A region of the galaxy where new stars are forming-contains a very tenuous gas with 100 `a t oms``/cm^3`. This gas is heated to `7500 K` by ultraviolet radiation from nearby stars. What is the Gas Pressure in ATM?

At this extremely low pressure, the ideal gas law gives an accurate estimate.

`PV=nRT`

where,

`P` is the gas pressure in atmospheres, `n` is the number of moles of gas in the sample, `T` is the Kelvin temperature, `V` is the volume of the sample in litres and `R` is the gas constant, which can have a variety of values, depending on the units chosen for the other variables. I used `R=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 `100` atoms per `cm^3` into moles per litre.

Since there are `6.02xx10^23` atoms in a mole, 100 atoms is only

`100-: 6.02xx10^23 = 1.66 xx10^(-22)` moles per `cm^3`.

Next, since there are `1000 cm^3` in a litre, we arrive at the result that

`100 "atoms"/(cm)^3 = 1.66xx10^(-19) "moles"/L`

In the ideal gas law, the above value would represent `n/V`, so if we write the law as

`P=(nRT)/V`

we need only multiply the above value by `R` and by `T` to find the answer

`P=(1.66xx10^(-19))(0.0821)(7500)=1.02xx10^(-16) atm`

The result is in atmospheres of pressure due to the choice made for the value of `R`.