Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 7, Problem 7.75QP

(a)

Interpretation Introduction

Interpretation: The root mean square speed and the wavelength of helium atom at the given temperature are to be calculated.

Concept introduction: Wavelength is defined as the equal distance between the two successive troughs and wave crests. It is denoted by a symbol lambda (λ) and is calculated in the unit of meter or nanometer. The wavelength of photons emitted is inversely proportional to the frequency of photon.

To determine: The root mean square speed of helium atom at 500K .

(a)

Expert Solution
Check Mark

Answer to Problem 7.75QP

Solution

The root mean square speed is 1.765×103m/s_ .

Explanation of Solution

Explanation

Given

The temperature is 500K .

The molar mass of helium is 4.0g/mol .

The conversion of g to kg is done as,

1g=103kg

Therefore the conversion of 4.0g to kg is done as,

4g=4×103kg

The molar mass is 4×103kg/mol

The speed of particles in gas phase is measured in terms of root mean square speed. The root mean square speed is given by the equation,

Crms=3RTM

Where,

  • Crms is the root mean square speed.
  • R is the universal gas constant (8.314J/mol.K) .
  • T is the temperature.
  • M is the molar mass.

Substitute the value of R , T and M in the above equation to calculate the root mean square speed of helium.

Crms,He=3×8.314J/mol.K×500K4.0×103kg/mol=12471J4.0×103kg=3117750J/kg=1.765×103m/s_

Since, J/kg=Nm/kg=kgm2/s2kg=m/s .

Hence, the root mean square speed is 1.765×103m/s_ .

(b)

Interpretation Introduction

To determine: The wavelength of helium atom at 500K .

(b)

Expert Solution
Check Mark

Answer to Problem 7.75QP

Solution

The wavelength of helium atom is 5.521×1031nm_ .

Explanation of Solution

Explanation

Given

The mass of helium is 4×103kg/mol

The wavelength is given by equation,

λ=hmu

Where,

  • h is the Planck’s constant (6.626×1034J.s) .
  • u is the velocity (3×108m/s) .
  • m is the mass of the particle in kg .
  • λ is the wavelength of a particle in nm.

Substitute the values of h , u and m in the above equation to calculate the wavelength of helium atom.

λ=6.626×1034kg.m2/s2.s(4.0×103kg)(3×108m/s)=6.626×1034kg.m2/s12×105kg.m/s=5.521×1040m

The conversion of m to nm is done as,

1m=109nm

Therefore, the conversion of 5.521×1040m to nm is done as,

5.521×1040m=5.521×1040×109nm=5.521×1031nm_

Hence, the wavelength of helium atom is 5.521×1031nm_ .

Conclusion

  1. a. The root mean square speed is 1.765×103m/s_ .
  2. b. The wavelength of helium atom is 5.521×1031nm_ .

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Chapter 7 Solutions

Chemistry

Ch. 7.8 - Prob. 11PECh. 7.9 - Prob. 12PECh. 7.9 - Prob. 13PECh. 7.10 - Prob. 14PECh. 7 - Prob. 7.1VPCh. 7 - Prob. 7.2VPCh. 7 - Prob. 7.3VPCh. 7 - Prob. 7.4VPCh. 7 - Prob. 7.5VPCh. 7 - Prob. 7.6VPCh. 7 - Prob. 7.7VPCh. 7 - Prob. 7.8VPCh. 7 - Prob. 7.9VPCh. 7 - Prob. 7.10VPCh. 7 - Prob. 7.11VPCh. 7 - Prob. 7.12VPCh. 7 - Prob. 7.13VPCh. 7 - Prob. 7.14VPCh. 7 - Prob. 7.15VPCh. 7 - Prob. 7.16VPCh. 7 - Prob. 7.17VPCh. 7 - Prob. 7.18VPCh. 7 - Prob. 7.19VPCh. 7 - Prob. 7.20VPCh. 7 - Prob. 7.21QPCh. 7 - Prob. 7.22QPCh. 7 - Prob. 7.23QPCh. 7 - Prob. 7.24QPCh. 7 - Prob. 7.25QPCh. 7 - Prob. 7.26QPCh. 7 - Prob. 7.27QPCh. 7 - Prob. 7.28QPCh. 7 - Prob. 7.29QPCh. 7 - Prob. 7.30QPCh. 7 - Prob. 7.31QPCh. 7 - Prob. 7.32QPCh. 7 - Prob. 7.33QPCh. 7 - Prob. 7.34QPCh. 7 - Prob. 7.35QPCh. 7 - Prob. 7.36QPCh. 7 - Prob. 7.37QPCh. 7 - Prob. 7.38QPCh. 7 - Prob. 7.39QPCh. 7 - Prob. 7.40QPCh. 7 - Prob. 7.41QPCh. 7 - Prob. 7.42QPCh. 7 - Prob. 7.43QPCh. 7 - Prob. 7.44QPCh. 7 - Prob. 7.45QPCh. 7 - Prob. 7.46QPCh. 7 - Prob. 7.47QPCh. 7 - Prob. 7.48QPCh. 7 - Prob. 7.49QPCh. 7 - Prob. 7.50QPCh. 7 - Prob. 7.51QPCh. 7 - Prob. 7.52QPCh. 7 - Prob. 7.53QPCh. 7 - Prob. 7.54QPCh. 7 - Prob. 7.55QPCh. 7 - Prob. 7.56QPCh. 7 - Prob. 7.57QPCh. 7 - Prob. 7.58QPCh. 7 - Prob. 7.59QPCh. 7 - Prob. 7.60QPCh. 7 - Prob. 7.61QPCh. 7 - Prob. 7.62QPCh. 7 - Prob. 7.63QPCh. 7 - Prob. 7.64QPCh. 7 - Prob. 7.65QPCh. 7 - Prob. 7.66QPCh. 7 - Prob. 7.67QPCh. 7 - Prob. 7.68QPCh. 7 - Prob. 7.69QPCh. 7 - Prob. 7.70QPCh. 7 - Prob. 7.71QPCh. 7 - Prob. 7.72QPCh. 7 - Prob. 7.73QPCh. 7 - Prob. 7.74QPCh. 7 - Prob. 7.75QPCh. 7 - Prob. 7.76QPCh. 7 - Prob. 7.77QPCh. 7 - Prob. 7.78QPCh. 7 - Prob. 7.79QPCh. 7 - Prob. 7.80QPCh. 7 - Prob. 7.81QPCh. 7 - Prob. 7.82QPCh. 7 - Prob. 7.83QPCh. 7 - Prob. 7.84QPCh. 7 - Prob. 7.85QPCh. 7 - Prob. 7.86QPCh. 7 - Prob. 7.87QPCh. 7 - Prob. 7.88QPCh. 7 - Prob. 7.89QPCh. 7 - Prob. 7.90QPCh. 7 - Prob. 7.91QPCh. 7 - Prob. 7.92QPCh. 7 - Prob. 7.93QPCh. 7 - Prob. 7.94QPCh. 7 - Prob. 7.95QPCh. 7 - Prob. 7.96QPCh. 7 - Prob. 7.97QPCh. 7 - Prob. 7.98QPCh. 7 - Prob. 7.99QPCh. 7 - Prob. 7.100QPCh. 7 - Prob. 7.101QPCh. 7 - Prob. 7.102QPCh. 7 - Prob. 7.103QPCh. 7 - Prob. 7.104QPCh. 7 - Prob. 7.105QPCh. 7 - Prob. 7.106QPCh. 7 - Prob. 7.107QPCh. 7 - Prob. 7.108QPCh. 7 - Prob. 7.109QPCh. 7 - Prob. 7.110QPCh. 7 - Prob. 7.111QPCh. 7 - Prob. 7.112QPCh. 7 - Prob. 7.113QPCh. 7 - Prob. 7.114QPCh. 7 - Prob. 7.115QPCh. 7 - Prob. 7.116QPCh. 7 - Prob. 7.117QPCh. 7 - Prob. 7.118QPCh. 7 - Prob. 7.119QPCh. 7 - Prob. 7.120QPCh. 7 - Prob. 7.121QPCh. 7 - Prob. 7.122QPCh. 7 - Prob. 7.123QPCh. 7 - Prob. 7.124QPCh. 7 - Prob. 7.125QPCh. 7 - Prob. 7.126QPCh. 7 - Prob. 7.127QPCh. 7 - Prob. 7.128QPCh. 7 - Prob. 7.129APCh. 7 - Prob. 7.130APCh. 7 - Prob. 7.131APCh. 7 - Prob. 7.132APCh. 7 - Prob. 7.133APCh. 7 - Prob. 7.134APCh. 7 - Prob. 7.135APCh. 7 - Prob. 7.136APCh. 7 - Prob. 7.137APCh. 7 - Prob. 7.138APCh. 7 - Prob. 7.139APCh. 7 - Prob. 7.140APCh. 7 - Prob. 7.141APCh. 7 - Prob. 7.142APCh. 7 - Prob. 7.143APCh. 7 - Prob. 7.144APCh. 7 - Prob. 7.145APCh. 7 - Prob. 7.146APCh. 7 - Prob. 7.147APCh. 7 - Prob. 7.148AP
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