Clausius inequality : the cyclic integral of $$\delta Q / T$$ is always less than or equal to zero
$$\oint \frac{\delta Q}{T} \leq 0$$
Kelvin–Planck statement of the second law: no system can produce a net amount of work while operating in a cycle and exchanging heat with a single thermal energy reservoir
$$\oint \frac{\delta Q}{T} \leq 0$$
$$d S=\left(\frac{\delta Q}{T}\right)_{\mathrm{int} \mathrm{rev}} \quad(\mathrm{kJ} / \mathrm{K})$$
$$\Delta S=S_{2}-S_{1}=\int_{1}^{2}\left(\frac{\delta Q}{T}\right)_{\mathrm{int,rev}} \quad(\mathrm{kJ} / \mathrm{K})$$
$$\Delta S=\int_{1}^{2}\left(\frac{\delta Q}{T}\right)_{\mathrm{intrev}}=\int_{1}^{2}\left(\frac{\delta Q}{T_{0}}\right)_{\mathrm{intrev}}=\frac{1}{T_{0}} \int_{1}^{2}(\delta Q)_{\mathrm{int} \text { rev }}$$
$$\Delta S=\frac{Q}{T_{0}} \quad(\mathrm{k} J / \mathrm{K})$$
where $$T_{0}$$ is the constant temperature of the system and $$Q$$ is the heat transferfor the internally reversible process
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