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Chapter 3: Problem 60

What is the ratio of kinetic energies of \(3 \mathrm{~g}\) of hydrogen and 4 grams of oxygen at \(\mathrm{T}(\mathrm{K})\) ? (a) \(12: 1\) (b) \(6: 1\) (c) \(1: 6\) (d) \(24: 1\)

Short Answer

Expert verified
The ratio of kinetic energies of hydrogen and oxygen is 12:1.

Step by step solution

01

Understand Kinetic Energy Formula

The kinetic energy of gas molecules at a particular temperature can be given by the formula \( KE = \frac{3}{2} k T \), where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin. However, here we're dealing with moles of gas, so we use molar kinetic energy as \( KE = \frac{3}{2} RT \), where \( R \) is the universal gas constant.
02

Calculate Moles of Gases

First, we need to calculate the moles of hydrogen and oxygen. The molar mass of hydrogen is approximately \(2 \mathrm{~g}/\mathrm{mol}\) and for oxygen is \(32 \mathrm{~g}/\mathrm{mol}\). Thus, moles of hydrogen = \( \frac{3}{2} = 1.5 \) moles, and moles of oxygen = \( \frac{4}{32} = 0.125 \) moles.
03

Calculate Kinetic Energy for Each Gas

Since the kinetic energy per mole for any ideal gas at a given temperature \( T \) is the same, total kinetic energy for a gas is given by \( KE = n \times \frac{3}{2} RT \). Therefore, kinetic energy for hydrogen is \( KE_H = 1.5 \times \frac{3}{2} RT \) and for oxygen is \( KE_O = 0.125 \times \frac{3}{2} RT \).
04

Ratio of Kinetic Energies

Now, we find the ratio of kinetic energies: \( \text{Ratio} = \frac{KE_H}{KE_O} = \frac{1.5 \times \frac{3}{2} RT}{0.125 \times \frac{3}{2} RT} = \frac{1.5}{0.125} \). Calculate this ratio to get \( \frac{1.5}{0.125} = 12 \).
05

Conclusion

The ratio of kinetic energies of the hydrogen to oxygen is therefore 12:1. Compare it with the given options.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry and physics that relates four key properties of gases: pressure, volume, temperature, and the number of moles. The formula is expressed as:\[ PV = nRT \]where:
  • P is the pressure of the gas
  • V is the volume of the gas
  • n is the number of moles of gas
  • R is the universal gas constant \(8.314 \, \text{J/(mol K)}\)
  • T is the temperature in Kelvin
This law assumes that the gas behaves ideally, meaning that the molecules do not interact with one another and occupy no volume themselves. In reality, gases approximate this behavior under high temperature and low pressure conditions, where intermolecular forces are negligible.The ideal gas law can be used to derive important information about the gas when other variables are known. For instance, if you know the pressure, volume, and temperature of a gas, you can calculate the amount in moles, and vice versa.
Moles Calculation
Calculating the moles of a substance is crucial for understanding how much of a compound or element is present in a given sample.Moles are calculated using the formula:\[ n = \frac{m}{M} \]where:
  • \( n \) is the number of moles
  • \( m \) is the mass of the substance in grams
  • \( M \) is the molar mass of the substance in \( \text{g/mol} \)
In the step-by-step solution, we calculate the moles of hydrogen and oxygen using their given masses and molar masses. For hydrogen, with a molar mass of approximately \(2 \, \text{g/mol}\), the moles are calculated as \( \frac{3}{2} = 1.5 \) moles. For oxygen, which has a molar mass of \(32 \, \text{g/mol}\), the moles are \( \frac{4}{32} = 0.125 \) moles.By understanding how to calculate moles, we can determine the quantity of a substance present and further use this in various equations, such as those involving kinetic energy.
Molar Mass
Molar mass is a key concept in chemistry that is used to convert between the mass of a substance and the number of moles. The molar mass is the mass of one mole of a given substance, typically expressed in grams per mole (\(\text{g/mol}\)). It is calculated by summing the atomic masses of all atoms in a molecule of the substance, as provided on the periodic table.For instance:
  • Hydrogen (H2) has a molar mass of approximately \(2 \, \text{g/mol}\) since the atomic mass of a single hydrogen atom is \(1 \, \text{u}\).
  • Oxygen (O2) has a molar mass of approximately \(32 \, \text{g/mol}\) because the atomic mass of a single oxygen atom is \(16 \, \text{u}\).
Molar mass is instrumental in converting mass to moles and vice versa. It provides the key link required to use the ideal gas law and perform calculations involving chemical reactions and kinetic energy, as demonstrated in the original exercise. Understanding molar mass allows you to accurately determine how a substance will behave under different chemical scenarios.

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Most popular questions from this chapter

One litre of gas \(\mathrm{A}\) at \(2 \mathrm{~atm}\) pressure at \(27^{\circ} \mathrm{C}\) and two litres of gas \(\mathrm{B}\) at \(3 \mathrm{~atm}\) pressure at \(127^{\circ} \mathrm{C}\) are mixed in a 4 litre vessel. The temperature of the mixture is maintained at \(327^{\circ} \mathrm{C}\). What is the total pressure of the gaseous mixture? (a) \(3.93 \mathrm{~atm}\) (b) \(3.25 \mathrm{~atm}\) (c) \(4.25 \mathrm{~atm}\) (d) \(6.25 \mathrm{~atm}\)

A gas has the van der Waals constants, \(a=1.49 \mathrm{~L}^{2}\) \(\mathrm{atm} \mathrm{mol}^{-2}\) and \(\mathrm{b}=0.04 \mathrm{~L} \mathrm{~mol}^{-1} .\) Its Boyle's temperature is nearly (a) \(50^{\circ} \mathrm{C}\) (b) \(354^{\circ} \mathrm{C}\) (c) \(454 \mathrm{~K}\) (d) \(408 \mathrm{~K}\)

Containers A and B have same gases. Pressure, volume and temperature of \(A\) are all twice that of \(B\), then the ratio of number of molecules of \(\mathrm{A}\) and \(\mathrm{B}\) are (a) \(1: 2\) (b) 2 (c) \(1: 4\) (d) 4

If two moles of an ideal gas at a temperature \(546 \mathrm{~K}\), occupy a volume of \(44.8\) litres its pressure must be (a) \(4 \mathrm{~atm}\) (b) \(3 \mathrm{~atm}\) (c) \(2 \mathrm{~atm}\) (d) \(1 \mathrm{~atm}\)

If \(\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3} \ldots \ldots \ldots\) represents the speed of \(\mathrm{n}_{1}\), \(\mathrm{n}_{2}, \mathrm{n}_{3}, \ldots\) molecules, then the root mean square of speed is (a) \(\left(\frac{\mathrm{n}_{1} \mathrm{C}_{1}^{2}+\mathrm{n}_{2} \mathrm{C}_{2}^{2}+\mathrm{n}_{3} \mathrm{C}_{3}^{2}+\ldots}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}+\ldots}\right)^{1 / 2}\) (b) \(\left(\frac{n_{1} C_{1}^{2}+n_{2} C_{2}^{2}+n_{3} C_{3}^{2}+\ldots}{n_{1}+n_{2}+n_{3}+\ldots}\right)^{2}\) (c) \(\frac{\left(\mathrm{n}_{1} \mathrm{C}_{1}^{2}\right)^{1 / 2}}{\mathrm{n}_{1}}+\frac{\left(\mathrm{n}_{2} \mathrm{C}_{2}^{2}\right)^{1 / 2}}{\mathrm{n}_{2}}+\frac{\left(\mathrm{n}_{3} \mathrm{C}_{3}^{2}\right)^{1 / 2}}{\mathrm{n}_{3}}+\ldots\) (d) \(\left[\frac{\left(n_{1} C_{1}+n_{2} C_{2}+n_{3} C_{3}+\ldots\right)^{2}}{n_{1}+n_{2}+n_{3}+\ldots}\right]^{1 / 2}\)
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