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General Mechanism of Electrophilic Aromatic Substitution

The entropy c...

Question

The entropy change involved in the isothermal reversible expansion of $2$ moles of an ideal gas from a volume of $10$ dm $^{3}$ to a volume of $100$ dm $^{3}$ at $27^{\circ} C$ is :

42.3 J m o l^{- 1} K^{- 1}

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38.3 J m o l^{- 1} K^{- 1}

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35.8 J m o l^{- 1} K^{- 1}

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32.3 J m o l^{- 1} K^{- 1}

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Solution

The correct option is C $38.3 J m o l^{- 1} K^{- 1}$
$Δ U = 0$ for an isothermal process so $Q = - W$

$W = - n R T l n (\frac{V_{2}}{V_{1}})$

$W = - 2 R (300) l n (10)$

$W = - 11486.21 J$

So, $Δ S = \frac{q}{T}$

$Δ S = \frac{11486.21}{300}$

$Δ S = 38.3 J m o l^{- 1} K^{- 1}$

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The entropy change involved in the isothermal reversible expansion of 2 mole of an ideal gas from a volume of $10 d m^{3}$ to a volume of $100 d m^{3}$ at $27^{\circ} C$ is :

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The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27∘C is :

The correct option is C 38.3 Jmol−1K−1ΔU=0 for an isothermal process so Q=−WW=−nRTln(V2V1)W=−2R(300)ln(10)W=−11486.21JSo, ΔS=qTΔS=11486.21300ΔS=38.3Jmol−1K−1

The entropy change involved in the isothermal reversible expansion of $2$ moles of an ideal gas from a volume of $10$ dm $^{3}$ to a volume of $100$ dm $^{3}$ at $27^{\circ} C$ is :

The correct option is C $38.3 J m o l^{- 1} K^{- 1}$
$Δ U = 0$ for an isothermal process so $Q = - W$

$W = - n R T l n (\frac{V_{2}}{V_{1}})$

$W = - 2 R (300) l n (10)$

$W = - 11486.21 J$

So, $Δ S = \frac{q}{T}$

$Δ S = \frac{11486.21}{300}$

$Δ S = 38.3 J m o l^{- 1} K^{- 1}$