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Thermodynamics question

2020 · 2 Sep · Shift 2 · Q5
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Thermodynamics question

2020 · 2 Sep · Shift 2 · Q5

JEE MainChemistryThermodynamicsNumerical+4 / −1
The heat of combustion of ethanol into carbon dioxide and water is – 327 kcal at constant pressure. The heat evolved (in cal) at constant volume and 27oC (if all gases behave ideally) is (R = 2 cal mol–1 K–1) ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 326400

  1. Write the combustion reaction of ethanol

For ethanol, C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)\mathrm{C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)}C2​H5​OH(l)+3O2​(g)→2CO2​(g)+3H2​O(l)

Since the problem gives heat of combustion at constant pressure, this is: qp=ΔH=−327 kcalq_p = \Delta H = -327\ \text{kcal}qp​=ΔH=−327 kcal

We need heat evolved at constant volume, i.e. qv=ΔUq_v = \Delta Uqv​=ΔU

  1. Use the relation between enthalpy and internal energy

For ideal gases, ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RTΔH=ΔU+Δng​RT

So, ΔU=ΔH−ΔngRT\Delta U = \Delta H - \Delta n_g RTΔU=ΔH−Δng​RT

where Δng\Delta n_gΔng​ is the change in moles of gaseous species.

  1. Calculate Δng\Delta n_gΔng​

From the balanced reaction:

  • Gaseous reactants: 333 mol O2O_2O2​
  • Gaseous products: 222 mol CO2CO_2CO2​

Thus, Δng=2−3=−1\Delta n_g = 2 - 3 = -1Δng​=2−3=−1

  1. Substitute into the formula

Given:

  • T=27∘C=300 KT = 27^\circ C = 300\ \text{K}T=27∘C=300 K
  • R=2 cal mol−1K−1R = 2\ \text{cal mol}^{-1}\text{K}^{-1}R=2 cal mol−1K−1
  • ΔH=−327 kcal=−327000 cal\Delta H = -327\ \text{kcal} = -327000\ \text{cal}ΔH=−327 kcal=−327000 cal

Now, ΔU=ΔH−ΔngRT\Delta U = \Delta H - \Delta n_g RTΔU=ΔH−Δng​RT ΔU=−327000−(−1)(2)(300)\Delta U = -327000 - (-1)(2)(300)ΔU=−327000−(−1)(2)(300) ΔU=−327000+600\Delta U = -327000 + 600ΔU=−327000+600 ΔU=−326400 cal\Delta U = -326400\ \text{cal}ΔU=−326400 cal

  1. Heat evolved

Since the question asks for heat evolved, we report the magnitude: 326400 cal326400\ \text{cal}326400 cal

  1. Comparison with stored answer

Derived answer = 326400326400326400

Stored correct answer = 326400326400326400

They match.

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