Problem 60

Question

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) \(24: 1\)

Step-by-Step Solution

Verified
Answer
The ratio of kinetic energies of hydrogen and oxygen is 12:1.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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 \( KE_O = 0.125 \times \frac{3}{2} RT \).
4Step 4: 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 \( \frac{1.5}{0.125} = 12 \).
5Step 5: Conclusion
The ratio of kinetic energies of the hydrogen to oxygen is therefore 12:1. Compare it with the given options.

Key Concepts

Ideal Gas LawMoles CalculationMolar Mass
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.