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Thermodynamics question

2022 · 30 Jun · Shift 1 · Q17
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Thermodynamics question

2022 · 30 Jun · Shift 1 · Q17

JEE MainChemistryThermodynamicsNumerical+4 / −1
1.0 mol of monoatomic ideal gas is expanded from state 1 to state 2 as shown in the figure. The magnitude of the work done for the expansion of gas from state 1 to state 2 at 300 K is ‾\underline{\hspace{2cm}}​ J. (Nearest integer) (Given : R = 8.3 J K −-− 1 mol −-− 1, ln10 = 2.3, log2 = 0.30) JEE Main 2022 (Online) 30th June Morning Shift Chemistry - Thermodynamics Question 89 English
Numerical answer
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Correct answer: 1718

  1. Interpret the process from the figure

    From the given PPP–VVV graph, the gas expands from state 1 to state 2 along an isotherm at T=300 KT=300\,\text{K}T=300K.

    For an isothermal expansion of an ideal gas, W=nRTln⁡(V2V1)W = nRT \ln\left(\frac{V_2}{V_1}\right)W=nRTln(V1​V2​​)

  2. Read volume ratio from the graph

    From the figure, the final volume is double the initial volume: V2V1=2\frac{V_2}{V_1}=2V1​V2​​=2

  3. Substitute the given values

    Given:

    • n=1.0 moln=1.0\,\text{mol}n=1.0mol
    • R=8.3 J mol−1K−1R=8.3\,\text{J mol}^{-1}\text{K}^{-1}R=8.3J mol−1K−1
    • T=300 KT=300\,\text{K}T=300K

    Hence, W=(1)(8.3)(300)ln⁡2W = (1)(8.3)(300)\ln 2W=(1)(8.3)(300)ln2

  4. Evaluate ln⁡2\ln 2ln2 from the given data

    We are given:

    • ln⁡10=2.3\ln 10 = 2.3ln10=2.3
    • log⁡2=0.30\log 2 = 0.30log2=0.30

    Using ln⁡2=(log⁡2)(ln⁡10)=0.30×2.3=0.69\ln 2 = (\log 2)(\ln 10) = 0.30 \times 2.3 = 0.69ln2=(log2)(ln10)=0.30×2.3=0.69

  5. Calculate work

    W=8.3×300×0.69W = 8.3 \times 300 \times 0.69W=8.3×300×0.69 W=2490×0.69W = 2490 \times 0.69W=2490×0.69 W=1718.1 JW = 1718.1\,\text{J}W=1718.1J

  6. Nearest integer

    W≈1718 JW \approx 1718\,\text{J}W≈1718J

Therefore, the magnitude of work done is 1718 J1718\,\text{J}1718J.

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