(9%) Problem 7: There are lots of examples of ideal gases in the universe, and t

Question

(9%) Problem 7: There are lots of examples of ideal gases in the universe, and they exist in many different conditions. In this problem we will examine what the temperature of these various phenomena are ? 25% Part (a) Give an expression for the temperature of an ideal gas in terms of pressure P, particle density per unit volume ? and fundamental constants Grade Summary 096 100% Potential 78 9 Submissions Attempts remaining:J5 (500 per attempt) detailed view kB BACKSPACE Submit Hint I give up! Hints: 1 % deduction per hint. Hints r Feedback: 0% deduction per feedback. 25% Part (b) Near the surface of Venus, its atmosphere has a pressure = 94 times the pressure of Earth's atmosphere, and a particle density of around Pv-0.98 × 10. m.. What is the temperature of Venus atmosphere (in C) near the surface? - 25% Part (c) 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 4.9 x 10 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 P 6.1x10 Pa, what is the average temperature (in K) of the nebula? Assume the mass of the sun is M.-1,.989 x 1030 kg and the mass of a hvdrogen atom is mH- 1.67 x 1027 kg 25% Part (d) Deep in space there is an average particle density ? = 1.01 cm3 and an extremely low pressure of Po-4.98 101 Nm What is the average temperature of (mostly) empty space? Give your answer in K.

Explanation / Answer

7.a . given pressure P

density rho

temperature = T

volume = V

gas sontant = R

form ideal gas equation

PV = nRT

N = number of moles

M = molar mass

now

rho = n/V

hence

P = rho(R)T

T = P/R*rho

b. P = 94 atm

rho = 0.98*10^27 /m^3

hence

T = 1.165*10^-21 K = -273.1566 C

c. r = 4.9*10^15 m

m = 4000*2*10^30 kg

gas is H2

P = 6.1*10^-9 Pa

mh = 1.67*10^-27 kg

T = PM/R*rho

M = mh*6.62*1023

rho = 3m/4*pi*r^3

hence

T = 49.99080545 K

d. rhoo = 1.01 / cm^3 = 1.01*10^6 / m^3

Po = 4.9*10^-17 N/m^2

T = 5.838*10^-24 K

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