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Q11.22(a):
Answer the following questions based on the P – T phase diagram of $CO_2$: (a) $CO_2$ at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase ?
Solution :
Step 1: Identifying the Phase States of $CO_2$
To determine the phase transition of $CO_2$ under isothermal compression, we must reference the standard Phase Diagram (Pressure-Temperature) for Carbon Dioxide. The key parameters are:
- Triple Point ($T_t, P_t$): $5.11 \text{ atm}$ and $-56.6 \text{ °C}$ ($216.55 \text{ K}$). [Per the phase equilibrium constant of $CO_2$].
- Critical Point ($T_c, P_c$): $73.0 \text{ atm}$ and $31.1 \text{ °C}$ ($304.25 \text{ K}$).
- Sublimation Curve: The boundary between the solid and gas phases, which exists below the triple point temperature ($-56.6 \text{ °C}$).
Step 2: Analyzing the Initial State
The system begins at:
Pressure ($P_i$) = $1 \text{ atm}$
Temperature ($T_i$) = $-60 \text{ °C}$
Comparing $T_i$ to the triple point temperature ($T_t$):
$-60 \text{ °C} < -56.6 \text{ °C}$
Since the operating temperature is lower than the triple point temperature, the system resides in the region of the $P-T$ diagram where the solid phase is in equilibrium with the vapor (gas) phase. [Per the Phase Rule for a one-component system: $F = C - P + 2$, where $F=1$ along the sublimation curve].
Step 3: Evaluating the Isothermal Compression Path
An isothermal compression implies that the temperature remains constant at $-60 \text{ °C}$ while the pressure increases from $1 \text{ atm}$.
As the pressure increases along the line $T = -60 \text{ °C}$, the path crosses the sublimation curve rather than the fusion or vaporization curves. This means that the substance transitions directly from the gaseous phase to the solid phase (deposition). It does not pass through the liquid phase because the entire isothermal process occurs at a temperature below the triple point temperature ($-56.6 \text{ °C}$).
Conclusion
For a substance to enter the liquid phase, the system must reach a temperature at least equal to the triple point temperature ($T \geq -56.6 \text{ °C}$). Because the isothermal process is fixed at $-60 \text{ °C}$, the liquid phase is thermodynamically inaccessible.
Final Solution: No, $CO_2$ does not go through a liquid phase; it undergoes a phase transition directly from gas to solid (deposition).
More Questions from Class 11 Physics Thermal Properties of Matter ADDITIONAL EXERCISES
- Q11.21(a): Answer the following questions based on the P-T phase diagram of carbon dioxide: (a) At what temperature and pressure can the solid, liquid and vapour phases of $CO_2$ co-exist in equilibrium ?
- Q11.21(b): Answer the following questions based on the P-T phase diagram of carbon dioxide: (b) What is the effect of decrease of pressure on the fusion and boiling point of $CO_2$ ?
- Q11.21(c): Answer the following questions based on the P-T phase diagram of carbon dioxide: (c) What are the critical temperature and pressure for $CO_2$ ? What is their significance ?
- Q11.21(d): Answer the following questions based on the P-T phase diagram of carbon dioxide: (d) Is $CO_2$ solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm, (c) 15 °C under 56 atm ?
- Q11.22(b): Answer the following questions based on the P – T phase diagram of $CO_2$: (b) What happens when $CO_2$ at 4 atm pressure is cooled from room temperature at constant pressure ?
- Q11.22(c): Answer the following questions based on the P – T phase diagram of $CO_2$: (c) Describe qualitatively the changes in a given mass of solid $CO_2$ at 10 atm pressure and temperature –65 °C as it is heated up to room temperature at constant pressure.
- Q11.22(d): Answer the following questions based on the P – T phase diagram of $CO_2$: (d) $CO_2$ is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do you expect to observe ?
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