance, you'll need to use its mass and its molar mass. The formula to determine the number of moles is:
  • \( ext{Moles} = \frac{\text{Mass (in grams)}}{\text{Molar mass (g/mol)}} \)
Calculating moles is very important because it allows you to relate the mass of a substance to the number of particles it contains鈥攃ritical for understanding reactions and gas behaviors. If you have a gas sample and know its mass, you can use its molar mass to find how many moles are present, which is directly tied to how gases behave under various conditions.
Molar Mass
The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is determined by summing the atomic masses of the elements in a compound, considering their respective quantities. For instance:
  • The molar mass of \( \text{O}_2 \) is 32 g/mol. This is because oxygen has an atomic mass of approximately 16, and \( \text{O}_2 \) has two oxygen atoms.
  • For \( \text{H}_2 \), the molar mass is 2 g/mol since hydrogen has an atomic mass of about 1, and \( \text{H}_2 \) has two atoms of hydrogen.
Molar mass helps in converting between the mass of a substance and the amount of substance in moles, facilitating calculations in reactions and formulas, especially in gas laws.
Gas Constant
The gas constant \( R \) is crucial in calculations involving gas laws. It bridges the relationship between moles, volume, temperature, and pressure in the ideal gas law. The value of the gas constant \( R \) is:
  • \( R = 0.0821 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}} \)
This constant is used in the ideal gas law equation \( PV = nRT \) where:
  • \( P \) is the pressure in atmospheres,
  • \( V \) is the volume in liters,
  • \( n \) is the number of moles,
  • \( T \) is the temperature in Kelvin.
Using the gas constant allows for solving one of these variables if the others are known, making it an essential tool for understanding gas behavior.
Temperature Conversion
To work with gas laws, converting temperature to Kelvin from Celsius is necessary because the Kelvin scale is absolute. The Kelvin scale is directly related to the energy of the particles; it starts at absolute zero, the point where particle motion stops. The conversion from Celsius to Kelvin is straightforward:
  • \( T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \)
This step is crucial when using the ideal gas law, because the formula \( PV = nRT \) relies on temperature in Kelvin to accurately describe the gas' behavior. In the exercise example, the temperature at \( 0^{\circ} \text{C} \) becomes \( 273.15 \text{K} \), which is then used in the final calculation of pressure.

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Most popular questions from this chapter

The ration between the r.m.s velocity of \(\mathrm{H}_{2}\) at \(50 \mathrm{~K}\) ad that of \(\mathrm{O}_{2}\) at \(800 \mathrm{~K}\) is

The ratio between the root mean square velocity of \(\mathrm{H}_{2}\) at \(50 \mathrm{~K}\) and that of \(\mathrm{O}_{2}\) at \(800 \mathrm{~K}\) is (a) 4 (b) 2 (c) 1 (d) \(\frac{1}{4}\)

In the given sample of a gas all molecules do not possess same speed. Due to frequent molecular collisions, the molecules move with ever changing speeds and also with changing direction. There are three types of velocities (i) root mean square velocity (ii) average velocity and (iii) most probable velocity. Oxygen has a density of \(1.429 \mathrm{gm} / \mathrm{L}\) at STP. The RMS velocity of \(\mathrm{O}_{2}\) molecules in \(\mathrm{cms}_{-1}\) (a) \(4.61 \times 10_{3}\) (b) \(4.16 \times 10_{3}\) (c) \(46.1 \times 10_{3}\) (d) \(6.41 \times 10_{3}^{3}\)

A gas, which obeys Boyle's law, Charle's law, Avogadro's law etc., or ideal gas equation \(\mathrm{PV}=\mathrm{nRT}\) under all conditions of temperature and pressure, is called ideal gas. No gas is ideal. All gases are real gases. The real gas obeys these gas laws only when the temperature is high or pressure is low. The extent of derivations of a real gas form ideal behaviour is expressed in terms of compressibility factor \(Z\) defined as \(\mathrm{Z}=\frac{\mathrm{PV}}{\mathrm{nRT}}\) Real gases have characteristic temperatures like critical temperature, inversion temperature and Boyle temperature. These temperatures can be calculated using van der Waal constants. Density of two gases of same molecular weight are in the ratio \(1: 3\) and their temperatures are in the ratio \(3: 2 .\) The ratio of respective pressures is (a) \(2: 1\) (b) \(2: 3\) (c) \(3: 2\) (d) \(1: 2\)

For an ideal gas, number of mole per litre in terms of its pressure \(\mathrm{P}\), temperature \(\mathrm{T}\) and gas constant \(\mathrm{R}\) is [2002] (a) \(\mathrm{PT} / \mathrm{R}\) (b) PRT (c) \(\mathrm{P} / \mathrm{RT}\) (d) \(\mathrm{RT} / \mathrm{P}\)
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