Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IIsothermal Process/Isobaric/IsochoricMedium2 minQB

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Passage / Comprehension

The second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transition and chemical equilibrium. 3.00 mol of $CO_{2}$ gas expands isothermally (in thermal contact with the surroundings; temperature $= 15. 0^{\circ} C$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are 10.0 L and 30.0 L , respectively.

Calculate the change in entropy of the universe :

Options:

Answer:

Solution:

For the system,

$Δ S_{sys} = n R ln \frac{V _{2}}{V _{1}} = 3 \times 8.3 \times ln 3 = 27.4 J K^{- 1}$

Work done:

$w = - P_{e x t} Δ V = - 1.00 (30 - 10) = - 20 L ba r = - 2000 J$

For an isothermal process of an ideal gas, $Δ U = 0$ , so

$q = + 2000 J, Δ S_{s u rr} = \frac{- q}{T} = \frac{- 2000}{288} = - 6.94 J K^{- 1}$

Hence,

$Δ S_{u ni v erse} = Δ S_{sys} + Δ S_{s u rr} = 27.4 - 6.94 = 20.5 J K^{- 1}$

$Δ S_{u ni v erse} = 20.5 J K^{- 1}$

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Isothermal Process/Isobaric/Isochoric

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