/*! 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} Q43P Question 43: (II) (a) What is th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Q43P (page 138)

Question 43: (II) (a) What is the average translational kinetic energy of a nitrogen molecule at STP? (b) What is the total translational kinetic energy of 1.0 mol of molecules at 25掳C?

Short Answer

Expert verified
  1. The average translational kinetic energy of a nitrogen molecule at STP is \(5.65 \times {10^{ - 21}}\;{\rm{J}}\).
  2. The total translational kinetic energy of 1.0 mole of molecules at \(25^\circ {\rm{C}}\) is \(3.{\rm{7}} \times {10^3}\;{\rm{J}}\).

Step by step solution

01

Given Data

The pressure of the nitrogen gas at STP is \(P = 1\;{\rm{atm}}. = 1.013 \times {10^{ - 5}}\;{\rm{Pa}}\).

The temperature of the nitrogen gas at STP is \(T = 0^\circ {\rm{C}} = 273\;{\rm{K}}\).

The number of moles of nitrogen gas is n = 1.0 mol.

The temperature is \(T' = 25^\circ {\rm{C}} = \left( {25 + 273} \right)\;{\rm{K}} = 29{\rm{8}}\;{\rm{K}}\).

02

Understanding average translational Kinetic energy of molecules

According to the Kinetic theory of gases,the average translational kinetic energy of the molecules in random motion in an ideal gas is directly proportional to the absolute temperature of the gas.

The average translational kinetic energy is given as

\(\overline {KE} = \frac{3}{2}kT\).

Here, k is the Boltzmann constant, whose value is \(1.38 \times {10^{ - 23}}\;{\rm{J/K}}\).

03

(a) Determination of average translational kinetic energy of the nitrogen molecule

The average translational kinetic energy of a nitrogen molecule at STP can be calculated in the following manner:

\(\begin{aligned}{c}\overline {KE} &= \frac{3}{2}kT\\ &= \frac{3}{2}\left( {1.38 \times {{10}^{ - 23}}\;{\rm{J/K}}} \right)\left( {273\;{\rm{K}}} \right)\\ &= 5.65 \times {10^{ - 21}}\;{\rm{J}}\end{aligned}\)

Thus, the average translational kinetic energy of a nitrogen molecule at STP is \(5.65 \times {10^{ - 21}}\;{\rm{J}}\).

04

(b) Determination of total translational kinetic energy of one mol of molecules

The average translational kinetic energy of a molecule at \(25^\circ {\rm{C}}\) can be calculated in the following manner:

\(\begin{aligned}{c}\overline {KE} &= \frac{3}{2}kT'\\ &= \frac{3}{2}\left( {1.38 \times {{10}^{ - 23}}\;{\rm{J/K}}} \right)\left( {298\;{\rm{K}}} \right)\\ &= 616.9 \times {10^{ - 23}}\;{\rm{J}}\end{aligned}\)

One mole of an ideal gas contains an Avogadro number of molecules, that is,\({N_{\rm{A}}} = 6.02 \times {10^{23}}\;{\rm{molecules}}\).

The total translational kinetic energy of one mole of molecules at \(25^\circ {\rm{C}}\) can be calculated in the following manner:

\(\begin{aligned}{c}K{E_{{\rm{Total}}}} &= {{\rm{N}}_{\rm{A}}}\left( {\overline {KE} } \right)\\ &= \left( {6.02 \times {{10}^{23}}} \right)\left( {616.9 \times {{10}^{ - 23}}\;{\rm{J}}} \right)\\ &= 3713.7\;{\rm{J}}\\ &= 3.{\rm{7}} \times {10^3}\;{\rm{J}}\end{aligned}\)

Thus, the total translational kinetic energy of 1.0 mole of molecules at \(25^\circ {\rm{C}}\) is \(3.{\rm{7}} \times {10^3}\;{\rm{J}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A hand exerts a constant horizontal force on a block that is free to slide on a frictionless surface (Fig. 6鈥�30). The block starts from rest at point A, and by the time it has traveled a distance d to point B it is traveling with speed \({v_{\rm{B}}}\). When the block has traveled another distance d to point C, will its speed be greater than, less than, or equal to \(2{v_{\rm{B}}}\)? Explain your reasoning.

FIGURE 6鈥�30 Question 8.

The space shuttle launches an 850-kg satellite by ejecting it from the cargo bay. The ejection mechanism is activated and is in contact with the satellite for 4.8 s to give it a velocity of\(0.30\;{\rm{m/s}}\)in the x direction relative to the shuttle. The mass of the shuttle is 92,000 kg. (a) Determine the component of velocity\({v_{\rm{f}}}\)of the shuttle in the minus x direction resulting from the ejection. (b) Find the average force that the shuttle exerts on the satellite during the ejection.

(II) A ski starts from rest and slides down a 28掳 incline 85 m long. (a) If the coefficient of friction is 0.090, what is the ski鈥檚 speed at the base of the incline? (b) If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level? Use energy methods.

A 950 kg car strikes a huge spring at a speed of 25 m/s (Fig. 11鈥�57), compressing the spring 4.0 m. (a) What is the spring stiffness constant of the spring? (b) How long is the car in contact with the spring before it bounces off in the opposite direction?

In what ways is the word 鈥渨ork鈥� as used in everyday language the same as it is defined in physics? In what ways is it different? Give examples of both.
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Rotational Dynamics

Read Explanation

Astrophysics

Read Explanation

Famous Physicists

Read Explanation

Dynamics

Read Explanation

Energy Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.