EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 19, Problem 26P

(a)

To determine

The average time before molecules reach in left half of the box.

(a)

Expert Solution
Check Mark

Answer to Problem 26P

The average time before molecules reach in left half of the box is 5sec .

Explanation of Solution

Given:

The number of molecules is 10 .

The volume of the box is V=1.0L .

Formula used:

The expression for average time is given as,

  t=2n2(100s 1)

Here, n is the number of molecules.

Calculation:

The average time for n=10 molecules can be calculated as,

  t=2n2( 100 s 1 )=2 102( 100 s 1 )=5.12s5s

Conclusion:

Therefore, the average time before molecules reach in left half of the box is 5sec .

(b)

To determine

The average time before molecules reach in left half of the box.

(b)

Expert Solution
Check Mark

Answer to Problem 26P

The average time before molecules reach in left half of the box is 2×1020yr .

Explanation of Solution

Given:

The number of molecules is 100 .

Calculation:

The average time for n=100 molecules can be calculated as,

  t=2n2( 100 s 1 )=2 1002( 100 s 1 )=6.34×1027s×1yr3.156× 107s2×1020yr

Conclusion:

Therefore, the average time before molecules reach in left half of the box is 2×1020yr .

(c)

To determine

The average time before molecules reach in left half of the box.

(c)

Expert Solution
Check Mark

Answer to Problem 26P

The average time before molecules reach in left half of the box is 1.5×10291yr .

Explanation of Solution

Given:

The number of molecules is 1000 .

Calculation:

The average time for n=1000 molecules can be calculated as,

  t=2n2( 100 s 1 )=2 10002( 100 s 1 ) ........(a)

The conversion of 21000 in power of 10 can be calculated as,

  10x=21000xln10=1000×ln2x=301

Substitution of the conversion in equation (a) can be given as,

  t= 10 3012( 100 s 1 )=0.5×10299×1yr3.156× 107s1.5×10291yr

Conclusion:

Therefore, the average time before molecules reach in left half of the box is 1.5×10291yr .

(d)

To determine

The average time before molecules reach in left half of the box.

(d)

Expert Solution
Check Mark

Answer to Problem 26P

The average time before molecules reach in left half of the box is 1.5×1015yr .

Explanation of Solution

Given:

The number of molecules is 1mol .

Calculation:

The average time for n=1mol , molecules can be calculated as,

  t=2n2( 100 s 1 )=2 1mol2( 100 s 1 )=2 6.022× 10 23 2( 100 s 1 ) ........(b)

The conversion of 26.022×1023 in power of 10 can be calculated as,

  10x=26.022× 10 23xln10=6.022×1023×ln2x1023

Substitution of the conversion in equation (2) can be given as,

  t=0.5×1023s×1yr3.156× 107s1.5×1015yr

Conclusion:

Therefore, the average time before molecules reach in left half of the box is 1.5×1015yr .

(e)

To determine

The physicist waiting time before all of the gas molecules in the vacuum chamber occupy only the left half of chamber and comparison with expected lifetime of the universe.

(e)

Expert Solution
Check Mark

Answer to Problem 26P

The average time is 103×109yr and it is 0.3 times of expected lifetime of universe.

Explanation of Solution

Given:

The pressure at which best vacuum created is P=1012torr .

The expected lifetime of the universe is 1010yr .

Formula used:

The expression for the ideal gas is given as,

  PV=NKT

Here, K is the Boltzmann constant, T is the temperature and N is the number of molecules.

The expression for the comparison is given as,

  ttuniverse

Calculation:

The number of moles in vacuum condition at 300K and 1012torr can be calculated as,

  PV=NKTN=

  PV=NKTN=PVKT=( 10 12 torr× 133.32Pa 1torr )( 1.0L)1.381× 10 23J/K( 330K)=1.33× 10 10PaL× 10 5 bar 1Pa1.381× 10 23J

On further solving the above equation,

  N=1.33× 10 15barL× 100J 1barL1.381× 10 23J=1.33× 10 15barL× 100J 1barL1.381× 10 23J=9.63×109

The average time of molecules in vacuum to occupy left half of the chamber can be calculated as,

  t=29.63× 1092(100s1)

The conversion of 29.63×109 in power of 10 can be calculated as,

  10x=29.63× 109xln10=9.63×109×ln2x3×109

Substitution of the conversion in equation (2) can be given as,

  t= 10 3× 10 9 2( 100 s 1 )=0.5×103× 109×1yr3.156× 107s103× 109yr

The comparison can be given as,

  tt universe= 10 3× 10 9 yr 10 10 10 yr0.3tuniverse

Conclusion:

Therefore, the average time is 103×109yr and it is 0.3 times of expected lifetime of universe.

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