Thermodynamics | JEE Main 2025 Solution

JEE MAIN 2025

23-01-2025 SHIFT-1

Question

Ice at $-{{5}^{{}^\circ }}\text{C}$ is heated to become vapor with temperature of ${{110}^{{}^\circ }}\text{C}$ at atmospheric pressure. The entropy change associated with this process can be obtained from

Select the correct option:

$\int_{{268K}}^{{383K}}{{{{C}_{p}}}}dT+\frac{{\Delta {{H}_{{\text{melting }}}}}}{{273}}+\frac{{\Delta {{H}_{{\text{boiling }}}}}}{{373}}$

$\int_{{268~\text{K}}}^{{273~\text{K}}}{{\frac{{{{\text{C}}_{\text{p}}},\text{m}}}{\text{T}}}}\text{dT}+\frac{{\Delta {{\text{H}}_{\text{m}}},\text{ fusion }}}{{{{\text{T}}_{\text{f}}}}}+\frac{{\Delta {{\text{H}}_{{\text{m},\text{ vaporisation }}}}}}{{{{\text{T}}_{\text{b}}}}}+\int_{{273~\text{K}}}^{{373~\text{K}}}{{\frac{{{{\text{C}}_{{\text{p},\text{m}}}}\text{dT}}}{\text{T}}}}+\int_{{373~\text{K}}}^{{383~\text{K}}}{{\frac{{{{\text{C}}_{{\text{p},\text{m}}}}\text{dT}}}{\text{T}}}}$

$\int_{{268~\text{K}}}^{{383~\text{K}}}{{{{\text{C}}_{\text{p}}}}}\text{dT}+\frac{{{{\text{q}}_{{\text{rev }}}}}}{\text{T}}$

$\int_{{268~\text{K}}}^{{273~\text{K}}}{{{{\text{C}}_{\text{p}}}}},\text{mdT}+\frac{{\Delta {{\text{H}}_{\text{m}}},\text{ fusion }}}{{{{\text{T}}_{\text{f}}}}}+\frac{{\Delta {{\text{H}}_{{\text{m},\text{ vaporisation }}}}}}{{{{\text{T}}_{\text{b}}}}}+\int_{{273~\text{K}}}^{{373~\text{K}}}{{{{\text{C}}_{\text{p}}}}},\text{mdT}+\int_{{373~\text{K}}}^{{383~\text{K}}}{{{{\text{C}}_{\text{p}}}}},\text{mdT}$

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Solution

Question Tags

JEE Main

Chemistry

Hard

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