At this extremely low pressure, the ideal gas law gives an accurate estimate.
PV=nRTPV=nRT
where,
PP is the gas pressure in atmospheres, nn is the number of moles of gas in the sample, TT is the Kelvin temperature, VV is the volume of the sample in litres and RR is the gas constant, which can have a variety of values, depending on the units chosen for the other variables. I used R=0.0821R=0.0821 to be consistent with the units given in the problem, and to produce an answer that would be a fraction of normal atmospheric pressure.
First, we must convert 100100 atoms per cm^3cm3 into moles per litre.
Since there are 6.02xx10^236.02×1023 atoms in a mole, 100 atoms is only
100-: 6.02xx10^23 = 1.66 xx10^(-22)100÷6.02×1023=1.66×10−22 moles per cm^3cm3.
Next, since there are 1000 cm^31000cm3 in a litre, we arrive at the result that
100 "atoms"/(cm)^3 = 1.66xx10^(-19) "moles"/L100atoms(cm)3=1.66×10−19molesL
In the ideal gas law, the above value would represent n/VnV, so if we write the law as
P=(nRT)/VP=nRTV
we need only multiply the above value by RR and by TT to find the answer
P=(1.66xx10^(-19))(0.0821)(7500)=1.02xx10^(-16) atmP=(1.66×10−19)(0.0821)(7500)=1.02×10−16atm
The result is in atmospheres of pressure due to the choice made for the value of RR.
