Question

(a) How much work is done on the steam when 3.52 mol of water at 100°C boils and becomes 3.52 mol of steam at 100°C at 1.00 atm pressure? (Assume the latent heat of vaporization of water is 2.26 106 J/kg.) kJ

(b) Assume the steam to behave as an ideal gas. Determine the change in internal energy of the system of the water and steam as the water vaporizes. kJ

Answer 1

Part A.

We know that work-done is given by:

W = -P*dV

W = -P*(Vf - Vi) = P*Vi - P*Vf

P = Pressure = 1.00 atm = 1.01325*10^5 N/m^2

Vi = Volume of water = ?

m = mass of water = 3.52 moles of water = 3.52 moles*(18.0 gm/mol) = 63.36 gm

Volume = mass/density = 63.36 gm/(1.0 gm/cm^3) = 63.36 cm^3 = 63.36*10^-6 m^3

Vf = Volume of steam

Using ideal gas law for steam,

P*Vf = n*R*T

T = final temperature = 100 C = 373 K

n = number of moles of steam = 3.52 moles

P*Vf = 3.52*8.314*373

P*Vf = 10915.95

So,

W = P*Vi - P*Vf

W = 1.01325*10^5*63.36*10^-6 - 10915.95

W = -10909.53 J

W = Work-done on the steam = -10.9*10^3 J = -10.9 kJ

Part B.

Energy absorbed by water during phase change will be:

Q = m*Lv

m = mass of water = 63.36 gm = 0.06336 kg

Lv = latent heat of vaporization = 2.26*10^6 J/kg

So,

Q = 0.06336*2.26*10^6

Q = 143193.6 J

Now using 1st law of thermodynamics:

dU = Q + W

dU = change in internal energy = 143193.6 + (-10909.53)

dU = 132284.07 J = 132.3*10^3 J

dU = 132.3 kJ

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