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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.

User Efx
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Answer:

T=183.21K

Step-by-step 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=(4)/(3)\pi r^3=(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=((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=(PV)/(nR)=((6.8*10^(-9)*9.85*10^(-6)atm)(8.86*10^(50)L))/((0.0820(atm*L)/(mol*K))(3.95*10^(36)mol))\\\\T=183.21K

hope this helps!!

User Sheryl
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