Question

(20% Problem 5: 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 p, and fundamental constants T (P)(pkg vCorrect! 25% Part (b) Near the surface of Venus, its atmosphere has a pressure f 88 times the pressure of Earth's atmosphere, and a particle density of around py 0.92 x 10- m3. What is the temperature of Venus' atmosphere (in C) near the surface? - 429 Correct! 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 = 4.1 x 101 m (around 6 light years) with a total mass 4000 times the mass of the Sun. Ifthe gas is all diatomic hydrogen and the pressure in the nebula is P 6.6 x 10-9 Pa, what is the average temperature (in K) of the nebula? Assume the mass of the sun is M 1.989 x 1050 kg and the mass of a hydrogen atom is mH= 1.67 x 102 kg. Grade Summary Deductions 0% Potential 100% ( 78 9 ET 45 6 23 sin cotan atan cosh tan НОME Submissions asin acotan acos0 Attempts remaining: 4 C0% per attempt) detailed view sinh cotanh 1 tanh0 0 END 0% 1 Degrees Radians 0BACKSPACE DEL CLEAR Hint I give up! Submit Feedback Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback 25% Part (d) Deep in space there is an average particle density po- 1.09 cm and an extremely low pressure of Po = 5.1 x 10-17N/m2. What is the average temperature of (mostly) empty space? Give your answer in K.

Answer 1

at Mass of nebula, Mo = 4000 mg Mo = 7.956% 1033 Kg Mo = 7.956 x 10369 Molecular mass of hydrogen, - M =2.0169 / molt No, of moles of hydrogen in Nebula, n Mo M m = 7.956 x 1036 2.016 n = 3.946 mot X10% moles By equation of Pn Vn = state, N. R To

m = 47783. (Volume) = A (41 X10's) = 2.925 x 10t6m3 = (6.6x109) (2.925x1046) (3.9464 10311 8.314) Tn = 5.884 °K

d we have from Cal Kp = 1.381810-23 m2 Kg 52 kr = 5.1 X 10-17 (1.89x10-6 (1.381x102) To = 3.389 x 1012 ºr