Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15 K. If pressure of the…
Chemical Equilibrium · Class 11 · JEE Main Previous Year Question
Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15 K. If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:
(a) Volume of system increases. (b) Liquid phase disappears completely. (c) The amount of ice decreases. (d) The solid phase (ice) disappears completely.
- a
Volume of system increases.
- b
Liquid phase disappears completely.
- c
The amount of ice decreases.
- d✓
The solid phase (ice) disappears completely.
The solid phase (ice) disappears completely.
Phase diagram of water — key feature:
The solid-liquid boundary (melting curve) of water has a negative slope (unlike most substances). This means increasing pressure at constant temperature lowers the melting point of ice.
Analysis at 273.15 K, pressure doubled to 2 atm:
At 273.15 K and 1 atm, ice and water coexist (on the melting curve). When pressure is increased to 2 atm:
- The melting point of ice decreases below 273.15 K
- At 273.15 K and 2 atm, we are now above the melting curve (in the liquid region)
- All ice melts → solid phase disappears completely
(a) Volume increases: FALSE — liquid water is denser than ice; melting ice → volume decreases.
(b) Liquid disappears: FALSE — pressure increase favors liquid (denser phase).
(c) Amount of ice decreases: Partially true but incomplete — ALL ice disappears.
(d) Solid phase disappears completely: TRUE ✓ — at 2 atm and 273.15 K, system is in liquid region.
Answer: Option (d)
Key Points: Water's melting curve has negative slope; increasing pressure melts ice (Le Chatelier: favors denser liquid phase); ice → water at 2 atm, 273.15 K.
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