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Standard XII

Chemistry

Entropy Change in Isothermal Process

One mole of a...

Question

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 surrounding $(Δ S_{m a x})$ in $J K^{- 1}$ is ( $1 L a t m = 101.3 J$ )

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
Since it is an isothermal process, $Δ U = 0$

$d q = - d W = P_{e x t} (V_{2} - V_{1}) = 3 L - a t m = 3 \times 101.3 J$

$Δ S_{s u r r o u n d i n g} - \frac{3 \times 101.3}{300} J K^{- 1} = - 1.013 J K^{- 1}$

$∴ Δ S_{s u r r} = - 1.013 J K^{- 1}$

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Similar questions

Q. 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 surrounding

(Δ S_{m a x})

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1 L a t m = 101.3 J

)

Q. One mole of an ideal gas at

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Pressure of 10 moles of an ideal gas is changed from 2 atm to 1 atm against constant external pressure without change in temperature. If surrounding temperature (300 K) and pressure (1 atm) always remains constant then calculate total entropy change ( $Δ S_{s y s t e m} + Δ S_{s u r r o u n d i n g})$ for given process. [Given: ℓn2 = 0.70 amd R=8.0 J/mol/K]

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Q. One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is:

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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 surrounding (ΔSmax) in JK−1 is (1L atm=101.3 J)

The correct option is C -1.013Since it is an isothermal process, ΔU=0 dq=−dW=Pext(V2−V1)=3L−atm=3×101.3 J ΔSsurrounding−3×101.3300 J K−1=−1.013 J K−1 ∴ΔSsurr=−1.013 J K−1

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 surrounding $(Δ S_{m a x})$ in $J K^{- 1}$ is ( $1 L a t m = 101.3 J$ )

The correct option is C -1.013
Since it is an isothermal process, $Δ U = 0$

$d q = - d W = P_{e x t} (V_{2} - V_{1}) = 3 L - a t m = 3 \times 101.3 J$

$Δ S_{s u r r o u n d i n g} - \frac{3 \times 101.3}{300} J K^{- 1} = - 1.013 J K^{- 1}$

$∴ Δ S_{s u r r} = - 1.013 J K^{- 1}$