Name: Gibbs Free Energy, EMF and Equilibrium Constant
Uploaded: 2019-07-19
Duration: 6 min 15 s
Description: There are various thermodynamic parameters that are to be considered while studying the working of an electrochemical cell. Let's learn about them here!

Question

Can

ΔU

and

ΔS

discriminate between irreversible and reversible process? Explain.

Accepted Answer

$ΔU$ does not discriminate between irreversible and reversible process. For isothermal process involving ideal gas $T$ is constant. Hence $ΔU = 0$ for both reversible and irreversible process.

$ΔS$ discriminates the irreversible and reversible process.

In an isothermal reversible process if the amount of heat $Q$ is absorbed from the surroundings at a temperature $T$ , the increase in the entropy of the system will be

$Δ_{sys} = + \frac{Q}{T}$

On the other hand surroundings lose the same amount of heat at the same temperature. The decrease in entropy of the surroundings will

${ΔS}_{surr} - \frac{Q}{T}$

${ΔS}_{Total} = Δ_{sys} + {ΔS}_{surr} = \frac{Q}{T} - \frac{Q}{T} = 0$

In the irreversible process the system is at higher temperature $T_{1}$ and its surroundings at lower temperature $T_{2}$ . 'Q' amount of heat goes irreversibly from system to surroundings $∴ {ΔS}_{sys} = \frac{- Q}{T_{1}}$

${ΔS}_{sur} = \frac{+ Q}{T_{2}}$

${ΔS}_{process} = {ΔS}_{sys} + {ΔS}_{sur}$

$= \frac{- Q_{}}{T_{1}} + \frac{Q_{}}{T_{2}} = Q [\frac{T_{1} - T_{2}}{T_{1} T_{2}}]$

But $T_{1} > T_{2}$

$∴ T_{1} - T_{2} = + ve$ or ${ΔS}_{process} > 0$

Hence, entropy increases in an irreversible process.

Can $ΔU$ and $ΔS$ discriminate between irreversible and reversible process? Explain.

Important Questions on Thermodynamics

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