This preview shows pages 1–2. 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: Problem Set #1 Solutions September 17, 2008 2.3 In a reversible, adiabatic process, PV γ = const. , so throughout the process we will have P = P 1 V γ 1 V γ With P 2 = P 1 V γ 1 V γ 2 Since the process is reversible, W by system = ˆ PdV = P 1 V γ 1 ˆ V 2 V 1 V γ dV = 1 1 γ P 1 V γ 1 [ V 1 γ 2 V 1 γ 1 ] = 1 γ 1 [ P 1 V 1 P 2 V 2 ] 2.5 Consider the process in small steps. At each small step, a small quantity of gas, n small , escapes from inside the container. And the amount of work done on it by the rest of the gas in the container is Δ W by gas = PV small , where V small = n small v is the volume the quantity of gas takes up in the conditions of the container. Since the amount of moles in the container decreases as the gas esacpes, Δ n = n small , and so Δ W by gas = Pv Δ n Since Pv = RT , and since the process occurs isothermally, we will get simply that W by gas = RT X steps Δ n = RT ( n initial n final ) Since we have RT = P v and n final = V/v , and since we are given that n initial = n , then W by gas = P v ( n V/v ) = P ( nv V ) Many students argued that we should have the result simply as...
View
Full
Document
This note was uploaded on 10/01/2008 for the course PHYS 2218 taught by Professor Wittich,p during the Fall '08 term at Cornell.
 Fall '08
 WITTICH,P

Click to edit the document details