Question

One mole of liquid water at 100°C is heated until the liquid is converted entirely to vapor at 100°C and 1 bar pressure. Calculate q, w, U, and H for each of the following:
A. The vaporization is carried out in a cylinder where the external pressure on the piston is maintained at 1 bar throughout.
B. The cylinder is first expanded against a vacuum (Pex=0) to the same volumn as in part (a), and then sufficient heat is added to vaporize the liquid completely to 1 atm pressure.

Answer 1

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$\Delta H^0_{vap}$ = 2260 kJ/kg

number of moles = 1 mole  of liquid water

mass of the water = 18 g = 0.018 kg

∴

Heat (q) required  = m$\Delta H^0_{vap}$

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 ($V_2-V_1$ )

where;

1 bar = 10⁵ Pa

$V_2$ = specific volume of steam = 0.0305 m³/mol

$V_1$ = 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 $\simeq$ 3.05 kJ

The enthalpy of reaction (H)  is the energy measured in the system = 40.68 kJ/mol

The internal energy U = $\Delta H - w$

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 $\Delta H = U+ w$

ΔH = 37.58 kJ/mol + 0

ΔH = 37.58 kJ/mol