Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15 K. If pressure of the syste

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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…

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Given: Ice and water are in equilibrium at $273.15 \, \text{K}$ and $1 \, \text{atm}$ .

Find: What happens when pressure is doubled at constant temperature.

At $273.15 \, \text{K}$ and $1 \, \text{atm}$ , ice and water are at the melting equilibrium.

The Clausius-Clapeyron relation for phase transition is:

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

For the ice-water transition, $\Delta H$ is positive and

\Delta V = V_{\text{water}} - V_{\text{ice}}

Since ice is less dense than water,

V_{\text{ice}} > V_{\text{water}}

Therefore,

\Delta V < 0

Hence,

\frac{dP}{dT} < 0

So, increasing pressure lowers the melting point of ice.

At constant temperature $273.15 \, \text{K}$ , when pressure is increased, the system becomes effectively above the new melting point. Therefore, some ice melts to form water.

Conclusion: The amount of ice decreases, so the correct option is B.

Given: Ice and water coexist in equilibrium at $1 \, \text{atm}$ and $273.15 \, \text{K}$ .

Find: The effect of doubling pressure while keeping temperature constant.

For water, solid ice has larger volume than liquid water. Therefore, the liquid phase occupies less volume than the solid phase.

By Le Chatelier's principle, when pressure is increased, the equilibrium shifts toward the phase with lower volume.

So the equilibrium shifts from ice to water.

Therefore, some ice melts and the amount of ice decreases. Hence, the correct option is B.

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