# 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?

Feb 9, 2017

#### Answer:

The pressure of the gas under these conditions would be $1.02 \times {10}^{- 16} a t m$

#### Explanation:

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

$P V = n R T$

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 $c {m}^{3}$ into moles per litre.

Since there are $6.02 \times {10}^{23}$ atoms in a mole, 100 atoms is only

$100 \div 6.02 \times {10}^{23} = 1.66 \times {10}^{- 22}$ moles per $c {m}^{3}$.

Next, since there are $1000 c {m}^{3}$ in a litre, we arrive at the result that

$100 \frac{\text{atoms"/(cm)^3 = 1.66xx10^(-19) "moles}}{L}$

In the ideal gas law, the above value would represent $\frac{n}{V}$, so if we write the law as

$P = \frac{n R T}{V}$

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

$P = \left(1.66 \times {10}^{- 19}\right) \left(0.0821\right) \left(7500\right) = 1.02 \times {10}^{- 16} a t m$

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