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?

Question

Deepak Patra · Answer

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 J
From the above observation we conclude that, the absorbed heat Q will be 1.87×10⁴ J.

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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?

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?

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