Homework Problem Set 5 Solutions ( )2
Homework Problem Set 5 Solutions ( )2
Homework Problem Set 5 Solutions ( )2
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
5. The Maxwell-Boltzmann distribution for speeds is<br />
12<br />
F(v) = 4π v 2<br />
3/ 2<br />
⎛⎛ m ⎞⎞<br />
⎜⎜ ⎟⎟<br />
⎝⎝ 2π k B T ⎠⎠<br />
⎛⎛<br />
exp − mv2 ⎞⎞<br />
⎜⎜<br />
⎝⎝ 2k B T<br />
⎟⎟ .<br />
⎠⎠<br />
In this equation, m is the mass of a particle, k B is the Boltzmann constant, T is<br />
temperature, and v € is the speed.<br />
velocity vector,<br />
where<br />
v x ,<br />
v y, and<br />
The most probable speed<br />
€ by € the equation €<br />
The speed v is defined to be the magnitude of the<br />
2<br />
v = v x +<br />
2 vy +<br />
2<br />
[ vz ] 1/ 2 ,<br />
v z are the components of the velocity vector.<br />
€<br />
v mp corresponding to the maximum in the distribution is given<br />
€<br />
v mp =<br />
⎛⎛<br />
⎜⎜<br />
⎝⎝<br />
2k B T<br />
m<br />
1/ 2<br />
⎞⎞<br />
⎟⎟ .<br />
⎠⎠<br />
a.) Plot the Maxwell-Boltzmann distribution function F(v) (y-axis) versus v (x-axis) for a<br />
collection of atoms at 300 € K. Assume that one mole of atoms weighs 12.0 g.<br />
b.)<br />
Using the same mass as in part (a), plot € the Maxwell-Boltzmann distribution function<br />
F(v) versus v for a collection of atoms at 1000 K. Discuss the effect of temperature<br />
on the distribution of atomic speeds.<br />
€<br />
Both questions (a) and (b) are answered below.<br />
The Maxwell-Boltzmann distribution function for the atoms at 300 K is given by<br />
⎢⎢ 0.012 kg/mol<br />
( ) = 4π v 2 ⎢⎢<br />
F v<br />
3/ 2<br />
⎡⎡<br />
⎛⎛ 1mol ⎞⎞ ⎤⎤<br />
( ) ⎜⎜<br />
⎝⎝ 6.02217 × 10 23 ⎟⎟<br />
⎠⎠<br />
⎥⎥<br />
⎥⎥<br />
⎢⎢ 2π( 1.38062 × 10 -23 JK −1 )( 300 K)<br />
⎥⎥<br />
⎢⎢<br />
⎥⎥<br />
⎣⎣<br />
⎦⎦<br />
⎧⎧<br />
⎛⎛ 1mol ⎞⎞<br />
( 0.012 kg/mol)<br />
⎜⎜<br />
⎝⎝ 6.02217 × 10<br />
× exp −<br />
23 ⎟⎟ v 2 ⎫⎫<br />
⎪⎪<br />
⎠⎠<br />
⎪⎪<br />
⎪⎪<br />
⎪⎪<br />
⎨⎨<br />
2( 1.38062 × 10 -23 JK −1 ⎬⎬<br />
⎪⎪<br />
)( 300 K)<br />
⎪⎪<br />
⎩⎩ ⎪⎪<br />
⎭⎭ ⎪⎪<br />
€<br />
F( v) = ( 8.41956 × 10 −9 v 2 ) exp{ −2.40549 × 10 −6 v 2 }.<br />
€