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Thermodynamics question

2023 · Shift 2 · Q15
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Thermodynamics question

2023 · Shift 2 · Q15

JEE AdvancedChemistryThermodynamicsNumerical+3 / −1
The entropy versus temperature plot for phases α\alphaα and β\betaβ at 1 bar pressure is given. STS_{\mathrm{T}}ST​ and S0S_0S0​ are entropies of the phases at temperatures T\mathrm{T}T and 0 K0 \mathrm{~K}0 K, respectively. JEE Advanced 2023 Paper 2 Online Chemistry - Thermodynamics Question 4 English Comprehension The transition temperature for α\alphaα to β\betaβ phase change is 600 K600 \mathrm{~K}600 K and Cp,β−Cp,α=1 J mol−1 K−1C_{\mathrm{p}, \beta}-C_{\mathrm{p}, \alpha}=1 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}Cp,β​−Cp,α​=1 J mol−1 K−1. Assume (Cp,β−Cp,α)\left(C_{\mathrm{p}, \beta}-C_{\mathrm{p}, \alpha}\right)(Cp,β​−Cp,α​) is independent of temperature in the range of 200 to 700 K.Cp,α700 \mathrm{~K} . C_{\mathrm{p}, \alpha}700 K.Cp,α​ and Cp,βC_{\mathrm{p}, \beta}Cp,β​ are heat capacities of α\alphaα and β\betaβ phases, respectively. The value of enthalpy change, Hβ−Hα (in Jmol−1 ), at 300 K is \text { The value of enthalpy change, } \mathrm{H}_\beta-\mathrm{H}_\alpha \text { (in } \mathrm{J} \mathrm{mol}^{-1} \text { ), at } 300 \mathrm{~K} \text { is } The value of enthalpy change, Hβ​−Hα​ (in Jmol−1 ), at 300 K is  ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 300

  1. Use the given entropy–temperature graph information

    For any phase at constant pressure,

    rac{dS}{dT} = \frac{C_p}{T}

    Hence, for the two phases,

    ddT(Sβ−Sα)=Cp,β−Cp,αT\frac{d}{dT}(S_\beta-S_\alpha)=\frac{C_{p,\beta}-C_{p,\alpha}}{T}dTd​(Sβ​−Sα​)=TCp,β​−Cp,α​​

    Given:

    Cp,β−Cp,α=1 J mol−1K−1C_{p,\beta}-C_{p,\alpha}=1\ \text{J mol}^{-1}\text{K}^{-1}Cp,β​−Cp,α​=1 J mol−1K−1

    so

    ddT(Sβ−Sα)=1T\frac{d}{dT}(S_\beta-S_\alpha)=\frac{1}{T}dTd​(Sβ​−Sα​)=T1​

  2. Integrate the entropy difference

    Therefore,

    Sβ−Sα=ln⁡T+constantS_\beta-S_\alpha=\ln T + \text{constant}Sβ​−Sα​=lnT+constant

    More usefully, between two temperatures T1T_1T1​ and T2T_2T2​,

    (Sβ−Sα)T2−(Sβ−Sα)T1=∫T1T21TdT=ln⁡(T2T1)(S_\beta-S_\alpha)_{T_2}-(S_\beta-S_\alpha)_{T_1} =\int_{T_1}^{T_2}\frac{1}{T}dT =\ln\left(\frac{T_2}{T_1}\right)(Sβ​−Sα​)T2​​−(Sβ​−Sα​)T1​​=∫T1​T2​​T1​dT=ln(T1​T2​​)

  3. Use transition temperature condition

    At the transition temperature Tt=600 KT_t=600\ \text{K}Tt​=600 K, the two phases are in equilibrium, so

    Gβ=GαG_\beta=G_\alphaGβ​=Gα​

    Hence,

    ΔG=ΔH−TΔS=0⇒ΔH600=600 ΔS600\Delta G = \Delta H - T\Delta S = 0 \quad\Rightarrow\quad \Delta H_{600}=600\,\Delta S_{600}ΔG=ΔH−TΔS=0⇒ΔH600​=600ΔS600​

    where

    ΔH=Hβ−Hα,ΔS=Sβ−Sα\Delta H = H_\beta-H_\alpha, \qquad \Delta S = S_\beta-S_\alphaΔH=Hβ​−Hα​,ΔS=Sβ​−Sα​

  4. Read the entropy graph values

    From the graph, the entropy difference at T=600 KT=600\ \text{K}T=600 K is

    ΔS600=1 J mol−1K−1\Delta S_{600}=1\ \text{J mol}^{-1}\text{K}^{-1}ΔS600​=1 J mol−1K−1

    Therefore,

    ΔH600=600×1=600 J mol−1\Delta H_{600}=600\times 1 = 600\ \text{J mol}^{-1}ΔH600​=600×1=600 J mol−1

  5. Relate enthalpy difference at 300 K and 600 K

    Since

    ddT(Hβ−Hα)=Cp,β−Cp,α=1\frac{d}{dT}(H_\beta-H_\alpha)=C_{p,\beta}-C_{p,\alpha}=1dTd​(Hβ​−Hα​)=Cp,β​−Cp,α​=1

    we get

    ΔH600−ΔH300=∫3006001 dT=300 J mol−1\Delta H_{600}-\Delta H_{300} =\int_{300}^{600}1\,dT =300\ \text{J mol}^{-1}ΔH600​−ΔH300​=∫300600​1dT=300 J mol−1

    Thus,

    ΔH300=ΔH600−300=600−300=300 J mol−1\Delta H_{300}=\Delta H_{600}-300=600-300=300\ \text{J mol}^{-1}ΔH300​=ΔH600​−300=600−300=300 J mol−1

  6. Final answer

    Hβ−Hα at 300 K=300 J mol−1H_\beta-H_\alpha \text{ at } 300\ \text{K} = 300\ \text{J mol}^{-1}Hβ​−Hα​ at 300 K=300 J mol−1
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