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Thermodynamics

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

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

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)