Theoretical Foundation: Phase Diagram of $CO_2$

The phase diagram of $CO_2$ (Pressure-Temperature) represents the equilibrium states of the substance. Unlike water, $CO_2$ exhibits a positive slope for the fusion (melting) curve, meaning the solid phase is denser than the liquid phase [Per the Clausius-Clapeyron relation: $\frac{dP}{dT} = \frac{L}{T(V_v - V_l)}$].

Step 1: Analyzing the Fusion Point (Solid-Liquid Equilibrium)

The fusion curve for $CO_2$ starts from the triple point ($5.11 \text{ atm}$, $216.55 \text{ K}$) and slopes upward to the right.

• The slope of the fusion line is positive, signifying that as pressure increases, the melting point increases.
• Conversely, a decrease in pressure results in a decrease in the fusion point (melting temperature).

Step 2: Analyzing the Boiling Point (Liquid-Vapor Equilibrium)

The vaporization curve extends from the triple point to the critical point ($73 \text{ atm}$, $304.15 \text{ K}$).

• As pressure decreases along the vaporization curve, the boiling point moves toward lower temperatures.
• A decrease in pressure leads to a decrease in the boiling point of the liquid phase.

Visual Representation of $P-T$ Phase Diagram ($CO_2$)

Summary of Thermodynamic Effects

Phase Change	Effect of Pressure Decrease	Thermodynamic Principle
Fusion	Decreases	Positive Clausius-Clapeyron Slope
Boiling	Decreases	Standard Volumetric Expansion ($V_v > V_l$)

Final Solution: For carbon dioxide, a decrease in pressure results in a decrease in both the fusion point (melting temperature) and the boiling point.

Initial State and Thermodynamic Coordinates

Given the initial state of the carbon dioxide ($CO_2$) sample:

• Pressure ($P$): $10\text{ atm}$
• Temperature ($T$): $-65\text{ °C}$
• Phase: Solid ($CO_2$ ice/dry ice)

[Per the $P-T$ phase diagram of $CO_2$, the triple point is located at $P = 5.11\text{ atm}$ and $T = -56.6\text{ °C}$]. Since the applied pressure ($10\text{ atm}$) is significantly above the triple point pressure, and the temperature ($-65\text{ °C}$) is well below the sublimation/fusion thresholds, the sample exists in the stable solid phase.

Phase Transition Analysis

As the system is heated at a constant pressure of $10\text{ atm}$, it traverses the following path in the $P-T$ coordinate plane:

1. Heating the Solid: The temperature increases from $-65\text{ °C}$. The solid $CO_2$ undergoes sensible heating, increasing in internal energy without a phase change until the fusion curve is intersected.
2. Melting (Solid to Liquid): At $10\text{ atm}$, the melting point of $CO_2$ is approximately $-56\text{ °C}$. Upon reaching this temperature, the solid absorbs latent heat of fusion [Per the First Law of Thermodynamics, $dQ = dU + PdV$]. The phase transition from solid to liquid occurs while the temperature remains constant during the phase change.
3. Heating the Liquid: Once the mass is entirely in the liquid phase, further heating causes the temperature of the liquid $CO_2$ to rise.
4. Vaporization (Liquid to Gas): As heating continues toward room temperature ($20\text{ °C}$–$25\text{ °C}$), the system approaches the boiling point. At $10\text{ atm}$, which is lower than the critical pressure of $CO_2$ ($73\text{ atm}$), the liquid will eventually cross the vaporization curve and undergo a phase change into the gaseous state (vapor).

Visual Representation of the Phase Transition Path

Summary of Thermodynamic Evolution

The progression of the system can be summarized by the following table:

Phase	Temperature Range	Mechanism
Solid	$-65\text{ °C}$ to $\approx -56\text{ °C}$	Sensible heating of solid
Solid + Liquid	$\approx -56\text{ °C}$	Latent heat of fusion
Liquid	$-56\text{ °C}$ to $T_{boiling}$	Sensible heating of liquid
Gas	$T_{boiling}$ to Room Temp	Expansion/Vaporization

Final Solution: The solid $CO_2$ first undergoes sensible heating, then melts into liquid $CO_2$ at approximately $-56\text{ °C}$, and finally, as heating continues toward room temperature at $10\text{ atm}$, the liquid will eventually vaporize into gas.

Step 1: Theoretical Basis of Phase Equilibrium

The state of a substance where three phases (solid, liquid, and gas/vapour) exist simultaneously in thermodynamic equilibrium is defined as the Triple Point. [Per the Gibbs Phase Rule, $F = C - P + 2$, where $F$ is degrees of freedom, $C$ is components (1 for $CO_2$), and $P$ is phases (3). At the triple point, $F = 1 - 3 + 2 = 0$, meaning the point is invariant and fixed at a specific temperature and pressure].

Step 2: Empirical Data for Carbon Dioxide

For Carbon Dioxide ($CO_2$), the triple point parameters are experimentally determined constants within the $P-T$ phase diagram. Unlike water, the triple point of $CO_2$ occurs at a pressure significantly higher than standard atmospheric pressure ($1 \text{ atm} \approx 1.013 \times 10^5 \text{ Pa}$).

Parameter	Value
Temperature ($T_{tp}$)	$216.55 \text{ K}$ (or $-56.6^\circ\text{C}$)
Pressure ($P_{tp}$)	$5.11 \text{ atm}$ (or $5.17 \times 10^5 \text{ Pa}$)

Step 3: Phase Diagram Representation

The following SVG illustrates the $P-T$ phase diagram for $CO_2$, highlighting the triple point where the sublimation, fusion, and vaporisation curves intersect.

Step 4: Analytical Conclusion

Because the triple point pressure ($5.11 \text{ atm}$) is greater than $1 \text{ atm}$, solid $CO_2$ (dry ice) at standard atmospheric pressure cannot transition into a liquid phase when heated. Instead, it undergoes sublimation directly from a solid to a gaseous state, as the $1 \text{ atm}$ isobar lies entirely below the triple point on the $P-T$ diagram. [Based on the Clausius-Clapeyron relation describing phase boundaries].

Final Solution: The solid, liquid, and vapour phases of $CO_2$ co-exist in equilibrium at a temperature of $216.55 \text{ K}$ and a pressure of $5.11 \text{ atm}$.

Initial State & Physical Context

The phase diagram of $CO_2$ (Carbon Dioxide) maps the state of matter as a function of Pressure ($P$) and Temperature ($T$). Key reference points for $CO_2$ include:

• Triple Point: $P_t = 5.11 \text{ atm}$, $T_t = 216.55 \text{ K}$ ($-56.6 ^\circ\text{C}$).
• Critical Point: $P_c = 73 \text{ atm}$, $T_c = 304.15 \text{ K}$ ($31.1 ^\circ\text{C}$).

Given the condition: $P = 4 \text{ atm}$ (Constant). Since $4 \text{ atm} < 5.11 \text{ atm}$ (the triple point pressure), $CO_2$ does not possess a liquid phase at this pressure.

Step 1: Visualization of the P-T Phase Path

Step 2: Thermodynamic Analysis of the Cooling Process

The cooling process starts at room temperature ($T \approx 293 \text{ K}$) at a constant pressure of $4 \text{ atm}$.

1. Initial Phase: At $4 \text{ atm}$ and room temperature, $CO_2$ exists in the gaseous (vapor) phase.
2. The Phase Boundary: As the temperature decreases, the state moves horizontally to the left on the phase diagram. Because the pressure is constant at $4 \text{ atm}$, the system will intersect the sublimation curve (the boundary separating the solid phase and the vapor phase).
3. The Phase Transition: Since the triple point pressure for $CO_2$ is $5.11 \text{ atm}$, and our operating pressure is below this, the liquid phase is skipped entirely [Per the Phase Rule: $F = C - P + 2$]. The $CO_2$ undergoes desublimation (or deposition).

Step 3: Conclusion on the Physical Outcome

When the temperature drops to the sublimation temperature corresponding to $4 \text{ atm}$, the gaseous $CO_2$ molecules transition directly into the solid state without passing through the liquid state. This solid $CO_2$ is commonly referred to as "dry ice."

Final Solution: The $CO_2$ will condense directly from the gaseous phase to the solid phase (dry ice) at the sublimation temperature, as the given pressure of $4 \text{ atm}$ is lower than the triple point pressure of $5.11 \text{ atm}$.

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).

Step 1: Analyzing the Critical Parameters of $CO_2$

To determine the state changes of carbon dioxide ($CO_2$) at $70\text{ }^\circ\text{C}$, we must first identify the critical temperature ($T_c$) of the substance.

• Critical Temperature of $CO_2$ ($T_c$): $31.1\text{ }^\circ\text{C}$
• Critical Pressure of $CO_2$ ($P_c$): $73.8\text{ atm}$
• Operating Temperature ($T$): $70\text{ }^\circ\text{C}$

[Per the Kinetic Theory of Gases, a substance above its critical temperature cannot be liquefied by pressure alone, regardless of the magnitude of the applied pressure.]

Step 2: Thermodynamic Behavior Analysis

Since the given temperature ($70\text{ }^\circ\text{C}$) is greater than the critical temperature ($31.1\text{ }^\circ\text{C}$), $CO_2$ exists in the gaseous phase (specifically as a supercritical fluid if pressure is increased sufficiently).
As the gas is compressed isothermally at $70\text{ }^\circ\text{C}$:

• The volume of the $CO_2$ sample decreases, adhering to Boyle's Law ($PV = \text{constant}$) for ideal gases, though it will show deviations as a real gas.
• The density of the fluid increases continuously.
• There is no sharp transition or phase change (i.e., no condensation into a distinct liquid phase).
• The substance transitions from a gaseous state to a dense, supercritical fluid state.

Step 3: Comparison of Physical Properties

Property	Observation at $T > T_c$
Phase Transition	None (remains a fluid)
Density	Increases monotonically
Appearance	Uniform; no meniscus formation

Summary of Observations

Because the temperature $70\text{ }^\circ\text{C} > 31.1\text{ }^\circ\text{C}$, the $CO_2$ will not liquefy. As it is compressed, it will undergo a continuous transition into a dense, supercritical state. The substance will retain the properties of a fluid, exhibiting high density, but will not undergo a phase separation into liquid and vapor phases typically associated with condensation.

Final Solution: $CO_2$ will not liquefy; it will exist as a gas that becomes increasingly dense and eventually enters a supercritical fluid state without any discontinuous change in state.

1. Theoretical Foundation: Phase Diagram of $CO_2$

The state of $CO_2$ at specific pressure ($P$) and temperature ($T$) conditions is determined by its phase diagram, which demarcates the stability regions of the solid, liquid, and gas phases. Key characteristic points for $CO_2$ are:

• Triple Point: $P_t = 5.11 \text{ atm}$, $T_t = -56.6 \text{ °C}$.
• Critical Point: $P_c = 73 \text{ atm}$, $T_c = 31.1 \text{ °C}$.

2. Analysis of Specific States

Condition	Given Point $(T, P)$	Phase Determination
(a)	$-70 \text{ °C}$, $1 \text{ atm}$	Since $T < T_t$ and $P < P_t$, the substance lies within the Solid region (dry ice).
(b)	$-60 \text{ °C}$, $10 \text{ atm}$	Since $T > T_t$ (slightly) and $P > P_t$, but below critical curves for liquid, and specifically for $CO_2$, the pressure $10 \text{ atm}$ at $-60 \text{ °C}$ lies in the Liquid region.
(c)	$15 \text{ °C}$, $56 \text{ atm}$	Since $T < T_c$ ($31.1 \text{ °C}$) and $P < P_c$ ($73 \text{ atm}$), this point resides in the Liquid region (as it is above the liquid-vapor equilibrium line).

3. Logical Justification

• For (a): At $-70 \text{ °C}$, the temperature is below the triple point temperature ($-56.6 \text{ °C}$). At $1 \text{ atm}$, which is below the triple point pressure ($5.11 \text{ atm}$), $CO_2$ undergoes sublimation and remains in the solid phase [Per the phase diagram of $CO_2$].
• For (b): At $-60 \text{ °C}$ and $10 \text{ atm}$, the pressure is higher than the triple point pressure ($5.11 \text{ atm}$), pushing the phase into the liquid regime [By state equilibrium analysis].
• For (c): Given $15 \text{ °C}$ and $56 \text{ atm}$, the system is below the critical temperature and pressure, placing it firmly in the liquid state [Per Clapeyron-Clausius relation for phase stability].

Final Solution: (a) Solid, (b) Liquid, (c) Liquid.

1. Identification of Critical Parameters for $CO_2$

The critical point on a $P-T$ phase diagram represents the terminal point of the liquid-gas equilibrium curve. Beyond this point, the physical distinction between liquid and gaseous phases vanishes. For carbon dioxide ($CO_2$), experimental measurements define the following critical values:

• Critical Temperature ($T_c$): $31.1 \, ^\circ\text{C}$ (or $304.1 \, \text{K}$)
• Critical Pressure ($P_c$): $73.0 \, \text{atm}$ (or approximately $7.39 \times 10^6 \, \text{Pa}$)

2. Conceptual Significance of the Critical Point

The critical point is a state of matter where the density of the gas phase equals the density of the liquid phase. The significance is categorized into two primary physical implications:

1. Disappearance of Phase Boundary: At temperatures above $T_c$, $CO_2$ exists as a supercritical fluid. In this state, it is impossible to liquefy the gas regardless of the amount of pressure applied [Per the Kinetic Molecular Theory of Gases].
2. Non-liquefaction Constraint: Below the critical temperature, $CO_2$ can be liquefied by applying sufficient pressure to move the state across the liquid-vapor equilibrium line. However, at $T > T_c$, the gas occupies a single, homogeneous state that possesses properties of both a gas (low viscosity) and a liquid (high solvent capability).

3. Visual Representation of the Phase Diagram

4. Practical Implications in Thermodynamics

Parameter	Physical Significance
$T > T_c$	The gas cannot be liquefied, regardless of compression.
$P > P_c$	The system behaves as a supercritical fluid with unique solvent properties.

Final Solution: The critical temperature of $CO_2$ is $31.1 \, ^\circ\text{C}$ and the critical pressure is $73.0 \, \text{atm}$. Their significance lies in the fact that above these values, $CO_2$ exists only as a supercritical fluid, and it is impossible to condense the gas into a liquid state by increasing pressure alone.