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An ideal undergoes a reversible isothermal expansion at 132ºC. The entropy of the gas increase by 46.2 J/K. How much heat is absorbed?
To obtain the absorbed heat Q, substitute 46.2 J/K for ΔS and 132̊ C for T in the equation Q = (ΔS) (T),Q = (ΔS) (T) = (46.2 J/K) (132̊ C) = (46.2 J/K) ((132 +273) K) = (46.2 J/K) (405 K) = 1.87×104 JFrom the above observation we conclude that, the absorbed heat Q will be 1.87×104 J.
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