Question

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K. If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from the following:

• A. Volume of system increases.
• B. The amount of ice decreases.
• C. Liquid phase disappears completely.
• D. The solid phase (ice) disappears completely.

Hint:

Le Chatelier's principle can help predict how pressure changes affect the phases of a substance in equilibrium.

Answer 1

The Correct Option is B

We are given that ice and water are in equilibrium at 273.15 K and 1 atm pressure. This is the melting point of ice.

The Clausius-Clapeyron equation describes the relationship between pressure and temperature for phase transitions:

$$ \frac{dP}{dT} = \frac{\Delta H}{T \Delta V} $$

For the ice-water transition, $ \Delta H $ is the enthalpy of fusion (positive), and $ \Delta V = V_{\text{water}} - V_{\text{ice}} $. Since ice is less dense than water, $ V_{\text{ice}} > V_{\text{water}} $, so $ \Delta V < 0 $.

Therefore, $ \frac{dP}{dT} < 0 $. This means that an increase in pressure will lower the melting point of ice.

Since the temperature is kept constant at 273.15 K and the pressure is increased, the ice will start to melt to form water. This is because at the higher pressure, the temperature is above the new melting point.

Conclusion: The amount of ice will decrease.

Final Answer:
The final answer is $ \text{The amount of ice decreases.} $