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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

For a reversible process at equilibrium, the entropy change $ \Delta S$ is given by

$\begin {array} {1 1} (A)\;\large\frac{\Delta H}{\Delta T} & \quad (B)\;\large\frac{T}{Q_{rev}} \\ (C)\;\large\frac{Q_{rev}}{T} & \quad (D)\;T \Delta H \end {array}$

 

1 Answer

Since, at equilibrium, $ \Delta G = \Delta H – T \Delta S = 0, \: i.e, \: \Delta H = T \Delta S$
For a reversible process, $ \Delta H = Q_{rev}$
So, $Q_{rev} = T \Delta S$
Ans : (C)
answered Mar 15, 2014 by thanvigandhi_1
 

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