Question

A nebula—a region of the galaxy where new stars are forming—contains a very tenuous gas with $100atoms/c{m}^3$ . This gas is heated to $7500K$ by ultraviolet radiation from nearby stars. What is the gas pressure in atm?

Short answer

The gas pressure is$1.02\times {10}^{-16}atm$

Step-by-step solution

: Given information and Formula used 

Given : A nebula contains gas with : $100\mathrm{atoms}/c{m}^3$.

Temperature : $7500K$

Heated by : Ultraviolet radiation

Theory used :
The number density is given by $\frac{N}{V}$

The ideal gas law is given by $PV=nRT$
Rewriting ideal gas law for $V\mathrm{in}\mathrm{the}\mathrm{form}:$localid="1648796492843" $V=\frac{nRT}{p}$
If we know the number of molecules $N$, we obtain the number of moles $n$by using Avogadro's number :

$$\begin{gathered} n=\frac{N}{N_A} \\ \Rightarrow n=n{N}_A \end{gathered}$$

Finding the gas pressure .

Now substitute the expressions for $N\mathrm{and}V$into number density formula to get the form of the number density $Numberdensity=\frac{N_A\rho }{RT}$
We obtain the number density by $100/cm³=0.01x{10}^{-6}/m³$.

We solve the above equation for$p$ to get the pressure in Pascal.
Now we substitute the values for N₁, P, R and T into the equation to get the number density
$p=\frac{RT}{N_A}$
We substitute the values for$R,T,N_A$and the number density to obtain
$$\begin{gathered} p=\frac{(8.314J/molK)(7500K)}{(6.023x1025molecules/mol)}(100x10m³) \\ =1.035x{10}^{-11}Pa \end{gathered}$$
Convert the pressure from pascal to atm into the form

$$\begin{gathered} p=(1.035\times {10}^{-11}Pa)\frac{1atm}{1.013x{10}^{5}Pa} \\ =1.02X{10}^{-16}atm \end{gathered}$$