Answer:
Explanation:
From the given information:
We are to calculate the heat (q), work done (w), Internal Energy (U), Enthalpy (H) for both cases.
Latent heat of vaporization is the heat (q) change when one mole of liquid is vaporized
Given that:
Pressure (constant) = 1 bar = 100 kPa
The latent heat of vaporization[tex]\Delta H^0_{vap}[/tex] = 2260 kJ/kg
number of moles = 1 mole of liquid water
mass of the water = 18 g = 0.018 kg
∴
Heat (q) required = m[tex]\Delta H^0_{vap}[/tex]
q = (0.018 kg) × 2260 kJ/Kg
q = 40.68 kJ/mol
The work done required is calculated by using the equation :
w = PdV
w = 1 bar ([tex]V_2-V_1[/tex] )
where;
1 bar = 10⁵ Pa
[tex]V_2[/tex] = specific volume of steam = 0.0305 m³/mol
[tex]V_1[/tex] = specific volume of water = 0.00001837 m³/mol
∴
w = 10⁵ Pa ( 0.0305 - 0.00001837 ) m³/mol
w = 10⁵ Pa (0.03048163) m³/mol
w = 3048.163 J
w [tex]\simeq[/tex] 3.05 kJ
The enthalpy of reaction (H) is the energy measured in the system = 40.68 kJ/mol
The internal energy U = [tex]\Delta H - w[/tex]
The internal energy U = 40.68 - 3.05
The internal energy U = 37.63 kJ/mol
B.
here;
the volume is constant which is given as 0
Pressure = 1 atm
Therefore;
w =PdV
w = P(0)
w = 0
we know that:
q = U+w
q = U+0
q = U
q = U = 37.58 kJ/mol
The enthalpy change [tex]\Delta H = U+ w[/tex]
ΔH = 37.58 kJ/mol + 0
ΔH = 37.58 kJ/mol
