A beaker containing one mole of liquid water––the “system”––at 325K and one atm

Question

A beaker containing one mole of liquid water––the “system”––at 325K and one atm pressure is placed in a hot water bath that is maintained at a constant temperature of 100°C. The beaker is left in the hot water bath until the temperature of the water in the beaker just reaches 350K. For this process, calculate: (a) H(syst) (b) W(syst) (c) U(syst) (d) S(syst) (e) S(univ) (f) G(syst) (g) if this is a spontaneous process. Why?

Data: Density of liquid water at 325K = 0.988 g cm–3; at 350K = 0.972 g cm–3 Molar mass of water = 18.015 g mol–1 Molar entropy of liquid water at 25°C and one atm pressure = 69.939 J K–1 mol–1 Molar heat capacity for liquid water: cP = 75.3 J mol–1 K–1

Explanation / Answer

a)

Enthalpy Definition

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Enthalpy is basically a thermodynamic quantity. It is net heat content of the body.
Mathematically, the enthalpy of a system H is defined as

H = U + pV

where,
H is the enthalpy of the system
U is the internal energy of the system
p is the pressure at the boundary of the system and its surrounding
V is the volume of the system.

Enthalpy = energy/mass

= 18.015/0.972

= 18.053 J/G

B)W(syst) =

w = d × f

Where:

w = work, in joules (N×m) (or calories, but we are using primarily SI units)
d = distance in meters
f = opposing force in Newtons (kg*m/s2)

In chemical reactions, work is generally defined as :

w = distance × (area × pressure)

The value of distance times area is actually the volume. If we imagine a reaction taking place in a container of some volume, we measure work by pressure times the change in volume.

w = V × P

Where:

V is the change in volume, in liters

Density = mass/volume here density = 0.988 gm/cm3 & Mass = 0.972 gm/cm3

So volume of water is 0.988=0.972/volume

Volume = 0.972/0.988

= 0.983 l

So W(System)= 0.983*1(there is no change in volume)

= 0.983 lit/atm

C)U(syst) : H+PV

= 18.053+1*(0.983)

=19.036

D)S(syst) : S= q/t here q=cp=75.3 J mol k-1

= 75.3/350

=0.2151

E)S(Universe)= S(system)+S(Surrounding)

=0.2151+0

= 0.2151

F)G(Syst) = H-TS

=18.053-[(325*0.2151)]

=18.053-69.9

= -51.847

G)Gibbs free energy is negetive so Process is Spontenous

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