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Standard XII

Chemistry

Molar and Specific Heat Capacity

If Δ Hvap o...

Question

If $Δ H_{v a p}$ of pure water at $100^{o} C$ is 40.627 kJmol $^{- 1}$ , then the value of $Δ S_{v a p}$ is:

108.91 kJmol

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108.91 JK

^{- 1}

mol

^{- 1}

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606.27 JK

^{- 1}

mol

^{- 1}

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808.27 JK

^{- 1}

mol

^{- 1}

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Solution

The correct option is C 108.91 JK $^{- 1}$ mol $^{- 1}$
For vapourisation of water at $100^{\circ} C$ ,

$Δ H = T Δ S$
$∴ Δ S = \frac{Δ H_{v a p}}{T}$
$Δ S = \frac{40.627}{373} = 0.1089 k J K^{- 1} m o l^{- 1} = 108.91 J K^{- 1} m o l^{- 1}$

Option B is correct.

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Similar questions

Q. Calculate

△_{f} G^{o}

k J m o l^{- 1}

for

(N H_{4} C l, s)

at 310 K.
Given :

△_{f} H^{o} (N H_{4} C l, s) = - 314.5 k J m o l^{- 1}; △_{r} C_{p} = 0

S_{N_{2} (g)}^{o} = 192 J K^{- 1} m o l^{- 1}; S_{H_{2} (g)}^{o} = 130.5 J K^{- 1} m o l^{- 1};

S_{C l_{2}}^{o} (g) = 233 J K^{- 1} m o l^{- 1}; S_{N H_{4} C l (s)}^{o} = 99.5 J K^{- 1} m o l^{- 1}

All given data are at 300 K.

Q. The

Δ_{f} G^{\circ}

for

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310 K

- x {kJ mol}^{- 1}

.
Given:

Δ_{f} H^{\circ} (N H_{4} C l, s) = - 314.5 {kJ mol}^{- 1}; Δ_{r} C_{P} = 0 S_{N_{2} (g)}^{\circ} = 192 {J K}^{- 1} {mol}^{- 1}; S_{H_{2} (g)}^{\circ} = 130.5 {J K}^{- 1} {mol}^{- 1} S_{C l_{2} (g)}^{\circ} = 223 {J K}^{- 1} {mol}^{- 1}; S_{N H_{4} C l (s)}^{\circ} = 99.5 {J K}^{- 1} {mol}^{- 1}

All given data at

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If ΔHvap of pure water at 100oC is 40.627 kJmol−1, then the value of ΔSvap is:

The correct option is C 108.91 JK−1 mol−1For vapourisation of water at 100∘ C, ΔH=TΔS∴ΔS=ΔHvapTΔS=40.627373=0.1089kJK−1mol−1=108.91 JK−1mol−1Option B is correct.

If $Δ H_{v a p}$ of pure water at $100^{o} C$ is 40.627 kJmol $^{- 1}$ , then the value of $Δ S_{v a p}$ is:

The correct option is C 108.91 JK $^{- 1}$ mol $^{- 1}$
For vapourisation of water at $100^{\circ} C$ ,

$Δ H = T Δ S$
$∴ Δ S = \frac{Δ H_{v a p}}{T}$
$Δ S = \frac{40.627}{373} = 0.1089 k J K^{- 1} m o l^{- 1} = 108.91 J K^{- 1} m o l^{- 1}$

Option B is correct.