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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...
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 Fall '08
 WITTICH,P
 Thermodynamics, P1 V1, Wby gas, Wby

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