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Standard XII

Chemistry

Entropy Change in Irreversible Process

Consider an i...

Question

Consider an isothermal reversible compression of 1 mole of an ideal gas in which pressure of the system increased from 5 atm to 30 atm at 300 K. The entropy change (in kJ) of the surrounding is:
( $R = 8.314 J m o l^{- 1} K^{- 1}$ )

1.5 k J / m o l

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15 k J / m o l

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0.07 k J / m o l

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0.7 k J / m o l

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Solution

The correct option is B $15 k J / m o l$
We know, $d W = - p d V$
Then
$W = - \int_{V_{1}}^{V_{2}} P d V = - \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V = - n R T l n (\frac{V_{1}}{V_{2}})$

or, $W = - n R T l n (\frac{P_{1}}{P_{2}})$ [ $∵ \frac{P_{1}}{P_{2}} = \frac{V_{1}}{V_{2}}$ ]

or, $W = - 1 \times 8.314 \times 300 \times l n (\frac{5}{30}) = - 8.314 \times 300 \times 1.79$

or, $W = - 4.46 k J$
Therefore, $Q = - W = 4.46 k J$ [ $∵ Δ U = 0$ ]

$Δ S = \frac{Q}{T} = \frac{4.46 \times 10^{3}}{3 \times 10^{2}} = 15 k J / m o l$

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Q. Consider an isothermal reversible compression of 1 mole of an ideal gas in which pressure of the system increased from 5 atm to 30 atm at 300 K. The entropy change (in kJ) of the surrounding is:
(

R = 8.314 J m o l^{- 1} K^{- 1}

)

Q. Consider an isothermal reversible compression of 1 mole of an ideal gas in which pressure of the system increased from 5 atm to 30 atm at 300 K. The entropy change (in kJ) of the surrounding is:
(

R = 8.314 J m o l^{- 1} K^{- 1}

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Q. The molar entropy content of 1 mole of oxygen

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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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Consider an isothermal reversible compression of 1 mole of an ideal gas in which pressure of the system increased from 5 atm to 30 atm at 300 K. The entropy change (in kJ) of the surrounding is: (R=8.314 Jmol−1K−1)

The correct option is B 15 kJ/molWe know, dW=−pdV Then W=−∫V2V1PdV=−∫V2V1nRTVdV=−nRT ln(V1V2) or, W=−nRT ln(P1P2) [∵P1P2=V1V2] or, W=−1×8.314×300×ln(530)=−8.314×300×1.79 or, W=−4.46 kJ Therefore, Q=−W=4.46 kJ [∵ΔU=0] ΔS=QT=4.46×1033×102=15 kJ/mol

Consider an isothermal reversible compression of 1 mole of an ideal gas in which pressure of the system increased from 5 atm to 30 atm at 300 K. The entropy change (in kJ) of the surrounding is:
( $R = 8.314 J m o l^{- 1} K^{- 1}$ )