latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10 .0 kcal/ mol . What will be the change in internal energy (`DeltaE`) of 3 mo

Question

latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10 .0 kcal/ mol . What will be the change in internal energy (`DeltaE`) of 3 mo

latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10 .0 kcal/ mol . What will be the change in internal energy (`DeltaE`) of 3 moles of liquid at same temperature ?
A. 30 kcal
B. `-54 kcal`
C. 27.0 kcal
D. 50 kcal

Monjur Alahi

6 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
`DeltaH=DeltaE+Deltan_(g) RT, 30 = 30 Delta + 3xx2xx500xx10^(-3)`
`DeltaE=27 kcal`

6 views Answered 2023-02-05 15:29