JEE Advanced 2023 — Paper 2 — Question 43

$\mathrm{S}=\mathrm{S}_{0}+\int \mathrm{C}_{\mathrm{p}} \frac{\mathrm{dT}}{\mathrm{T}}$ $\mathrm{S}_{\alpha}=\mathrm{S}_{0}+\int\left(\mathrm{C}_{\mathrm{p}}\right)_{\alpha} \frac{\mathrm{dT}}{\mathrm{T}}$ $S_{\beta}=S_{0}+\int\left(C_{p}\right)_{\beta} \frac{d T}{T}$ $S_{\beta}-S_{\alpha}=\left[\left(C_{p}\right)_{\beta}-\left(C_{p}\right)_{\alpha}\right] \int \frac{d T}{T}$ Given $\mathrm{C}_{\mathrm{P}_{\beta}}-\mathrm{C}_{\mathrm{P}_{\alpha}}=1$ $\mathrm{S}_{\beta}-\mathrm{S}_{\alpha}=\ln \mathrm{T}+\mathrm{C}$ at any temperature T . $\left(\mathrm{S}_{\beta}-\mathrm{S}_{\alpha}\right)_{\mathrm{T}_{2}}-\left(\mathrm{S}_{\beta}-\mathrm{S}_{\alpha}\right)_{\mathrm{T}_{1}}=\ln \mathrm{T}_{2}-\ln \mathrm{T}_{1}$ $\mathrm{T}_{2}=600 \mathrm{~K}, \mathrm{~T}_{1}=300 \mathrm{~K}$, from the graph $\mathrm{S}_{\beta}-\mathrm{S}_{\alpha}$ at $600^{\circ} \mathrm{C}=1$ (1) $-\left(\mathrm{S}_{\beta}-\mathrm{S}_{\alpha}\right)_{300}=\ln 600-\ln 300$ $1-\left(\mathrm{S}_{\beta}-\mathrm{S}_{\alpha}\right)_{300}=\ln 2=0.69$ $\Rightarrow\left(\mathrm{S}_{\beta}-\mathrm{S}_{\alpha}\right)_{300}=1-0.69$ $=0.31$