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Thermodynamics question

2021 · 16 Mar · Shift 2 · Q17
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Thermodynamics question

2021 · 16 Mar · Shift 2 · Q17

JEE MainChemistryThermodynamicsNumerical+4 / −1
At 25 ∘^\circ∘ C, 50 g of iron reacts with HClHClHCl to form FeCl2FeCl_2FeCl2​. The evolved hydrogen gas expands against a constant pressure of 1 bar. The work done by the gas during this expansion is ‾\underline{\hspace{2cm}}​ J. (Round off to the Nearest Integer). [Given : R = 8.314 J mol −-− 1 K −-− 1. Assume, hydrogen is an ideal gas] [Atomic mass of FeFeFe is 55.85 u]
Numerical answer
View written solutionFree

Correct answer: 2218

  1. Write the reaction

    Iron reacts with hydrochloric acid as: Fe+2HCl→FeCl2+H2Fe + 2HCl \rightarrow FeCl_2 + H_2Fe+2HCl→FeCl2​+H2​

    From the equation, 1 mol Fe→1 mol H21\text{ mol Fe} \rightarrow 1\text{ mol } H_21 mol Fe→1 mol H2​

  2. Calculate moles of iron

    Given mass of iron = 50 g50\,\text{g}50g

    Molar mass of iron = 55.85 g mol−155.85\,\text{g mol}^{-1}55.85g mol−1

    nFe=5055.85=0.8953 moln_{Fe} = \frac{50}{55.85} = 0.8953\,\text{mol}nFe​=55.8550​=0.8953mol

    Therefore, nH2=0.8953 moln_{H_2} = 0.8953\,\text{mol}nH2​​=0.8953mol

  3. Use work done in expansion against constant pressure

    For expansion work, w=−PextΔVw = -P_{ext}\Delta Vw=−Pext​ΔV

    Since hydrogen is ideal gas and is evolved at 25∘C=298 K25^\circ C = 298\,\text{K}25∘C=298K, ΔV=nRTPext\Delta V = \frac{nRT}{P_{ext}}ΔV=Pext​nRT​

    Hence, w=−Pext(nRTPext)=−nRTw = -P_{ext}\left(\frac{nRT}{P_{ext}}\right) = -nRTw=−Pext​(Pext​nRT​)=−nRT

    Magnitude of work done by gas: ∣w∣=nRT|w| = nRT∣w∣=nRT

  4. Substitute values

    ∣w∣=(0.8953)(8.314)(298)|w| = (0.8953)(8.314)(298)∣w∣=(0.8953)(8.314)(298)

    First, 8.314×298=2477.5728.314 \times 298 = 2477.5728.314×298=2477.572

    Then, ∣w∣=0.8953×2477.572≈2218.43 J|w| = 0.8953 \times 2477.572 \approx 2218.43\,\text{J}∣w∣=0.8953×2477.572≈2218.43J

  5. Round to nearest integer

    2218 J\boxed{2218\,\text{J}}2218J​

  6. Compare with stored correct answer

    Stored correct answer = 2218

    Our derived answer matches it exactly.

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