At 373.15K and 1 atm, the molar volume of liquid water and steamare 1.88 X 10-5 m3 and 3.06 X 10-2m3, respectively. Given that the heat of vaporization ofwater is 40.79 kJ/mol, calculate the values of ?H and ?Ufor 1 mole in the following process:
H2O (l, 373.15 K, 1 atm) ---> H2O(g, 373.15 K, 1 atm)

Answer The value of and is, 40.79 kJ and 37.7 kJ respectively.

Explanation :

Heat released at constant pressure is known as enthalpy.

The formula used for change in enthalpy of the gas is:

Now we have to calculate the work done.

Formula used :

where,

w = work done  = ?

P = external pressure of the gas = 1 atm

= initial volume =

= final volume =

Now put all the given values in the above formula, we get:

Now we have to calculate the change in internal energy.

Thus, the value of and is, 40.79 kJ and 37.7 kJ respectively.

the term hydride is commonly named after binary compounds that hydrogen forms with other elements of the periodic table. hydride compounds in general form with almost any element, except a few noble gases. the trends and properties vary according to the type of intermolecular force that bonds the elements together, the temperature, its molecular masses, and other components. hydrides are classified into three major groups, depending on what elements the hydrogen bonds to. the three major groups are covalent, ionic, and metallic hydrides. formally, hydride is known as the negative ion of a hydrogen, h-, also called a hydride ion. because of this negative charge, hydrides have reducing, or basic properties. its special characteristics will be further discussed.

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