You visited us 4 times! Enjoying our articles? Unlock Full Access!

Entropy of Phase Change

At 373 K, ste...

Question

At 373 K, steam and water are in equilibrium and $△_{v a p} H = 40.98 k J m o l^{- 1} .$ What will be entropy change for conversion of water into steam?
$H_{2} O_{(l)} \to H_{2} O_{(g)}$

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $109.8 J K^{- 1} m o l^{- 1}$
The change in Entropy for the vaporisation process is given as:

$△ S_{v a p} = \frac{△ H_{v a p}}{T_{b}} = \frac{40.98 \times 1000 J m o l^{- 1}}{373 K}$

$= 109.8 J K^{- 1} m o l^{- 1}$

Suggest Corrections

Similar questions

△_{v a p} H = 40.98 k J m o l^{- 1} .

H_{2} O_{(l)} \to H_{2} O_{(g)}

Δ H

m o l^{- 1}

Δ S

H_{2} O_{(l)} \to H_{2} O_{(g)}

△_{v a p} H = 40.98 k J m o l^{- 1} .

H_{2} O_{(l)} \to H_{2} O_{(g)}

0^{o} C

H_{2} O (s) ⟶ H_{2} O (l)

6 k J {m o l}^{- 1}

0^{\circ} C

H_{2} O_{(s)} ⇌ H_{2} O_{(l)} i s 6.0 k J m o l^{- 1}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program