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:
Show Hint
- Volume of system increases.
- The amount of ice decreases.
- Liquid phase disappears completely.
- The solid phase (ice) disappears completely.
The Correct Option is B
Approach Solution - 1
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.} $
Approach Solution -2
We have a system containing ice and water in equilibrium at 1 atm and 273.15 K. When the pressure is doubled while keeping the temperature constant, we must determine what happens to the amount of ice.
Step 2: Apply the Clausius–Clapeyron equation.
The relationship between pressure and temperature at the phase boundary is given by:
\[ \frac{dT}{dP} = \frac{T(V_{liquid} - V_{solid})}{L} \] where:
- \( V_{liquid} \) and \( V_{solid} \) are the molar volumes of water and ice, respectively.
- \( L \) is the latent heat of fusion.
Step 3: Analyze the sign of the slope.
For water, the volume of ice is greater than that of liquid water, i.e. \( V_{solid} > V_{liquid} \).
Hence, \( (V_{liquid} - V_{solid}) \) is negative.
Thus, \( \frac{dT}{dP} \) becomes negative, meaning that as pressure increases, the melting point of ice decreases.
Step 4: Apply this to the given condition.
At constant temperature (273.15 K), when the pressure is increased, the equilibrium will shift in the direction that reduces the system’s volume — i.e., from solid (ice) to liquid (water).
Step 5: Conclusion.
Therefore, under increased pressure and constant temperature, some ice melts into water to reduce the overall volume of the system.
Final Answer:
The amount of ice decreases.
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