/*! 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 26 \(\bullet$$\bullet\) At what tem... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\bullet$$\bullet\) At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at \(20.0^{\circ} \mathrm{C} ?\) (Hint: The periodic table in Appendix \(C\) shows the molar mass (in \(\mathrm{g} / \mathrm{mol} )\) of each element under the chemical symbol for that element. The molar mass of \(\mathrm{H}_{2}\) is twice the molar mass of hydrogen atoms, and similarly for \(\mathrm{N}_{2}.)\)

Short Answer

Expert verified
The temperature is \( 4071.93 \text{ K} \).

Step by step solution

01

Understanding the Root-Mean-Square Speed Formula

The root-mean-square (RMS) speed of a gas molecule is given by the formula \( v_{rms} = \sqrt{\frac{3k_B T}{m}} \), where \( k_B \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the mass of a gas molecule. For molar calculations, this is often expressed in terms of molar mass \( M \) as \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the ideal gas constant and \( M \) is the molar mass in kg/mol.
02

Equating RMS Speeds of Nitrogen and Hydrogen

To find the temperature at which the RMS speed of nitrogen is equal to the RMS speed of hydrogen at \( 20.0^{\circ} \mathrm{C} \), set the formulas for both gases equal: \( \sqrt{\frac{3RT_{N_2}}{M_{N_2}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}} \). This implies that \( \frac{T_{N_2}}{M_{N_2}} = \frac{T_{H_2}}{M_{H_2}} \).
03

Substituting Known Values

Convert the temperature of hydrogen to Kelvin: \( T_{H_2} = 20.0 + 273.15 = 293.15 \text{ K} \). The molar masses are \( M_{H_2} = 2 \times 1.008 = 2.016 \text{ g/mol} = 0.002016 \text{ kg/mol} \) and \( M_{N_2} = 2 \times 14.01 = 28.02 \text{ g/mol} = 0.02802 \text{ kg/mol} \).
04

Solving for Temperature of Nitrogen

Using the equation \( \frac{T_{N_2}}{0.02802} = \frac{293.15}{0.002016} \), solve for \( T_{N_2} \):\[T_{N_2} = 293.15 \times \frac{0.02802}{0.002016} = 4071.93 \text{ K}.\]
05

Conclusion

The temperature at which the RMS speed of nitrogen equals the RMS speed of hydrogen at \( 20.0^{\circ} \mathrm{C} \) is \( 4071.93 \text{ K} \).

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Key Concepts

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

Root-Mean-Square Speed
The root-mean-square speed (often abbreviated as RMS speed) is a way of quantifying the typical speed of gas molecules in thermal motion. In the context of the "kinetic theory of gases," RMS speed is particularly useful because it links the microscopic properties of gas molecules to macroscopic observable characteristics like temperature and pressure. The formula for RMS speed is \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( M \) represents the molar mass of the gas molecules in kg/mol.
This formula indicates that the RMS speed is not only dependent on temperature but also on the molar mass of the molecules. Intuitively, at the same temperature, lighter molecules will move faster than heavier ones. Hence, understanding RMS speed helps us appreciate the dynamic interactions and collisions that occur between molecules in a gas.
Remember, RMS speed is an average measure, which means that while it represents a common speed, individual molecules in the gas may have velocities that are both higher and lower than this average.
Molar Mass
Molar mass is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance, measured in moles. It is usually expressed in units of grams per mole (g/mol). For gases, getting the molar mass right is crucial because it heavily influences the subsequent calculations, such as those for RMS speed.
In our exercise, the molecules in question are hydrogen (\(H_2\)) and nitrogen (\(N_2\)). The molar mass of \(H_2\) is double the atomic mass of a single hydrogen atom, and similarly, \(N_2\) is twice the atomic mass of a nitrogen atom. With \(H_2\), the molar mass is around 2.016 g/mol, and for \(N_2\), it's approximately 28.02 g/mol.
Understanding molar mass helps connect the microscopic scale of molecules and atoms with the macroscopic scale of mass we observe and use in calculations for chemical reactions and gas behaviors.
Temperature Conversion
Temperature conversion is a key step in thermodynamic calculations, especially when working with gas laws and kinetic theory. In science, temperatures in calculations are most commonly expressed in Kelvin. This is because Kelvin is absolute temperature scale; it starts at absolute zero, making it suitable for measuring thermal energy.
To convert Celsius to Kelvin, add 273.15 to the Celsius temperature:\( T_{K} = T_{°C} + 273.15 \).
For the root-mean-square speed calculations in the exercise, it was essential to convert the temperature of hydrogen from Celsius (20.0°C) to Kelvin, resulting in 293.15 K. This conversion ensures that our calculations are consistent with the properties of the Ideal Gas Law, where temperature needs to be absolute to accurately reflect the thermal energy.
Ideal Gas Constant
The ideal gas constant, symbolized as \( R \), is a fundamental constant in thermodynamics and appears in the ideal gas law \( PV = nRT \). Its value is approximately 8.314 J/(mol·K), making it useful in connecting energy considerations with temperature (Kelvin) and moles of substance.
The appearance of \( R \) in the root-mean-square speed formula \( v_{rms} = \sqrt{\frac{3RT}{M}} \) emphasizes its role in linking the large-scale behavior of gases (in terms of temperature and molar quantity) to the small-scale kinetic energies of molecules. Knowing and using the ideal gas constant correctly is essential to solving problems within the realm of kinetic theory and gas law applications.
Each occurrence of the ideal gas constant in equations involving gases serves to ensure that the energy considerations accounted for are accurate, based on the scale (moles) and energy (temperature) of the system in question.

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Most popular questions from this chapter

\(\bullet$$\bullet\) If a certain amount of ideal gas occupies a volume \(V\) at \(S T P\) on earth, what would be its volume (in terms of \(V )\) on Venus, where the temperature is \(1003^{\circ} \mathrm{C}\) and the pressure is 92 atm?

\(\bullet\) An experimenter adds 970 \(\mathrm{J}\) of heat to 1.75 \(\mathrm{mol}\) of an ideal gas to heat it from \(10.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C}\) at constant pressure. The gas does \(+223 \mathrm{Jof}\) work during the expansion. (a) Calculate the change in internal energy of the gas. (b) Calculate \(\gamma\) for the gas.

As the temperature of a monatomic gas increases, we expect its specific heat to A. remain the same B. increase C. decrease

\(\bullet$$\bullet\) A player bounces a basketball on the floor, compressing it to 80.0\(\%\) of its original volume. The air (assume it is essentially \(\mathrm{N}_{2}\) gas) inside the ball is originally at a temperature of \(20.2^{\circ} \mathrm{C}\) and a pressure of 2.00 atm. The ball's diameter is 23.9 \(\mathrm{cm} .\) (a) What temperature does the air in the ball reach at its maximum compression? (b) By how much does the internal energy of the air change between the ball's original state and its maximum compression?

\(\bullet\) A cylinder contains 0.250 mol of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas at a temperature of \(27.0^{\circ} \mathrm{C}\) . The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to \(127.0^{\circ} \mathrm{C}\) . Assume that the \(\mathrm{CO}_{2}\) may be treated as an ideal gas. (a) Draw a \(p V\) diagram of this process. (b) How much work is done by the gas in the process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

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