One mole of an ideal monatomic gas expands isothermally against a constant external pressure of $1$ atm from an initial volume of $1$ L to a state where its final pressure becomes equal to external pressure. If the initial temperature of the gas is $300$ K then the total entropy change of system in the given process is:
$[R = 0.082 L a t m m o l^{- 1} K^{- 1} = 8.3 J m o l^{- 1} K^{- 1}]$

0

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R l n (24.6)

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R l n (2490)

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\frac{3}{2}

l n (24.6)

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Solution

The correct option is B $R l n (24.6)$
General formula of entrophy changes,

$△ S = C_{v} l n (\frac{T_{2}}{T_{1}}) + R l n (\frac{V_{2}}{V_{1}})$

given {isothermal process $T_{1} = T_{2}$ }

$P_{v} = n R T$

$V_{f} = \frac{n R T}{P} = \frac{1 \times 0.082 \times 300}{1}$

$V_{f} = 24.6 L$

$△ S = R l n (24.6)$

Hence, the correct option is $B$

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Similar questions

Q. One mole of an ideal monoatomic gas expends isothermally against constant external pressure of

1

atm from initial volume of

1 L

to a state where its final pressure become equal to external pressure. If initial temperature of gas is

300 K

then total entropy change of system in the above process is:

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

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 :
(

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

)

Q. One mole of an ideal gas expands isothermally from initial state ( 1 atm, 22.4 L, 273 K ) to the final state (

0.25 a t m, 89.6 L

) in reversible manner. One mole of the same gas with the same initial state in a different cylinder expands against a constant external pressure of

0.25 a t m

to final volume

89.6 L

. The difference in work done in J is

(T a k e 1 L a t m = 101 J, R = 8.3 J K^{- 1} m o l^{- 1}, l o g 2 = 0.3)

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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The correct option is B R ln(24.6)General formula of entrophy changes,△S=Cvln(T2T1)+Rln(V2V1)given {isothermal process T1=T2}Pv=nRTVf=nRTP=1×0.082×3001Vf=24.6L△S=R ln (24.6)Hence, the correct option is B