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ChemHardAdv 2023 • Paper 2

Calculate Entropy Change for Phase Transition at 300 K

Comprehension Passage

The entropy versus temperature plot for phases α\alpha and β\beta at 1 bar1\text{ bar} pressure is given. STS_{\text{T}} and S0S_0 are entropies of the phases at temperatures T\text{T} and 0 K0\text{ K}, respectively.

The transition temperature for α\alpha to β\beta phase change is 600 K600\text{ K} and Cp,β−Cp,α=1 J mol−1 K−1C_{\text{p},\beta} - C_{\text{p},\alpha} = 1\text{ J mol}^{-1}\text{ K}^{-1}. Assume (Cp,β−Cp,α)(C_{\text{p},\beta} - C_{\text{p},\alpha}) is independent of temperature in the range of 200200 to 700 K700\text{ K}. Cp,αC_{\text{p},\alpha} and Cp,βC_{\text{p},\beta} are heat capacities of α\alpha and β\beta phases, respectively.

The value of entropy change, Sβ−SαS_\beta - S_\alpha (in J mol−1 K−1\text{J mol}^{-1}\text{ K}^{-1}), at 300 K300\text{ K} is _____.

[Use: ln⁡2=0.69\ln 2 = 0.69 Given: Sβ−Sα=0S_\beta - S_\alpha = 0 at 0 K0\text{ K}]

Question Diagram 1
Official Numerical Answer0.31

Topics & Concepts

ThermodynamicsEntropy and Second Law

Step-by-Step Solution

To find the value of the entropy change ΔS=Sβ−Sα\Delta S = S_\beta - S_\alpha at T=300 KT = 300\text{ K}, we analyze the information given in the graph and problem statement:

  1. Entropy change at 600 K600\text{ K}: From the provided entropy versus temperature plot at T=600 KT = 600\text{ K}: (S600−S0)β=6 J mol−1 K−1(S_{600} - S_0)_\beta = 6\text{ J mol}^{-1}\text{ K}^{-1} (S600−S0)α=5 J mol−1 K−1(S_{600} - S_0)_\alpha = 5\text{ J mol}^{-1}\text{ K}^{-1}

    Given that Sβ−Sα=0S_\beta - S_\alpha = 0 at 0 K0\text{ K}, i.e., S0,β=S0,α=S0S_{0,\beta} = S_{0,\alpha} = S_0, the entropy difference at 600 K600\text{ K} is: ΔS(600)=Sβ,600−Sα,600=(S600−S0)β−(S600−S0)α=6−5=1 J mol−1 K−1\Delta S(600) = S_{\beta, 600} - S_{\alpha, 600} = (S_{600} - S_0)_\beta - (S_{600} - S_0)_\alpha = 6 - 5 = 1\text{ J mol}^{-1}\text{ K}^{-1}

  2. Temperature dependence of entropy difference: The rate of change of entropy difference with respect to temperature is given by: d(ΔS)dT=ΔCpT=Cp,β−Cp,αT\frac{d(\Delta S)}{dT} = \frac{\Delta C_p}{T} = \frac{C_{p,\beta} - C_{p,\alpha}}{T}

    Integrating this equation between T=300 KT = 300\text{ K} and T=600 KT = 600\text{ K}: ΔS(600)−ΔS(300)=∫300600ΔCpTdT\Delta S(600) - \Delta S(300) = \int_{300}^{600} \frac{\Delta C_p}{T} dT

    Since ΔCp=Cp,β−Cp,α=1 J mol−1 K−1\Delta C_p = C_{p,\beta} - C_{p,\alpha} = 1\text{ J mol}^{-1}\text{ K}^{-1} is constant in the range 200 K200\text{ K} to 700 K700\text{ K}: ΔS(600)−ΔS(300)=ΔCpln⁡(600300)=1⋅ln⁡2\Delta S(600) - \Delta S(300) = \Delta C_p \ln\left(\frac{600}{300}\right) = 1 \cdot \ln 2

  3. Calculating ΔS(300)\Delta S(300): Using the given value ln⁡2=0.69\ln 2 = 0.69: 1−ΔS(300)=0.691 - \Delta S(300) = 0.69 ΔS(300)=1−0.69=0.31 J mol−1 K−1\Delta S(300) = 1 - 0.69 = 0.31\text{ J mol}^{-1}\text{ K}^{-1}

The value of the entropy change Sβ−SαS_\beta - S_\alpha at 300 K300\text{ K} is 0.310.31.

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Calculate Entropy Change for Phase Transition at 300 K | Chemistry PYQ Solution - JEE Challenger