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Interstellar space, far from any stars, is filled with a very low density of hydrogen atoms ,HnotH2. The number density is about1atom/cm3and the temperature is about3K.

a. Estimate the pressure in interstellar space. Give your answer in Paand in localid="1648635470965" atm.

b. What is the rms speed of the atoms?

c. What is the edge lengthlocalid="1648635477647" Lof anlocalid="1648637118909" L×L×Lcube of gas withlocalid="1648635499087" 1.0Jof thermal energy?

Short Answer

Expert verified

a. Pressure in interstellar space inpascalis4.14×10-17Paand in atmis 4.1×10-22atm.

b. RMS speed of atom is270m/s.

c. Edge length is 2.5×105m.

Step by step solution

01

Calculation of pressure inpascalandatm(part a)

a.

The number density is,

number density=NV.............1

Ideal gas law is given by ,

pV=NkBT

p=NVkBT

Where kBis Boltzmann's constant .

In Slunit its value is,

kB=1.38×10-23J/K

The number of density is ,

N/V=1atom/cm3=106atom/m3.

The values of(N/V),TandkBinto equation1to getpinPascal

P=NVkBT...........2

=106atom/m31.38×10-23J/K(3K)

=4.14×10-17Pa

Pressure of atom in localid="1648636270485" atm,

localid="1648636286605" p=4.14×10-17Pa1atm1.013×105Pa=4.1×10-22atm

02

Calculation for root mean square speed (part b)

b.

The average translational kinetic energy is,

ϵavg=32kBT.....3

ϵavg=12mvrms2....4

As shown, both equations3and4have the same left side,

Use these expressions to get an equation for root mean square velocity, 12mvrms2=32kBT

vrms2=3kBTm

vrms=3kBTm............5

The mass of the hydrogen atom is the same for the proton is,mp=1.67×10-27kg.

Substitute the values forkB,Tand mpinto equation 5to get vrms,

vrms=3kBTmp

localid="1648637031505" =31.38×10-23J/K(3K)1.67×10-27kg

localid="1648637058825" vrms=270m/s

03

Calculation for length (part c)

c.

Translational kinetic energy is the kinetic energy of a monatomic gas.

Eth=32NkBT...........6

Where Nis the number of particles

Ris the universal gas constant.

NV=106atom/m3

N=106atom/m3(V)

The volume is given by V=L×L×L=L3.

So, the number of molecules is,

N=106L3

Eth=32NkBT

Eth=32106L3kBT

L=2Eth3×106kBT3

=2(1.0J)3×1061.38×10-23J/K(3K)3

L=2.5×105m

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