Question

The Orion nebula is one of the brightest diffuse nebulae in the sky (look for it in the winter, just below the three bright stars in Orion's belt). It is a very complicated mess of gas, dust, young star systems, and brown dwarfs, but let's estimate its temperature if we assume it is a uniform ideal gas. Assume it is a sphere of radius r = 5.9 × 1015 m (around 6 light years) with a total mass 4000 times the mass of the Sun. If the gas is all diatomic hydrogen and the pressure in the nebula is Pn = 6.8 × 10-9 Pa, what is the average temperature (in K) of the nebula? Assume the mass of the sun is Ms = 1.989 × 1030 kg and the mass of a hydrogen atom is mH = 1.67 × 10-27 kg.

Answer 1

Answer:

T=183.21K

Explanation:

We have to take into account that the system is a ideal gas. Hence, we have the expression

$PV=nRT$

where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the ideal gas constant.

Thus, it is necessary to calculate n and V

V is the volume of a sphere

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (5.9*10^{15}m)^3=8.602*10^{47}m^3$

V=8.86*10^{50}L

and for n

$n=\frac{(4000M_s)/(2*mH)}{6.022*10^{23}mol^{-1}}=3.95*10^{36}mol$

Hence, we have (1 Pa = 9.85*10^{-9}atm)

$T=\frac{PV}{nR}=\frac{(6.8*10^{-9}*9.85*10^{-6}atm)(8.86*10^{50}L)}{(0.0820\frac{atm*L}{mol*K})(3.95*10^{36}mol)}\\\\T=183.21K$

hope this helps!!