This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Theoretical Question 2 / Solutions Page 1/6 Theoretical Question 2: Rising Balloon 1. Answers (a) P P P ng M F A B ∆ + = (b) = γ P g z ρ = 5.5 (c) = ∆ P − 7 1 1 4 λ λ κ r RT 2 3 4 5 0.1 0.2 0.3 0.4 0.5 0.6 (d) a =0.110 (e) f z =11 km, f λ =2.1. Theoretical Question 2 / Solutions Page 2/6 2. Solutions [Part A] (a) [1.5 points] Using the ideal gas equation of state, the volume of the helium gas of n moles at pressure P P ∆ + and temperature T is ) /( P P nRT V ∆ + = (a1) while the volume of ' n moles of air gas at pressure P and temperature T is P RT n V / ' = . (a2) Thus the balloon displaces P P P n n ∆ + = ' moles of air whose weight is g n M A ' . This displaced air weight is the buoyant force, i.e., P P P ng M F A B ∆ + = . ( a 3 ) (Partial credits for subtracting the gas weight.) (b) [2 points] The pressure difference arising from a height difference of z is gz ρ − when the air density ρ is a constant. When it varies as a function of the height, we have g T P P T g dz dP ρ ρ − = − = (b1) where the ideal gas law P T / ρ = constant is used. Inserting Eq. (2.1) and / 1 / z z T T − = on both sides of Eq. (b1), and comparing the two, one gets 52 . 5 10 01 . 1 8 . 9 10 9 . 4 16 . 1 5 4 = × × × × = =...
View Full Document
This note was uploaded on 11/08/2011 for the course PHYS 0000 taught by Professor Na during the Spring '11 term at Rensselaer Polytechnic Institute.
- Spring '11