18.0 g of water completely vaporises at ...

18.0 g of water completely vaporises at `100^(@)C` and 1 bar pressure and the enthalpy change in the process is `40.79 kJ mol^(-1)`. What will be the enthalpy change for vaporising two moles of water under the same conditions ? What is the standard enthalpy of vaporisation for water ?

Text Solution

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`18.0` g of water `=` 1 mole
`:. ` Standard enthalpy of vaporisation at`100^(@)C` and 1 bar pressure `=40.79 kJ mol^(-1)`
Enthalpy change in vaporisationof 2 moles of water `= 2 xx 40.79 kJ = 81. 58 kJ`

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The internal energy change for the vapourisation of one mole water at 1 atm and 100°C will be (Enthalpy of vapourization of water is 40.66 kJ

Zero

37.56 kJ

-37.56 kJ

40.6 kJ

If the enthalpy change for the transition of liquid water to steam is 300kJ mol^(-1)" at "27^(@)C , the entropy change for the proces would be

`1000 JK^(-1)mol^(-1)`

`10 JK^(-1)mol^(-1)`

`1.0 JK^(-1)mol^(-1)`

`0.1JK^(-1)mol^(-1)`

If the enthalpy change for the transition of liquid water to steam is 30 kJ mol^(-1) at 27^(@)C , the entropy change for the process would be

`100 J mol^(-1)K^(-1)`

`10J mol^(-1) K^(-1)`

`1.0 J mol^(-1) K^(-1)`

`0.1 J mol^(-1) K^(-1)`