JEE Main 2025 Jan 23 Shift 1, Chemistry Q9: Basic Concepts of Thermodynamic

JEE Main2025Jan 23, Shift 1Chemistry

Ice at $- 5^{\circ}$ C is heated to become vapor with temperature of $11 0^{\circ}$ C at atmospheric pressure. The entropy change associated with this process can be obtained from :

A
$\int_{268 K}^{383 K} C_{p} d T + \frac{Δ H _{melting}}{273} + \frac{Δ H _{boiling}}{373}$
B
$\int_{268 K}^{273 K} \frac{C _{p, m}}{T} d T + \frac{Δ H _{m, fusion}}{T _{f}} + \frac{Δ H _{m, vaporisation}}{T _{b}} + \int_{273 K}^{373 K} \frac{C _{p, m}}{T} d T + \int_{373 K}^{383 K} \frac{C _{p, m}}{T} d T$
C
$\int_{268 K}^{383 K} C_{p} d T + \frac{q _{r e v}}{T}$
D
$\int_{268 K}^{273 K} C_{p, m} d T + \frac{Δ H _{m, fusion}}{T _{f}} + \frac{Δ H _{m, vaporisation}}{T _{b}} + \int_{273 K}^{373 K} C_{p, m} d T + \int_{373 K}^{383 K} C_{p, m} d T$

Solution

The total process has five sub-steps; the total entropy change is the sum.

(i) Heating ice from 268 K to 273 K: $Δ S_{1} = \int_{268}^{273} \frac{C _{p, m}}{T} d T$ .

(ii) Fusion at 273 K (constant T): $Δ S_{2} = \frac{Δ H _{fusion}}{273}$ .

(iii) Heating water from 273 K to 373 K: $Δ S_{3} = \int_{273}^{373} \frac{C _{p, m}}{T} d T$ .

(iv) Vaporisation at 373 K: $Δ S_{4} = \frac{Δ H _{vap}}{373}$ .

(v) Heating vapour from 373 K to 383 K: $Δ S_{5} = \int_{373}^{383} \frac{C _{p, m}}{T} d T$ .

Crucially, the heating terms must use $\frac{C _{p}}{T} d T$ , not $C_{p} d T$ . Only option (2) gets this right.

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JEE Main 2025 Jan 23 Shift 1, Chemistry Q9: Basic Concepts of Thermodynamic

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