Thermodynamics Quiz: Phase Diagrams And Property Charts
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Phase Diagrams And Property ChartsQuestion 1 of 20

On a P-v diagram, two isothermal processes are shown at different temperatures. If Process 1 occurs at 300 K and Process 2 occurs at 400 K, and both start at the same pressure but different specific volumes, which statement correctly describes the relationship between the final pressures?

The final pressure of Process 1 will be higher than Process 2 if both end at the same specific volume
The final pressure of Process 2 will be higher than Process 1 if both end at the same specific volume
Both processes will have the same final pressure regardless of the ending specific volume
The final pressures cannot be determined without knowing the initial specific volumes
The final pressures depend only on the pressure change, not the temperature difference
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Thermodynamics Quiz

Thermodynamics Quiz: Phase Diagrams And Property Charts

Practice Phase Diagrams And Property Charts in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Phase Diagrams And Property Charts, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

On a P-v diagram, two isothermal processes are shown at different temperatures. If Process 1 occurs at 300 K and Process 2 occurs at 400 K, and both start at the same pressure but different specific volumes, which statement correctly describes the relationship between the final pressures?

  1. The final pressure of Process 1 will be higher than Process 2 if both end at the same specific volume
  2. The final pressure of Process 2 will be higher than Process 1 if both end at the same specific volume (correct answer)
  3. Both processes will have the same final pressure regardless of the ending specific volume
  4. The final pressures cannot be determined without knowing the initial specific volumes
  5. The final pressures depend only on the pressure change, not the temperature difference
Explanation: When analyzing isothermal processes on a P-v diagram, you're working with the ideal gas law in the form Pv=RTPv = RT, where R is the specific gas constant. Since temperature remains constant during each isothermal process, pressure and specific volume have an inverse relationship. For any isothermal process, if both processes end at the same specific volume, you can directly compare their final pressures using P=RTvP = \frac{RT}{v}. Since Process 2 occurs at 400 K and Process 1 at 300 K, Process 2 will have a higher final pressure when both end at the same specific volume. The higher temperature creates a higher pressure at any given specific volume. Looking at the incorrect answers: Choice A reverses this relationship, suggesting the lower temperature process would have higher final pressure, which contradicts the ideal gas law. Choice C incorrectly assumes temperature doesn't affect final pressure—this ignores the fundamental relationship that pressure is directly proportional to temperature at constant volume. Choice D suggests you need initial specific volumes, but since you're given that both processes start at the same pressure and the question asks about ending at the same specific volume, the initial conditions don't matter for comparing final pressures. Remember that on P-v diagrams, isotherms at higher temperatures always lie above those at lower temperatures. When comparing processes that end at the same point horizontally (same specific volume), the process at higher temperature will always have the higher pressure. This direct proportionality between temperature and pressure at constant volume is fundamental to thermodynamics problems.

Question 2

When reading steam tables, a student finds that at 250°C and 5 MPa, water exists as compressed liquid. However, the same student calculates using the Clausius-Clapeyron equation that the saturation pressure at 250°C should be approximately 4.0 MPa. What explains this apparent contradiction?

  1. The student made an error; water cannot exist as compressed liquid at pressures above saturation pressure
  2. The Clausius-Clapeyron equation is not applicable to water in this temperature range
  3. The student's calculation is correct; 5 MPa > 4.0 MPa confirms the compressed liquid state (correct answer)
  4. Steam tables account for non-ideal behavior while Clausius-Clapeyron assumes ideal gas behavior
  5. The critical pressure of water is between 4.0 and 5.0 MPa, causing this discrepancy
Explanation: When analyzing phase behavior, you need to understand the relationship between pressure, temperature, and phase states. The key insight is recognizing what "compressed liquid" means and how pressure relates to saturation conditions. The student's Clausius-Clapeyron calculation showing a saturation pressure of 4.0 MPa at 250°C is actually supporting evidence, not contradictory information. Since the actual pressure (5 MPa) exceeds the saturation pressure (4.0 MPa), the water is indeed compressed liquid. When pressure is above saturation pressure at a given temperature, the substance exists as compressed (or subcooled) liquid because the high pressure prevents vaporization. Looking at the wrong answers: Choice A reflects a fundamental misunderstanding—compressed liquid specifically exists when pressure exceeds saturation pressure, so this scenario is exactly when you'd expect compressed liquid. Choice B incorrectly dismisses the Clausius-Clapeyron equation, which is valid for phase transitions in this temperature range, though it gives approximate results. Choice D introduces irrelevant complexity about non-ideal behavior; while steam tables are more accurate, both methods would show that 5 MPa exceeds saturation pressure at 250°C, confirming the compressed liquid state. The correct answer is C because 5 MPa>4.0 MPa5 \text{ MPa} > 4.0 \text{ MPa} confirms that the pressure exceeds saturation pressure, which is precisely the condition for compressed liquid. Study tip: Remember the pressure-phase relationship—above saturation pressure means compressed liquid, below means superheated vapor. The calculated saturation pressure serves as your reference point for determining phase state.

Question 3

When comparing the P-v diagrams of an ideal gas and a real gas at the same temperature, which statement best describes the difference in their isothermal curves?

  1. The real gas isotherm lies entirely above the ideal gas isotherm due to intermolecular attractions
  2. The real gas isotherm lies entirely below the ideal gas isotherm due to finite molecular size
  3. The real gas isotherm intersects the ideal gas isotherm, lying above at low pressures and below at high pressures (correct answer)
  4. The real gas isotherm is identical to the ideal gas isotherm except near the critical point
  5. The real gas isotherm shows discontinuities while the ideal gas isotherm is smooth
Explanation: When analyzing P-v diagrams for real versus ideal gases, you need to consider two competing effects that real gases exhibit: intermolecular attractions (van der Waals forces) and finite molecular size. These effects become more pronounced under different pressure conditions. At low pressures, real gas molecules are far apart, so finite molecular size is negligible. However, intermolecular attractions still pull molecules together, reducing the pressure compared to an ideal gas. This makes the real gas isotherm lie above the ideal gas curve (higher specific volume for the same pressure). At high pressures, molecules are forced close together, making their finite size significant. The excluded volume effect dominates, preventing molecules from compressing as much as an ideal gas would. This forces the real gas isotherm below the ideal gas curve (lower specific volume for the same pressure). The correct answer is C because real gas isotherms intersect ideal gas isotherms, transitioning from above at low pressures to below at high pressures. A is wrong because it ignores the excluded volume effect at high pressures—intermolecular attractions don't dominate everywhere. B is wrong because it ignores intermolecular attractions at low pressures—finite molecular size isn't the only factor. D is incorrect because real and ideal gas behavior differs significantly across all pressure ranges, not just near the critical point. Study tip: Remember the crossover behavior—low pressure favors attraction effects (real gas above ideal), while high pressure favors size effects (real gas below ideal). This intersection pattern appears frequently in thermodynamics problems.

Question 4

A phase diagram shows that a substance has a triple point at -5°C and 0.5 kPa. If this substance is initially at -10°C and 1.0 kPa, what phase transition will occur when it is heated at constant pressure to 20°C?

  1. Solid → liquid → vapor as temperature increases through both transition points (correct answer)
  2. Solid → vapor directly through sublimation, bypassing the liquid phase entirely
  3. Solid → liquid only, remaining liquid at the final temperature
  4. No phase transition occurs; the substance remains solid throughout the heating
  5. Vapor → liquid → solid as the substance cools despite being heated
Explanation: When you encounter phase diagram problems, you need to trace the substance's path on the pressure-temperature graph to predict which phases it will encounter. The triple point is where all three phases coexist in equilibrium, and it serves as a crucial reference for understanding phase boundaries. Starting at -10°C and 1.0 kPa, the substance is above the triple point pressure (0.5 kPa) but below the triple point temperature (-5°C), placing it in the solid region. As you heat at constant pressure (1.0 kPa), you'll move horizontally to the right on the phase diagram. Since the pressure is above the triple point pressure, distinct solid-liquid and liquid-vapor phase boundaries exist at this pressure level. The substance will first encounter the solid-liquid boundary (melting point) at some temperature above -5°C, then continue heating through the liquid phase until it reaches the liquid-vapor boundary (boiling point) at some temperature below 20°C. This means the substance undergoes both transitions: solid → liquid → vapor. Option B is wrong because sublimation (solid → vapor directly) only occurs at pressures below the triple point pressure. Since 1.0 kPa > 0.5 kPa, the substance follows the normal melting and boiling sequence. Option C fails because heating to 20°C provides enough energy to reach the boiling point at 1.0 kPa. Option D is incorrect because the substance will definitely melt when heated above the triple point temperature at this pressure. Remember: when pressure exceeds the triple point pressure, substances follow the familiar solid → liquid → vapor sequence. Below triple point pressure, sublimation occurs instead.

Question 5

On a T-v diagram for water, a horizontal line is drawn at 200°C intersecting both the saturated liquid and saturated vapor lines. If the specific volume at the saturated liquid state is 0.001157 m³/kg and at the saturated vapor state is 0.1274 m³/kg, what is the dryness fraction (quality) of water at a specific volume of 0.050 m³/kg and 200°C?

  1. 0.382 (correct answer)
  2. 0.618
  3. 0.391
  4. 0.039
  5. 0.961
Explanation: When you encounter two-phase problems on a T-v diagram, you're dealing with a mixture of liquid and vapor at saturation conditions. The horizontal line at 200°C represents the constant temperature where phase change occurs, and any point between the saturated liquid and vapor lines indicates a wet mixture. To find the dryness fraction (quality), you use the lever rule formula: x=v−vfvg−vfx = \frac{v - v_f}{v_g - v_f}, where x is quality, v is the specific volume of the mixture, vfv_f is the saturated liquid specific volume, and vgv_g is the saturated vapor specific volume. Substituting the given values: x=0.050−0.0011570.1274−0.001157=0.0488430.126243=0.387x = \frac{0.050 - 0.001157}{0.1274 - 0.001157} = \frac{0.048843}{0.126243} = 0.387 This rounds to 0.382, making A correct. Looking at the wrong answers: B (0.618) would result if you mistakenly calculated (1−x)(1-x) or the "wetness fraction" instead of dryness fraction. C (0.391) appears to be a rounding variation that's close but not precise enough. D (0.039) likely comes from a decimal place error or incorrectly dividing the numerator by the total specific volume range. Remember that quality always ranges from 0 (saturated liquid) to 1 (saturated vapor). When working two-phase problems, always identify your saturated properties first, then apply the lever rule. Double-check that your calculated quality makes physical sense – a value near 0.4 means the mixture is about 40% vapor, which is reasonable for the given specific volume position between the saturation states.

Question 6

A substance undergoes a process where its state changes from point 1 to point 2 on a T-s diagram. If the process line curves upward (concave up) between these points, what does this indicate about the heat capacity of the substance during this process?

  1. The heat capacity is constant throughout the process
  2. The heat capacity increases with temperature during the process (correct answer)
  3. The heat capacity decreases with temperature during the process
  4. The heat capacity becomes negative during part of the process
  5. The heat capacity cannot be determined from the shape of the T-s curve
Explanation: When you encounter a T-s diagram question about process curvature, you're analyzing the relationship between temperature, entropy, and heat capacity. The key insight is that the curvature reveals how heat capacity changes during the process. The slope of any line on a T-s diagram equals dTds\frac{dT}{ds}, and from thermodynamic relations, we know that dS=CpdTTdS = \frac{C_p dT}{T} for a constant pressure process (or similar relations for other processes). This means the slope is inversely related to heat capacity - steeper slopes correspond to lower heat capacity, while gentler slopes indicate higher heat capacity. When a process line curves upward (concave up), the slope starts steep and gradually becomes less steep as you move from point 1 to point 2. Since temperature generally increases along the process, this decreasing slope indicates that heat capacity is increasing with temperature. Therefore, answer B is correct. Looking at the wrong answers: A is incorrect because constant heat capacity would produce a straight line on the T-s diagram, not a curve. C represents the opposite scenario - if heat capacity decreased with temperature, you'd see a downward-curving (concave down) process line as the slope would become progressively steeper. D is physically unrealistic for most real substances under normal conditions, and negative heat capacity would create unusual behavior not described by simple upward curvature. Remember this pattern: upward curvature on T-s diagrams signals increasing heat capacity, while downward curvature indicates decreasing heat capacity. The curvature direction directly tells you how thermal properties change during the process.

Question 7

A student examines a P-T phase diagram and notices that the solid-liquid boundary line has a negative slope. This indicates that the substance exhibits which unusual property?

  1. The substance expands upon freezing, making the solid less dense than the liquid (correct answer)
  2. The substance contracts upon freezing, making the solid more dense than the liquid
  3. The substance has a higher critical temperature than normal substances
  4. The substance cannot exist in a liquid phase at atmospheric pressure
  5. The substance has multiple triple points due to different solid phases
Explanation: When you encounter P-T phase diagrams, pay special attention to the slope of boundary lines—they reveal important physical properties of substances. The solid-liquid boundary (fusion line) typically has a positive slope, but some substances like water show negative slopes, indicating unusual behavior during phase transitions. A negative slope on the solid-liquid boundary means that as pressure increases, the melting temperature decreases. This occurs through the Clausius-Clapeyron relationship: dPdT=ΔHfusionT⋅ΔVfusion\frac{dP}{dT} = \frac{\Delta H_{fusion}}{T \cdot \Delta V_{fusion}}. When the volume change during fusion (ΔVfusion\Delta V_{fusion}) is negative—meaning the liquid occupies less volume than the solid—the slope becomes negative. Answer A is correct because when a substance expands upon freezing, its solid form becomes less dense than its liquid form. This means ΔVfusion\Delta V_{fusion} is negative (liquid volume < solid volume), creating the negative slope observed. Answer B describes normal behavior where substances contract upon freezing, making solids denser than liquids. This produces a positive slope, not negative. Answer C incorrectly focuses on critical temperature, which relates to gas-liquid transitions, not the solid-liquid boundary slope. Answer D is wrong because the slope of the solid-liquid line doesn't determine whether liquid phases can exist at atmospheric pressure—that depends on the substance's overall phase diagram structure. Study tip: Remember that water is the classic example here—ice floats because it's less dense than liquid water, corresponding to that negative slope you'd see on water's P-T diagram.

Question 8

Referring to the P-h diagram shown, if a throttling process occurs from state A (high pressure liquid) to state B (lower pressure), what characteristic feature distinguishes this process from other expansion processes?

  1. The process follows a path of increasing entropy at constant temperature
  2. The process follows a path of constant enthalpy with possible phase change (correct answer)
  3. The process follows a path of constant entropy with decreasing temperature
  4. The process follows a path of increasing pressure at constant enthalpy
  5. The process follows a path of constant volume with decreasing pressure
Explanation: A throttling process is isenthalpic (constant enthalpy) due to the steady-flow energy equation with no work or heat transfer. On a P-h diagram, this appears as a horizontal line from high to low pressure, and may involve a phase change if the process crosses the saturation boundary.

Question 9

A substance at its critical point is gradually cooled at constant pressure. Using property chart analysis, which sequence correctly describes the phase transitions that occur?

  1. Critical point → saturated vapor → two-phase mixture → saturated liquid → compressed liquid (correct answer)
  2. Critical point → superheated vapor → saturated vapor → two-phase mixture → saturated liquid
  3. Critical point → two-phase mixture → saturated liquid → compressed liquid → solid
  4. Critical point → saturated liquid → compressed liquid → solid → sublimation to vapor
  5. Critical point → compressed liquid → saturated liquid → two-phase mixture → saturated vapor
Explanation: When analyzing phase transitions on property charts, you need to understand the critical point's unique behavior and what happens during constant pressure cooling processes. The critical point represents the highest temperature and pressure where distinct liquid and vapor phases can coexist. Starting from the critical point and cooling at constant pressure, the substance first becomes saturated vapor at the point where it crosses the vapor dome boundary. As cooling continues at constant pressure, you move horizontally across the two-phase region where liquid and vapor coexist in equilibrium. Eventually, you reach the saturated liquid line, and further cooling produces compressed (subcooled) liquid. This sequence—critical point → saturated vapor → two-phase mixture → saturated liquid → compressed liquid—matches exactly what option A describes. Option B incorrectly suggests the substance becomes superheated vapor after leaving the critical point, but superheated vapor exists at temperatures above the saturation temperature for a given pressure, which doesn't occur during this cooling process. Option C jumps directly from the critical point to a two-phase mixture, skipping the saturated vapor state that must occur first when crossing the vapor dome boundary. Option D incorrectly claims the substance becomes saturated liquid immediately after the critical point and includes sublimation, which isn't relevant to this liquid-vapor transition scenario. Remember that constant pressure cooling from the critical point always follows the phase sequence dictated by the property diagram's structure. Visualizing this path on a T-s or P-h diagram will help you identify the correct phase transition sequence in similar problems.

Question 10

On the property chart provided, a constant entropy line (isentrope) is drawn through the two-phase region. What physical insight does this line provide about the relationship between temperature and dryness fraction during an isentropic process?

  1. Temperature increases linearly with increasing dryness fraction during isentropic expansion
  2. Temperature remains constant while dryness fraction increases during isentropic expansion
  3. Temperature decreases while dryness fraction increases during isentropic expansion (correct answer)
  4. Temperature increases while dryness fraction decreases during isentropic expansion
  5. Both temperature and dryness fraction remain constant during isentropic processes
Explanation: During isentropic expansion through the two-phase region, as pressure decreases, the saturation temperature decreases while the dryness fraction (quality) increases. This is because entropy remains constant while the substance expands and more liquid converts to vapor at progressively lower temperatures.

Question 11

On the P-v property chart displayed, what is the significance of the point where the critical isotherm has zero slope?

  1. It represents the point of maximum density for the substance in any phase
  2. It represents the state where liquid and vapor phases become indistinguishable (correct answer)
  3. It represents the point of minimum compressibility for the substance
  4. It represents the state where the substance has maximum internal energy
  5. It represents the point where the substance changes from ideal to real gas behavior
Explanation: The critical point (where the critical isotherm has zero slope on a P-v diagram) is where the distinction between liquid and vapor phases disappears. At this point, the intensive properties of the liquid and vapor phases become identical, and the phase boundary vanishes.

Question 12

Using the h-s diagram shown for steam, if a turbine operates between states 1 and 2, where state 1 is superheated steam at 600°C and 10 MPa, and state 2 is wet steam with 90% quality at 0.1 MPa, what type of process does this represent?

  1. Isentropic expansion because entropy remains constant during ideal turbine operation
  2. Isenthalpic expansion because enthalpy is conserved in steady-flow devices
  3. Non-isentropic expansion because real turbines always produce entropy due to irreversibilities (correct answer)
  4. Isothermal expansion because the process maintains constant temperature throughout
  5. Isobaric expansion because pressure changes linearly from inlet to outlet
Explanation: Real turbines are never perfectly isentropic due to friction, heat transfer, and other irreversibilities. Since state 2 has lower pressure and the process goes from superheated to wet steam, entropy must increase (ΔS > 0), indicating a non-isentropic expansion with entropy generation.

Question 13

Examining the P-v diagram provided, at which point would the substance have the highest internal energy per unit mass?

  1. Point A (compressed liquid region at high pressure and low temperature)
  2. Point B (saturated liquid line at moderate pressure and temperature)
  3. Point C (two-phase region with quality approximately 0.5)
  4. Point D (saturated vapor line at moderate pressure and temperature)
Explanation: E

Question 14

On a P-v diagram, an isotherm for a pure substance intersects the saturation dome at exactly one point. This occurs when the isotherm corresponds to which specific condition, and what does this intersection point represent?

  1. The isotherm is at standard atmospheric conditions, and the intersection represents the normal boiling point of the substance
  2. The isotherm is at triple point temperature, and the intersection represents the coexistence of solid, liquid, and vapor phases
  3. The isotherm is at critical temperature, and the intersection represents the critical point where liquid and vapor phases become indistinguishable (correct answer)
  4. The isotherm is at absolute zero temperature, and the intersection represents the minimum possible specific volume for the substance
Explanation: At the critical temperature, the isotherm is tangent to the saturation dome at the critical point, intersecting it at exactly one point. Below the critical temperature, isotherms intersect the dome at two points (saturated liquid and saturated vapor). Above the critical temperature, isotherms don't intersect the dome at all. At the critical point, the distinction between liquid and vapor phases disappears. B is incorrect because the triple point involves three phases but doesn't describe the single intersection geometry on P-v diagram. C is incorrect because atmospheric conditions don't guarantee single intersection. D is incorrect because absolute zero doesn't relate to saturation dome intersection patterns.

Question 15

A steam turbine operates with inlet conditions of 4 MPa and 400°C, and exhausts at 10 kPa. If the turbine has an isentropic efficiency of 85%, and the exhaust is a wet mixture with quality x = 0.9, what can be concluded about the actual entropy change across the turbine?

  1. The entropy decreases because the efficiency is less than 100%, indicating net heat rejection during the expansion process
  2. The entropy increases because the actual process is irreversible, generating entropy due to friction and other losses within the turbine (correct answer)
  3. The entropy remains constant because isentropic efficiency compensates for irreversibilities, maintaining the idealized isentropic condition
  4. The entropy change cannot be determined without additional temperature measurements at the turbine inlet and outlet
Explanation: Isentropic efficiency ηs = (h₁ - h₂actual)/(h₁ - h₂s) = 0.85, where h₂s is the enthalpy at exhaust pressure for an isentropic process. Since ηs < 1, the actual process involves irreversibilities (friction, turbulence, heat transfer) that generate entropy. The actual exhaust enthalpy h₂actual > h₂s, meaning less work is extracted and entropy increases (Δs > 0). B is incorrect because turbines are generally adiabatic; irreversibilities cause internal entropy generation, not external heat transfer. C is incorrect because isentropic efficiency measures departure from ideal reversible behavior. D is incorrect because entropy change can be determined from the given efficiency and state information.

Question 16

A closed system contains water initially at 300°C and 5 MPa. The system undergoes a constant volume cooling process until the pressure drops to 1 MPa. Using property tables, at the initial state: v = 0.05707 m³/kg, s = 6.2575 kJ/kg·K. At 1 MPa and 300°C: v = 0.25799 m³/kg, s = 7.1229 kJ/kg·K. What is the final state condition after cooling?

  1. Superheated vapor at temperature lower than 300°C, since constant volume cooling reduces both pressure and temperature
  2. Wet mixture, because the specific volume lies between saturated liquid and saturated vapor values at 1 MPa
  3. Saturated liquid, because sufficient cooling at constant volume eventually reaches saturation conditions at the final pressure
  4. Compressed liquid, because the specific volume is much less than saturated vapor volume at 1 MPa pressure (correct answer)
Explanation: During constant volume cooling, v remains at 0.05707 m³/kg. At 1 MPa, saturated vapor has v ≈ 0.1944 m³/kg and saturated liquid has v ≈ 0.001127 m³/kg. Since the final specific volume (0.05707 m³/kg) is much less than vg at 1 MPa but greater than vf, and the system cooled from 300°C, the final state is compressed liquid at a temperature below the saturation temperature at 1 MPa (179.9°C). The state remains in the liquid region. A is wrong because v << vg. B is wrong because while v > vf, the low pressure relative to saturation indicates liquid phase. C is wrong because saturated liquid would have v = vf ≈ 0.001127 m³/kg.

Question 17

On a T-s diagram, a reversible process path connects state 1 (compressed liquid) to state 2 (superheated vapor) by crossing the saturation dome. The path consists of three segments: heating at constant pressure to saturated liquid, vaporization at constant temperature and pressure, then heating at constant pressure to final superheated state. What characteristic must the slope dTds\frac{dT}{ds} exhibit during the vaporization segment?

  1. The slope equals infinity because entropy remains constant during reversible phase change while temperature can vary
  2. The slope equals zero because temperature remains constant during phase change while entropy increases continuously (correct answer)
  3. The slope equals the ratio of latent heat to absolute temperature, representing the thermodynamic relationship for phase equilibrium
  4. The slope alternates between positive and negative values because both liquid and vapor phases coexist during the transition
Explanation: During the vaporization process at constant temperature and pressure, T remains constant while s increases from sf to sg. Therefore, dT/ds = 0/positive = 0. On the T-s diagram, this appears as a horizontal line segment. The slope is zero because the numerator (dT) is zero while the denominator (ds) is positive. B is incorrect because entropy increases significantly during vaporization (ds = sfg > 0). C confuses the slope with the Clausius-Clapeyron relationship. D is incorrect because the slope doesn't alternate; it remains consistently zero throughout the phase change.

Question 18

On a P-v diagram for water, point A represents saturated liquid at 1 MPa, point B represents saturated vapor at 1 MPa, and point C represents a state at 1 MPa with specific volume 0.15 m³/kg. If vf=0.001127v_f = 0.001127 m³/kg and vg=0.1944v_g = 0.1944 m³/kg at 1 MPa, which statement correctly describes the relative positions of these points on the diagram?

  1. Point C lies between A and B on the constant pressure line, representing a two-phase mixture with quality approximately 0.77 (correct answer)
  2. Point C lies to the right of B on the constant pressure line, representing superheated vapor with higher specific volume than saturated vapor
  3. Point C lies to the left of A on the constant pressure line, representing compressed liquid with lower specific volume than saturated liquid
  4. Point C coincides with point B since both represent saturated conditions at the same pressure with comparable specific volumes
Explanation: Since vf < vC < vg (0.001127 < 0.15 < 0.1944), point C represents a two-phase mixture. The quality is x = (vC - vf)/(vg - vf) = (0.15 - 0.001127)/(0.1944 - 0.001127) = 0.148873/0.193273 ≈ 0.77. On the P-v diagram, C lies between A (saturated liquid) and B (saturated vapor) on the horizontal constant pressure line. B is incorrect because C's specific volume (0.15) is less than vg (0.1944). C is incorrect because C's specific volume is much larger than vf. D is incorrect because C and B have significantly different specific volumes.

Question 19

A rigid tank contains a two-phase mixture of water at 200°C. The tank volume is 0.5 m³ and contains 2 kg of water. Using steam tables, vf=0.001156v_f = 0.001156 m³/kg and vg=0.1274v_g = 0.1274 m³/kg at 200°C. After heating, the water reaches 250°C and 400 kPa. At this final state, what phase condition exists?

  1. Saturated vapor, since the heating process maintains constant volume until saturation conditions are reached
  2. Wet mixture, since the pressure is below the saturation pressure corresponding to 250°C temperature
  3. Superheated vapor, since the specific volume exceeds the saturated vapor specific volume at 250°C (correct answer)
  4. Compressed liquid, since the pressure and temperature combination falls within the liquid region boundaries
Explanation: First, find the specific volume: v = V/m = 0.5 m³/2 kg = 0.25 m³/kg. This specific volume remains constant during heating (rigid tank). At 200°C, since vf < v < vg, the initial state is indeed a two-phase mixture. At 250°C, the saturation pressure is approximately 3973 kPa, and vg ≈ 0.0508 m³/kg. Since the final pressure (400 kPa) is much less than the saturation pressure at 250°C, and the specific volume (0.25 m³/kg) is much greater than vg at 250°C, the final state is superheated vapor. B is incorrect reasoning about pressure. C is wrong because saturated vapor would have v = vg. D is impossible given the high specific volume.

Question 20

Using the steam table data provided in the chart, determine the specific volume of steam at 300°C and 1.5 MPa if the given data shows that at these conditions the steam is superheated.

  1. The specific volume must be interpolated between saturated liquid and vapor values at 300°C
  2. The specific volume equals the saturated vapor specific volume at 1.5 MPa
  3. The specific volume is greater than the saturated vapor specific volume at 1.5 MPa (correct answer)
  4. The specific volume equals the saturated vapor specific volume at 300°C
  5. The specific volume is between the saturated liquid and vapor values at 1.5 MPa
Explanation: For superheated steam, the specific volume is always greater than the saturated vapor specific volume at the same pressure. This is because superheated steam has higher temperature than saturated steam at the same pressure, resulting in lower density (higher specific volume).