P21_018 - 18(a In an adiabatic process Q = 0 This can be...

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Unformatted text preview: 18. (a) In an adiabatic process Q = 0. This can be done by placing the gas in a thermally insulated container whose volume can be adjusted (say, by means of a movable piston). If the volume is slowly increased from V i to V x , then the process is reversible. To realize the reversible, constant- volume process from x to f , we would place the gas in a rigid container, which has a fixed volume V f and is in thermal contact with a heat reservoir. If we gradually increase the temperature of the reservoir from T x to T f , the gas will undergo the desired reversible process from x to f . (b) For the two states i and x we have p i V i /T i = p x V x /T x and p i V γ i = p x V γ x . We eliminate p i and p x from these equations to obtain T x T i = µ V i V x ¶ γ − 1 . For monatomic ideal gases γ = 5 / 3 (see § 20-8 and § 20-11), so γ − 1 = 2 / 3. Also V x = V f . Substituting these into the equation above, we obtain T x = T i ( V i /V f ) 2 / 3 ....
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