You visited us 2 times! Enjoying our articles? Unlock Full Access!

Entropy of Phase Change

If Hvapourisa...

Question

If $△ H_{v a p o u r i s a t i o n}$ of substance $X (l)$ (molar mass = 30 g/mol) is 300 J/g at its boiling point 300 K, molar entropy change for reversible condensation process is:

30 J / m o l K^{- 1}

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

- 60 J / m o l K^{- 1}

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

- 30 J / m o l K^{- 1}

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

None of these

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

Open in App

Solution

The correct option is C $- 30 J / m o l K^{- 1}$
Vapourisation and condensation are reversible process,
Enthalpy of vapourisation = $300 J / g$
Enthalpy of condensation = $- 300 J / g$
Molar enthalpy of condensation = $- 300 \times 30 g / m o l$
Molar entropy, $△ S = \frac{△ H_{c o n d e n s a t i o n}}{T}$
$△ S = \frac{- 300 \times 30}{300}$
$= - 30 J / m o l K^{- 1}$

Suggest Corrections

Similar questions

Q. If

△ H_{v a p o r i s a t i o n}

of substance

X (l)

(molar mass:

30 g / m o l)

300 J / g

at it's boiling point

300 K

, then molar entropy change for reversible condensation process is____________.

If $Δ H_{v a p o r i s a t i o n}$ of substance X(1) (molar mass: 30 g/mol) is 300 J/g at its boiling point 300 K, then molar entropy change for reversible condensation process is

Q. If

Δ H_{v a p o r i s a t i o n}

of substance

X (l)

(molar mass :

30 g / m o l

) is

300 J / g

at its boiling point

300 K

, then molar entropy change for reversible condensation process is:

Q. If

△ H_{v a p o r i s a t i o n}

of substance

X (l)

(molar mass : 30 g/mol) is 300 J/g at it's boiling point 300 K, then molar entropy change for reversible condensation process is

Q. If

△ H_{v a p o r i s a t i o n}

of substance

X (l)

(molar mass :

30 g / m o l

) is

300 K

, then molar entropy change for reversible condensation process is:

Join BYJU'S Learning Program

If △Hvapourisation of substance X(l) (molar mass = 30 g/mol) is 300 J/g at its boiling point 300 K, molar entropy change for reversible condensation process is:

The correct option is C −30 J/molK−1Vapourisation and condensation are reversible process, Enthalpy of vapourisation = 300 J/g Enthalpy of condensation = −300 J/g Molar enthalpy of condensation = −300×30 g/mol Molar entropy, △S=△HcondensationT △S=−300×30300 =−30 J/molK−1

If $△ H_{v a p o u r i s a t i o n}$ of substance $X (l)$ (molar mass = 30 g/mol) is 300 J/g at its boiling point 300 K, molar entropy change for reversible condensation process is: