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Chapter 21: Problem 4

A \(2.00-\) mol sample of oxygen gas is confined to a \(5.00-\mathrm{L}\) vessel at a pressure of 8.00 atm. Find the average translational kinetic energy of an oxygen molecule under these conditions.

Short Answer

Expert verified
The average translational kinetic energy of an oxygen molecule under these conditions is approximately \(5.05 x 10^{-21}\) J.

Step by step solution

01

Solve for Temperature using the Ideal Gas Law

Using the Ideal Gas Law \( PV = nRT \), where P = 8.00 atm (but in SI units 8.00*1.013x10^5 Pa), V = 5.00 L (but in SI units 5.00/1000 = 0.005 m鲁), n = 2.00 mol, and R = 8.31 J/molK, isolate T: -> \( T = \frac{PV}{nR} \)
02

Substitute the values

Substitute the values into the equation: -> \( T = \frac {8.00*1.013x10^5 Pa * 0.005 m鲁}{2.00 mol * 8.31 J/(mol*K)}\)
03

Solve for Temperature

Solve for T: T is approximately 243.91 K
04

Solve for Average Translational Kinetic Energy

Now, solve for average translational kinetic energy using this formula \( KE_{avg} = \frac{3}{2} kT \), where k = 1.38 x 10^-23 J/K. Substituting the values gives: -> \( KE_{avg} = \frac{3}{2} * 1.38 x 10^-23 J/K * 243.91 K \)
05

Solve for Average Translational Kinetic Energy

Solve for KE_{avg}, which is approximately 5.05 x 10^-21 J.

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Key Concepts

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

Average Translational Kinetic Energy
The average translational kinetic energy refers to the average energy that molecules possess due to their motion. In gases, this energy is particularly significant because the molecules are in constant motion. When discussing gases at the molecular level, kinetic energy is calculated using the formula:
  • \( KE_{avg} = \frac{3}{2} kT \)
Here, \( k \) is the Boltzmann constant, which has a value of \( 1.38 \times 10^{-23} \) J/K. The variable \( T \) represents the absolute temperature in Kelvin.- **Translational Motion:** This is the motion by which a molecule changes its position from one point to another. It is crucial in defining how collision dynamics work in gases.- **Direct Proportionality With Temperature:** As the temperature increases, the average translational kinetic energy of the gas molecules also increases. This means that temperature has a direct effect on how fast or energetically the gas molecules move.
Temperature Calculation
Temperature is a measure of the average kinetic energy of the molecules in a substance. To find the temperature in this exercise, we use the Ideal Gas Law. This law helps relate pressure, volume, and temperature for a given amount of gas.The Ideal Gas Law is expressed as:
  • \( PV = nRT \)
Here, \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant (8.314 J/mol K), and \( T \) is the temperature in Kelvin.In the exercise, we have:- **Pressure (P)**: 8.00 atm, converted to Pascal for SI units, which equals \( 8.00 \times 1.013 \times 10^5 \) Pa.- **Volume (V)**: 5.00 L, which equals 0.005 m鲁 in SI units.- **Number of Moles (n)**: 2.00 mol.To find the temperature, rearrange the formula to solve for \( T \):\[ T = \frac{PV}{nR} \]Substitute the values, and you'll find that the temperature is approximately 243.91 K. This temperature is crucial for calculating the average kinetic energy.
Oxygen Gas Properties
Oxygen is a diatomic molecule, meaning its molecules consist of two oxygen atoms. This property affects its physical behavior as a gas. Oxygen is a vital component for combustion and is crucial for life on Earth. Here are some key properties: - **Molecular Weight:** Approximately 32 g/mol, as it consists of two oxygen atoms each with an atomic weight of 16. - **Diatomic Nature:** Influences oxygen to have rotational and vibrational modes of motion in addition to translational motion. However, in the context of ideal gas and translational kinetic energy, only translational motion is considered. Oxygen gas molecules behave ideally under many conditions, allowing us to use the Ideal Gas Law to make approximations. Additionally, properties like average translational kinetic energy are calculated, considering their molecular motion under specified conditions of pressure, volume, and temperature. This is useful in understanding how temperature and energy relate to the functional properties of gases.

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Most popular questions from this chapter

(a) Show that the speed of sound in an ideal gas is $$v=\sqrt{\frac{\gamma R T}{M}}$$ where \(M\) is the molar mass. Use the general expression for the speed of sound in a fluid from Section \(17.1,\) the definition of the bulk modulus from Section \(12.4,\) and the result of Problem 59 in this chapter. As a sound wave passes through a gas, the compressions are either so rapid or so far apart that thermal conduction is prevented by a negligible time interval or by effective thickness of insulation. The compressions and rarefactions are adiabatic. (b) Compute the theoretical speed of sound in air at \(20^{\circ} \mathrm{C}\) and compare it with the value in Table \(17.1 .\) Take \(M=\) \(28.9 \mathrm{g} / \mathrm{mol} .\) (c) Show that the speed of sound in an ideal gas is $$v=\sqrt{\frac{\gamma k_{\mathrm{B}} T}{m}}$$ where \(m\) is the mass of one molecule. Compare it with the most probable, average, and rms molecular speeds.

A vertical cylinder with a movable piston contains 1.00 mol of a diatomic ideal gas. The volume of the gas is \(V_{i},\) and its temperature is \(T_{i} .\) Then the cylinder is set on a stove and additional weights are piled onto the piston as it moves up, in such a way that the pressure is proportional to the volume and the final volume is \(2 V_{i} .\) (a) What is the final temperature? (b) How much energy is transferred to the gas by heat?

A cylinder containing \(n\) mol of an ideal gas undergoes an adiabatic process. (a) Starting with the expression \(W=-\int P \, d V\) and using the condition ,\(P V^{\gamma}=\) constant show that the work done on the gas is $$W=\left(\frac{1}{\gamma-1}\right)\left(P_{f} V_{f}-P_{i} V_{i}\right)$$ (b) Starting with the first law of thermodynamics in differential form, prove that the work done on the gas is also equal to \(n C_{V}\left(T_{f}-T_{i}\right) .\) Show that this result is consistent with the equation in part (a).

If you can't walk to outer space, can you at least walk halfway? Using the law of atmospheres from Problem \(43,\) we find that the average height of a molecule in the Earth's atmosphere is given by $$\bar{y}=\frac{\int_{0}^{\infty} y n_{V}(y) d y}{\int_{0}^{\infty} n_{V}(y) d y}=\frac{\int_{0}^{\infty} y e^{-m g / k_{B} T} d y}{\int_{0}^{\infty} e^{-m g / k_{B} T} d y}$$ (a) Prove that this average height is equal to \(k_{\mathrm{B}} T / m g\). (b) Evaluate the average height, assuming the temperature is \(10^{\circ} \mathrm{C}\) and the molecular mass is \(28.9 \mathrm{u}\)

A sample of monatomic ideal gas occupies \(5.00 \mathrm{L}\) at atmospheric pressure and \(300 \mathrm{K}\) (point \(A\) in Figure \(\mathrm{P} 21.67\) ). It is heated at constant volume to 3.00 atm (point \(B\) ). Then it is allowed to expand isothermally to 1.00 atm (point \(C\) ) and at last compressed isobarically to its original state. (a) Find the number of moles in the sample. (b) Find the temperature at points \(B\) and \(C\) and the volume at point \(C\). (c) Assuming that the molar specific heat does not depend on temperature, so that \(E_{\text {int }}=3 n R T / 2,\) find the internal energy at points \(A, B,\) and \(C\) (d) Tabulate \(P, V, T,\) and \(E_{\text {int }}\) for the states at points \(A, B,\) and \(C .\) (e) Now consider the processes \(A \rightarrow B, B \rightarrow C\), and \(C \rightarrow A\). Describe just how to carry out each process experimentally. (f) Find \(Q, W,\) and \(\Delta E_{\text {int }}\) for each of the processes. (g) For the whole cycle \(A \rightarrow B \rightarrow C \rightarrow A\) find \(Q, W,\) and \(\Delta E_{\text {int }}.\)
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