Question

Part A

Determine the work done (W) by 2.00 kg of water when it is all boiled to steam at 100 ?C. Assume a constant pressure of 1.00 atm.

Part B

Determine the change in internal energy (?U) of 2.00 kg of water when it is all boiled to steam at 100 ?C. Assume a constant pressure of 1.00 atm.

Answer 1

Given

Mass of water m = 2.00 kg

Temperature T = 100^oC = 373.15 K

Pressure P = 1 atm = 101325 Pa

Known

Density of water D = 1000 kg/m³

Molar mass of water M = 18.0 g/mol

Latent heat of evaporation of water L = 2.257 x 10⁶ J/Kg

Universal Gas constant R = 8.314 J/mol.K

Solution

A)

Initial volume

V_i = mass/density

V_i = 2 /1000

V_i = 0.002 m³

Number of moles n = m/M

n =2000/18.0

n = 111.11 mole

Final Volume

PV_f = nRT

V_f = nRT/P

V_f  = 111.1 x 8.314 x 373.15 / 101325

V_f  = 3.40 m³

W = P x delV

W = P(V_f - V_i)

W = 101325 (3.40-.002)

W = 344302.35 J

W = 3.44 x 10⁵ J

B)

The energy absorbed by water to become vapour

Q = mL

Q = 2 x 2.257 x 10⁶

Q = 45.14 x 10⁵ J

U = Q - W

U = 45.14 x 10⁵- 3.44 x 10⁵

U = 41.7 x 10⁵ J

U = 4.17 x 10⁶ J