Question

For the following change, \[ \mathrm{H_2O(l)} \xrightarrow[100^\circ C]{5^\circ C} \mathrm{H_2O(g)} \] Select the correct answer:

• A. \( q = +ve,\ w = +ve,\ \Delta H = +ve \)
• B. \( q = -ve,\ w = -ve,\ \Delta H = +ve \)
• C. \( q = +ve,\ w = -ve,\ \Delta H = +ve \)
• D. \( q = -ve,\ w = -ve,\ \Delta H = -ve \)

Hint:

Vaporisation processes are always endothermic and involve expansion work done by the system.

Answer 1

The Correct Option is C

Concept:
Heat absorbed by the system is taken as positive (\(q>0\)).
Work done by the system on surroundings is taken as negative (\(w<0\)). 
Enthalpy change (\(\Delta H\)) is positive for endothermic processes. 
Step 1: Nature of the process. The change involves: 
Heating water from \(5^\circ C\) to \(100^\circ C\) 
Phase change from liquid to gas Both processes require absorption of heat. \[ \Rightarrow q = +ve \] 
Step 2: Sign of work done. During vaporisation, water expands against atmospheric pressure. \[ \Rightarrow \text{System does work on surroundings} \] \[ \Rightarrow w = -ve \] 
Step 3: Enthalpy change. Since heat is absorbed at constant pressure: \[ \Delta H = q_p>0 \] Conclusion: \[ q = +ve,\quad w = -ve,\quad \Delta H = +ve \]