Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIAdiabatic Process/PolytropicMedium2 minPYQ_2020

ChemistryMediumsingle choice

One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings $(Δ S_{surr})$ in $J K^{- 1}$ is (1 L atm = 101.3 J)

Options:

Answer:

Solution:

$Δ E = q + w$

$0 = q - P_{e x t} Δ V$

$q = P_{e x t} Δ V = 3$ atm (2 - 1) L = 3 atm L

$= (3 \times 101.3)$ Joule

$Δ S_{s u r r} = - \frac{q}{T} = \frac{3 \times 101.3}{300} = - 1.013$ Joule/K

Stream:NTA_ABHYASSubject:ChemistryTopic:Thermodynamics - IISubtopic:Adiabatic Process/Polytropic

⏱ 2mℹ️ Source: PYQ_2020

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