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Reversible path for Entropy change on crystallisation

Part (b) The supercooled liquid is not in equilibrium with the solid and therefore $\Delta G$ is not zero, so you cannot calculate the entropy for the phase change from $\Delta H(265)/265$. You can't ...
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1 vote

Can anyone help me with the derivation for the formula of entropy?

For an ideal gas undergoing a reversible process, the 1st law of thermodynamics tells us that $$cU=nC_vdT=dq-PdV=dq-\frac{nRT}{V}dV$$Dividng both sides of this equation by T then gives: $$\frac{dq}{T}...
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4 votes

Can anyone help me with the derivation for the formula of entropy?

For an ideal gas, you can derive it this way: \begin{align} \Delta S &= \int_{T_1,V_1}^{T_2,V_2}{\frac{\delta Q_\mathrm{rev}}{T}} \\ &= \int_{T_1}^{T_2}{\left(\frac{\delta Q_\mathrm{rev}}{T}\...
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