Latent heat of vaporisation of a liquid at `500K` and `1` atm pressure is `10.0 kcal//mol`. What will be the change in internal energy `(DeltaE)` of `

Question

Latent heat of vaporisation of a liquid at `500K` and `1` atm pressure is `10.0 kcal//mol`. What will be the change in internal energy `(DeltaE)` of `

Latent heat of vaporisation of a liquid at `500K` and `1` atm pressure is `10.0 kcal//mol`. What will be the change in internal energy `(DeltaE)` of `3` mol of liquid at same temperature?
A. `13.0 kcal`
B. `-13.0 kcal`
C. `27.0 kcal`
D. `-27.0 kcal`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - C
Vaporisation of `3` moles of water is
`3H_(2)O(l)to3H_(2)O(g)`
Here `Deltan=3-0=3`
Heat change for `3` moles of water to vapours
`=3xx10=30 kJ`
`:. DeltaE=DeltaH-DeltanRT`
`=30-(3)(0.002)(500)=27 kcal`

5 views Answered 2023-02-05 15:29