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A cylinder containing an ideal gas is heated at constant pressure from 300K to 350K by immersion in a bath of hot water. Is this process reversible or irreversible? A] reversible B] irreversible

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What is the work done by the gas in the reversible isothermal expansion shown? A] p 0 V 0 ln(2) B] p 0 V 0 C] 2 p 0 V 0 D] 0 E] none of these What is the heat added, Q?

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No change in internal energy, so W=Q= p 0 V 0 ln(2). What is the entropy change of the gas? A] p 0 V 0 ln(2) B] nRln(2) C] nRln(1/2) D] 0 E] cannot determine What is the entropy change in the hot reservoir which is adding heat to the gas? S = Q/T for an isothermal process. Use p 0 V 0 =nRT along with Q= p 0 V 0 ln(2) to find S = nRln(2).

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A] p 0 V 0 ln(2) B] nRln(2) C] nRln(1/2) D] 0 E] cannot determine What is the entropy change in the hot reservoir which is adding heat to the gas? In a reversible process, S = 0. So the entropy change in the hot reservoir (which is at the same temperature T as the gas) is -nRln(2). Answer C.

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We showed, for a Carnot cycle, that Q H /T H = |Q c |/T C = -Q c /T c What is the change in entropy of the gas around the entire Carnot cycle? A] p 0 V 0 ln(2) B] nRln(2) C] nRln(1/2) D] 0 E] cannot determine

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Any reversible process consists of “adjoining” Carnot cycles. S for adjoining segments cancels. So: Entropy, like Internal Energy, is a “state” variable, and depends only on the state of a system (p, V for a gas). -> You can calculate entropy changes for irreversible processes by taking a reversible path to the same endpoint. Example: free expansion to double the volume. T f = T i.

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Entropy changes in non-isothermal processes Example 1: heating water Example 2a/b: heating a gas at constant V/p

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