Question

The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic.

H2​O(l)+40.7kJ H2​O(g)

Assume at exactly 100.0°C and 1.00 atm total pressure, 1.00 mole of liquid water and 1.00 mole of water vapor occupy 18.80 mL and 30.62 L, respectively.

A)Calculate the work done on or by the system when 2.45 mol of liquid H2O vaporizes.

__________J

Calculate the water's change in internal energy.

____________kj

Answer 1

From the question ,

The liquid water is evaporated to vapors from the following reaction -

H₂O (l) + 40.7kJ ----> H₂O(g)

i.e.,

The energy required for the conversion of 1 mole liquid water to 1mole water in gaseous state is 40.7kJ.

The evaporation process occurs at 1 atm ,

According to the question ,

For 1 mole of the water ,

The volume of liquid water ( initial volume ) = 18.80 mL = 0.01880 L ( Since , 1mL = 10^-3 L)

The volume of gaseous water ( final volume ) = 30.62 L

Calculating the value of work done ,

Work done can be calculated as the product of the pressure with the change in volume , i.e. , final volume – initial volume.

Since ,

the water is converting into gas , as this is the process of expansion ,

Hence ,

Work is done by the system , the formula becomes ,

W = - P Δ V

W = - P * ( Final volume – initial volume )

W = - 1 atm * ( 30.62 L – 0.01880 L )

W = - 30.601 L atm mol^-1

Since ,

1 L atm mol ^-1 = 101.325 J / mol .

W = - 30.601 * 101.325

W = - 3100.646J /mol

Since , - 3100.646 J /mol is the work done for 1 mol .

From the question ,

The work done for 2.45 mol , can be calculated as ,

W = - 3100.646 J /mol * 2.45 mol

W = - 7596.58 J

Since ,

1J = 10 ^-3 kJ

W = - 7596.58 J = - 7596.58 * 10 ^-3 kJ = - 7.59 kJ

According the first law of thermodynamics ,

The change in internal energy is given as

ΔH = ΔU + PΔV

ΔU = ΔH – PΔV

ΔU = ΔH – W ( Since , W = PΔV )

W = - 7.59 kJ ( as calculated above )

ΔH = 40.7 kJ ( Given in the question )

Putting the respective values in the formula ,

ΔU = ΔH – W

ΔU = 40.7 kJ - (– 7.59 kJ)

ΔU = 40.7 kJ + 7.59 kJ

ΔU = 48.29 kJ

Hence ,

The value of change in internal energy of water = 48.29 kJ.