Phase Change of a Pure Substance MCQ Quiz - Objective Question with Answer for Phase Change of a Pure Substance - Download Free PDF

Concept:

For a mixture of saturated liquid and vapor, the specific volume is calculated using the dryness fraction formula:

\( v = v_f + x(v_g - v_f) \)

Then, total volume is found using: \( V = m \cdot v \)

Given:

Mass, m = 4 kg

Dryness fraction, x = 0.25

Specific volume of saturated liquid, \(v_f = 0.0075~m^3/kg\)

Specific volume of saturated vapor, \(v_g = 0.1~m^3/kg\)

Calculation:

Specific volume of mixture: \( v = v_f + x(v_g - v_f) = 0.0075 + 0.25(0.1 - 0.0075) = 0.030625~m^3/kg \)

Total volume: \( V = m \cdot v = 4 \cdot 0.030625 = 0.1225~m^3 = 122.5~\text{liters} \)

Concept:

Saturation temperature:

• The saturation temperature is the temperature for a corresponding saturation pressure at which a liquid boils into its vapour phase.
• According to Mayer’s equation, PV = mRT, pressure is directly proportional to temperature.
• In microscopic point of view if pressure increases, the intermolecular forces increases and so the temperature. Therefore for each pressure steam has a specific saturation temperature.

Explanation:

Critical point:

• The point at which the saturated liquid line and saturated vapour line of a pure substance meet is called the critical point.
• At a critical point, the liquid is directly converted into vapour without having a two-phase transition. So, enthalpy of vaporization at a critical point is zero i.e. At the critical point, saturated liquid and saturated vapour phases are identical.
• The figure below represents the P-V diagram for a pure substance (water).

Below critical point, latent heat is required to convert water into vapour but at this point, there is no need for latent heat i.e. at this point latent heat equals zero.

For water critical point parameters are:

Pressure (Pc) = 22 MPa, Temperature (Tc) = 374.15°C

latent heat of vaporisation refers to the amount of heat energy required to transform a substance from the liquid phase into the vapor phase at a constant temperature and pressure. With that definition, let's evaluate all the options:

• Heat required for complete conversion of saturated liquid into dry saturated vapour: This statement essentially describes latent heat of vaporization. It refers to the necessary heat input to convert a substance from the liquid state (saturated liquid) into the gaseous state (dry saturated vapour), without a change in temperature.
• Heat required for complete conversion of ice into water: This does not describe latent heat of vaporisation. Instead, it illustrates the concept of the latent heat of fusion, which is the amount of heat required to convert a solid (in this case, ice) into a liquid (water) at its melting point.
• Heat added at a constant temperature of 100°C to convert water into steam: This statement also describes the latent heat of vaporisation. At a constant temperature of 100°C (at standard atmospheric pressure), water changes from its liquid phase to its vapor phase -- a process that calls for input of heat energy, i.e., the latent heat of vaporisation.
• Sum of internal latent heat and external work of evaporation: This is a more systematic look at the latent heat of vaporisation. The energy needed for the phase transformation from liquid to vapor can be divided into two components: the internal latent heat, which is the heat energy used to break the intermolecular forces within the liquid state; and the external work of evaporation, which is the energy used to expel the vapor created and make space for it (essentially, the work done against the atmospheric pressure).

In conclusion, the statement that does not correlate correctly with latent heat of vaporisation is option 2, "heat required for complete conversion of ice into water," as that defines latent heat of fusion, not vaporisation.

The correct answer is Solidifies directly

Concept:-

• Phase Diagram: A phase diagram is a graphical representation of pressure-temperature conditions under which various phases of a substance are in equilibrium. It delineates the relative areas of stability for gas, liquid, and solid phases, and shows the points of phase transitions, such as melting (solid to liquid), boiling (liquid to gas), sublimation (solid to gas), deposition/desublimation (gas to solid), etc.
• Triple Point: The triple point of a substance is a particular condition (specific pressure and temperature) at which all three phases - solid, liquid, and gas - coexist in equilibrium. Every substance has a unique triple point. For example, the triple point of water is exactly 273.16 Kelvin (0.01 Celsius) and 611.657 pascals of pressure.
• Deposition/Desublimation: This is the phase transition from the gas phase directly to the solid phase without passing through an intermediate liquid phase. It normally occurs at conditions of low temperature and pressure, such as under conditions less than the triple point in a phase diagram.
• Sublimation: It is the direct transition of a substance from the solid to the gaseous state, bypassing the liquid state. Dry ice, or solid carbon dioxide, is a common example of a substance that undergoes sublimation at room temperature and pressure.​​​​​

Explanation:-

• Substances can exist in three states: solid, liquid, and gas. These are determined by the intensity of molecular interactions and the thermal energy of the molecules, both of which are influenced by external conditions such as temperature and pressure.
• A phase diagram is used to represent the states of matter of a substance under different temperature and pressure conditions. This diagram highlights the equilibrium boundaries between the different states of matter.
• One important feature of a phase diagram is the triple point, which is the specific set of conditions (at a particular temperature and pressure) where all three states of matter (solid, liquid, and gas) are in equilibrium - meaning they co-exist without changing into each other.
• Now, when the pressure conditions are less than the triple point pressure, interesting things happen. Under this condition, the liquid state of the substance does not exist. This means there's no equilibrium boundary between the gas and liquid phases, which therefore makes the gas unable to transition into a liquid even if we cool it down.
• Instead, when we cool the gas under this pressure condition, it transforms directly into a solid. This process is known as desublimation or deposition, which is the transition from the gaseous state directly to the solid state, skipping the liquid state.

conclusion:-

So ,the vapour of a pure substance, when cooled under a pressure less than its triple-piont pressure Solidifies directly

\(v = {v_f} + x\;\left( {{v_g} - {v_f}} \right)\)

v = V/m = 0.5 m³/kg

\(0.5 = 0.00106 + x\;\left( {0.8908 - 0.00106} \right)\)

\(x = 0.56\)

\(0 \le x \le 1\)

x = 0, Saturated liquid

x = 1, Saturated vapor

As x value lies between 0 and 1, therefore it is a mixture of saturated liquid and saturated vapor.

Explanation:

Since, the properties like internal energy, enthalpy and entropy of a system cannot be directly measured. They are related to change in the energy of the system.

Hence, we can determine Δu, Δh, Δs but not the absolute values of these properties.

⇒ Therefore, It is necessary to choose a reference state to which, these properties are arbitrary assigned some numerical values.  

So for water, the triple point (T = 0.01°C & P = 611 Pa) is selected as reference a state, where the “Internal energy” (u) and “Entropy” (s) of saturated liquid are assigned a zero value.

#Note: h = u + Pv

At triple point, u = 0, but p × ν ≠ 0

Explanation

The phase diagram of water is given below.

A substance sublimates on heating If it is kept below its triple point pressure. Hence the lowest pressure at which water can exist in liquid phase in stable equilibrium is triple point pressure.

The triple point properties of water are given below.

Triple point pressure, P­_tp = 4.58 mm of Hg = 0.611 kPa

Triple point temperature, T_tp = 273.16 K = 0.01°C

Concept:

Joule – Thomson coefficient:- When the gas in steady flow passes through a constriction, e.g. in an orifice or valve, it normally experiences a change in temperature. From the first law of thermodynamics, such a process is isenthalpic and one can usefully define a Joule – Thomson coefficient as

\(\mu = {\left( {\frac{{\partial T}}{{\partial P}}} \right)_H}\)

As a measure of the change in temperature which results from a drop in pressure across the construction.

• For an ideal gas, μ = 0, because ideal gases neither warm not cool upon being expanded at constant enthalpy.
• If μ is +ve, then the temperature will fall during throttling.
• If μ is -ve, then the temperature will rise during throttling.

Explanation:

Critical point:

• The point at which the saturated liquid line and saturated vapour line of a pure substance meet is called the critical point.
• At a critical point, the liquid is directly converted into vapour without having a two-phase transition. So, enthalpy of vaporization at a critical point is zero i.e. At the critical point, saturated liquid and saturated vapour phases are identical.
• The figure below represents the P-V diagram for a pure substance (water).

Below critical point, latent heat is required to convert water into vapour but at this point, there is no need for latent heat i.e. at this point latent heat equals zero.

For water critical point parameters are:

Pressure (Pc) = 22 MPa, Temperature (Tc) = 374.15°C

Explanation:

Critical point:

• The point at which the saturated liquid line and saturated vapour line of a pure substance meet is called the critical point.
• At a critical point, the liquid is directly converted into vapour without having a two-phase transition. So, enthalpy of vaporization at a critical point is zero i.e. At the critical point, saturated liquid and saturated vapour phases are identical.
• The figure below represents the P-V diagram for a pure substance (water).

Below critical point, latent heat is required to convert water into vapour but at this point, there is no need for latent heat i.e. at this point latent heat equals zero.

For water critical point parameters are:

Pressure (P_c) = 22 MPa, Temperature (T_c) = 374°C

Triple Point

• The Triple point is a point on the P-T diagram where all the three phases solid, liquid and gases exist in equilibrium.
• At a pressure below the triple point line, the substance cannot exist in the liquid phase and the substance when heated, transforms from solid to vapour by absorbing the latent heat of sublimation from the surroundings.
• The triple point is merely the point of intersection of the sublimation and vaporization curves.
• It has been found that on a ‘P-T’ diagram the triple point is represented by a point and on a ‘P-V’ diagram it is a line, and on a ‘U-V’ diagram it is a triangle. In the case of ordinary water, the triple point is at a pressure of 4.58 mm Hg and a temperature of 0.01°C.

Explanation:

At a critical point, the liquid is directly converted into vapour without having a two-phase transition. So, the enthalpy of vaporization at a critical point is zero. The figure below represents the P-T diagram for a pure substance.

Because at critical point liquid directly convents into the vapour phase. Ice is converted directly into vapour if it is heated at constant pressure which is less than triple point pressure. Hence statement 3 is incorrect.

Concept:

Specific Latent Heat (L):

It is the heat required or liberated to change the phase of a substance at constant temperature. Unit - kJ/kg

Example- Latent heat of fusion or melting, Latent heat of Vaporization

Q = Mass × Specific Latent heat

Q = mL

Calculation:

Given:

Q = 10 kW

It is given that 10 kW heat is removed till the molten metal is completely solidified i.e. it is the latent heat which is required for its phase change i.e. temperature remains constant.

\(Q = \frac{{mL}}{t} \Rightarrow 10 = \frac{{2 × L}}{{\left( {20 - 10} \right)}}\)

L → latent heat of fusion L = 50 kJ/kg.

Note: Here Q is given into kW, so to convert kW into J, we need to divide it by time which is already given into the graph.

Explanation:

In the question, critical specific volume V_c = 0.003155 m³/kg is given.

Mixture specific volume v = \(\frac{{0.025\;}}{{10}}\) = 0.0025 m³/kg.

As V < V_c and due to the water mixture's heating temperature will rise and hence pressure will rise.

As pressure rises, the mixture's state will change from 1 to 2 as the rigid tank is given in the question so the mixture's total volume will be constant.

At state 1 quality is X₁ and at state 2 quality of the mixture will be X₂.

From figure X₂ < X₁. It means the liquid at state 2 will be more. Since liquid at state 2 is more therefore level of liquid inside the tank will rise.

Explanation:

Triple point:

• The temperature and pressure at which a substance can exist in equilibrium in the liquid, solid, and gaseous states.
• The triple point of pure water is at 0.01 °C (273.16K, 32.01°F) and 4.58 mm (611.2Pa) of mercury and is used to calibrate thermometers and compare critical points. 

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Important Points

• The matter solidifies on the left side of the triple point, not at the triple point.
• The matter liquifies on the upper side of the triple point, not at the triple point.
• The matter sublimates on the right side of the triple point, not at the triple point.

Additional Information
Critical point:

• The limiting equilibrium state of a two-phase system, in which the properties of both coexisting phases become identical.
• In water, the critical point occurs at 647.096 K (373.946 C, 705.103 F) and 22.064 megapascals (3,200.1 psi, 217.75 atm).