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Standard XI

Chemistry

Entropy

One mole of a...

Question

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

5.763

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1.013

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-1.013

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-5.763

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Solution

The correct option is C -1.013
We know,
From $1^{s t}$ law of thermodynamics
$q_{s y s} = Δ U - w$
As the process is isothermal,so
$q_{s y s} = 0 - [- P_{e x t} . Δ V]$
$= 3.0 a t m \times (2.0 L - 1.0 L) = 3.0 L a t m = 3.0 \times 101.3 J$ .

Putting the relevent values in formula of entropy,
$∴ Δ S_{surr} = \frac{q_{s u r r}}{T} = - \frac{q_{s y s}}{T}$

$= \frac{- 3.0 \times 101.3 J}{300 K}$

$= - 1.013 J / K$

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Q. One mole of an ideal gas at

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1.0 L t o 2.0 L

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In this process, the change in entropy of surroundings

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(G i v e n : 1 L a t m = 101.3 J)

Q. One mole of an ideal gas at

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In this process, the change in entropy of surroundings

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[R = 0.082 L a t m m o l^{- 1} K^{- 1} = 8.3 J m o l^{- 1} K^{- 1}]

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Entropy

Standard XI Chemistry

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One mole of an ideal gas at 300 K in thermal contact with surrounding 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 (ΔSsurr) in JK−1 is: (Given:1 Latm=101.3J)

The correct option is C -1.013We know, From 1st law of thermodynamics qsys=ΔU−w As the process is isothermal,so qsys=0−[−Pext.ΔV] =3.0 atm×(2.0L−1.0 L)=3.0 Latm=3.0×101.3J. Putting the relevent values in formula of entropy, ∴ΔSsurr=qsurrT=−qsysT =−3.0×101.3J300K =−1.013 J/K

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