Question

At 1 atm, liquid water is heated above 100°C. For this process which of the following choices (i–iv) is correct for ΔS_surr? ΔS? ΔS_univ? Explain each answer.

1. greater than zero
2. less than zero
3. equal to zero
4. cannot be determined.

Short answer

For the given process(evaporation) ΔS_surr is less than zero while ΔS_system and ΔS_univare greater than zero.

Step-by-step solution

ΔSsystem for the given process

When water is heated above 100°C (boiling point of water), it will undergo the following change:

${\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

As the liquid coverts into a gas, the volume of the system increases, this causes an increase in positional probability, hence ΔS_system has positive sign i.e., it is greater than zero.

ΔSsurr for the given process 

Vaporization of water is an endothermic process and during an endothermic process heat flows from surrounding to the system causing a decrease in randomness of the atoms in surroundings so entropy of surroundings decreases.

ΔS_surr = -ve

Or we can say that ΔS_surris lessthan zero.

ΔSuniv for the given process

ΔS_system is positive for vaporization while ΔS_surr is negative and ΔS_univ is positive(i.e. more than 0) above 100°C because above 100 °C reaction is spontaneous an ΔS_univdecides about the spontaneity of the reaction.