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    Laws of Thermodynamics

    Flashcards for topic Laws of Thermodynamics

    Intermediate42 cardsphysics

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    Card 1

    Front

    How do you calculate the work done by a gas in a thermodynamic process involving changes in pressure and volume?

    Back

    The work done by a gas in a thermodynamic process is:

    W=∫V1V2p dVW = \int_{V_1}^{V_2} p\,dVW=∫V1​V2​​pdV

    Key principles:

    • Work equals the area under the curve in a p-V representation
    • For constant pressure processes: W=p(V2−V1)W = p(V_2 - V_1)W=p(V2​−V1​)
    • Positive work occurs when a system expands (gas does work on surroundings)
    • Negative work occurs when a system is compressed (surroundings do work on gas)

    Example: For a gas expanding at constant pressure of 120 kPa from 200 cc to 450 cc, the work done is: W=(120×103 Pa)(450−200)×10−6 m3=30 JW = (120 \times 10^3 \text{ Pa})(450 - 200) \times 10^{-6} \text{ m}^3 = 30 \text{ J}W=(120×103 Pa)(450−200)×10−6 m3=30 J

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    Card 2

    Front

    State the First Law of Thermodynamics mathematically and explain its physical meaning.

    Back

    The First Law of Thermodynamics is expressed as: ΔU=ΔQ−ΔW\Delta U = \Delta Q - \Delta WΔU=ΔQ−ΔW

    Where:

    • ΔU\Delta UΔU = change in internal energy of the system
    • ΔQ\Delta QΔQ = heat added to the system
    • ΔW\Delta WΔW = work done by the system

    Physical meaning:

    • Represents conservation of energy
    • Internal energy changes only through heat transfer or work
    • A system's internal energy increases when heat is added (+ΔQ+\Delta Q+ΔQ) or work is done on it (−ΔW-\Delta W−ΔW)
    • When a system returns to its initial state after a cyclic process, ΔU=0\Delta U = 0ΔU=0

    Note: When work is done by the system, ΔW\Delta WΔW is positive; when work is done on the system, ΔW\Delta WΔW is negative.

    Card 3

    Front

    For a quasi-static process, what is the mathematical relationship for calculating work done by a gas? Show how this differs for specific process types and how it can be represented graphically.

    Back

    Work done by a gas in a quasi-static process: W=∫V1V2p dVW = \int_{V_1}^{V_2} p\,dVW=∫V1​V2​​pdV

    Specific process types:

    1. Isothermal (constant T): W=nRTln⁡V2V1W = nRT\ln\frac{V_2}{V_1}W=nRTlnV1​V2​​

    2. Isobaric (constant p): W=p(V2−V1)W = p(V_2-V_1)W=p(V2​−V1​)

    3. Isochoric (constant V): W=0W = 0W=0 (no work done)

    4. Adiabatic (no heat transfer): W=p1V1−p2V2γ−1W = \frac{p_1V_1-p_2V_2}{\gamma-1}W=γ−1p1​V1​−p2​V2​​ where γ=CpCv\gamma = \frac{C_p}{C_v}γ=Cv​Cp​​

    Graphical representation:

    • Work equals the area under the p-V curve between initial and final volumes
    • Positive work (system expands): area under curve when moving right on p-V diagram
    • Negative work (system compressed): area under curve when moving left
    • Path-dependent: different paths between same endpoints yield different work values

    In cyclic processes, work equals the area enclosed by the complete p-V curve.

    Card 4

    Front

    How does heat transfer occur between gas molecules and container walls when the walls are at a higher temperature than the gas?

    Back

    Heat transfer occurs through molecular collisions:

    1. Gas molecules strike the higher-temperature wall
    2. During collision, energy transfers from higher kinetic energy wall molecules to lower kinetic energy gas molecules
    3. These energized gas molecules share kinetic energy with other gas molecules through subsequent collisions
    4. This increases the total internal energy of the gas system
    5. The process continues until thermal equilibrium is reached

    Example: Air in a metal container placed on a hot surface receives energy through this collision-based transfer mechanism, gradually increasing the overall gas temperature.

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    Card 5

    Front

    In a molecular view of thermodynamics, how does a temperature difference (T₂ > T₁) between container walls and gas affect the velocities of gas molecules after collision?

    Back

    When T₂ > T₁ (wall temperature exceeds gas temperature):

    • Before collision: Gas molecules have average velocity corresponding to temperature T₁
    • During collision: Gas molecules rebound with increased velocity
    • Mechanism: Molecules in hotter walls have higher average kinetic energy, which transfers partially to gas molecules during elastic collisions
    • Result: The post-collision velocity (v₂) of a gas molecule is greater than its pre-collision velocity (v₁)
    • System effect: The average kinetic energy of all gas molecules increases, raising the gas temperature

    This explains thermodynamic equilibration at the molecular level - individual molecular collisions drive the macroscopic process of heat transfer until the gas and wall reach the same temperature.

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    Card 6

    Front

    What is the total work done in a three-step thermodynamic cycle where:

    1. Gas expands isothermally at 120 kPa from 200 cc to 450 cc
    2. Volume remains constant while pressure increases to 200 kPa
    3. Gas is compressed at varying pressure back to initial state

    Calculate the net work and explain its significance.

    Back

    The total work done in this cycle:

    Step 1 (a→b): Wab=pΔV=(120 kPa)(250 cc)=30 JW_{ab} = p\Delta V = (120 \text{ kPa})(250 \text{ cc}) = 30 \text{ J}Wab​=pΔV=(120 kPa)(250 cc)=30 J (positive)

    Step 2 (b→c): Wbc=0W_{bc} = 0Wbc​=0 (zero work as volume is constant)

    Step 3 (c→a): Wca=−40 JW_{ca} = -40 \text{ J}Wca​=−40 J (negative, calculated as area under curve)

    Net work: Wnet=30 J+0 J+(−40 J)=−10 JW_{net} = 30 \text{ J} + 0 \text{ J} + (-40 \text{ J}) = -10 \text{ J}Wnet​=30 J+0 J+(−40 J)=−10 J

    Significance:

    • Negative net work means the cycle operates as a refrigerator/heat pump (work is done ON the system)
    • The absolute value of net work equals the area enclosed by the cycle
    • This represents the energy converted between work and heat during the complete cycle
    • According to the First Law, this difference must be accounted for by heat transfer
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    Card 7

    Front

    How does a piston-cylinder system demonstrate the thermodynamic concept of PV work, and what sign convention applies?

    Back

    A piston-cylinder system demonstrates PV work through:

    1. Mechanism: Gas molecules collide with the piston surface, exerting force over a distance
    2. Energy transfer: Internal energy of the gas is converted to mechanical work when the piston moves

    Sign convention:

    • When gas expands (ΔV>0\Delta V > 0ΔV>0): ΔW>0\Delta W > 0ΔW>0 (positive work, energy leaves the system)
    • When gas compresses (ΔV<0\Delta V < 0ΔV<0): ΔW<0\Delta W < 0ΔW<0 (negative work, energy enters the system)

    Application in First Law: ΔU=Q−W\Delta U = Q - WΔU=Q−W

    • During expansion, internal energy decreases as work is done by the system
    • During compression, internal energy increases as work is done on the system

    Real-world example: In an internal combustion engine, expanding gases from fuel combustion push a piston, converting thermal energy to mechanical energy that drives the vehicle.

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    Card 8

    Front

    What is the fundamental difference between how work is calculated in thermodynamics versus mechanics, and why is the path-dependence of work significant?

    Back

    In thermodynamics, work is path-dependent, unlike mechanical work which depends only on initial and final positions.

    Key differences:

    • Mechanical work: W=F⃗⋅d⃗W = \vec{F} \cdot \vec{d}W=F⋅d (depends only on net displacement)
    • Thermodynamic work: W=∫V1V2p dVW = \int_{V_1}^{V_2} p \, dVW=∫V1​V2​​pdV (depends on specific path taken)

    Path-dependence significance:

    • Different processes between the same initial and final states yield different work values
    • The area under different p-V curves connecting the same endpoints differs
    • This allows for designing cyclic processes (heat engines) that produce net work
    • It's why you can extract different amounts of work from a system by choosing different expansion paths

    Example: More work is done by a gas expanding isothermally than by the same gas expanding adiabatically between identical volume states.

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    Card 9

    Front

    What is the purpose of the eccentric mechanism in a steam engine and how does it synchronize the D-valve with the piston movement?

    Back

    The eccentric mechanism in a steam engine:

    1. Primary purpose: Converts the rotary motion of the crankshaft into the linear motion needed to operate the D-valve

    2. Components:

      • Eccentric: An off-center disc fixed to the main shaft
      • Eccentric-rod: Connects the eccentric to the valve-rod
      • Crosshead (C₂): Provides a hinge point for the valve-rod
    3. Synchronization function:

      • Creates a 90° phase difference between piston and valve movements
      • When piston is at top/bottom positions, valve is at mid-stroke
      • When piston is at mid-stroke, valve is at extreme positions
    4. Operation sequence:

      • As crankshaft rotates, eccentric rotates with it
      • Eccentric's offset position creates reciprocating motion in the eccentric-rod
      • This motion transfers to the valve-rod, moving the D-valve up and down
      • The D-valve position is automatically synchronized with piston position, ensuring steam enters the cylinder at precisely the right moment

    This timing mechanism creates the self-regulating cycle essential for continuous engine operation without requiring external valve adjustment.

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    Card 10

    Front

    How does the Clausius statement of the Second Law of Thermodynamics specifically relate to refrigeration processes? Why is this principle significant?

    Back

    The Clausius statement of the Second Law states:

    • "It is impossible to design a refrigerator which works in a cyclic process and whose only result is to transfer heat from a body to a hotter body."

    Key implications:

    • Heat naturally flows from hot to cold; reversing this direction requires work input
    • A refrigerator/heat pump must consume energy (W) to move heat (Q₂) from a cold region to a hot region
    • The total heat delivered to the hot reservoir (Q₁) equals Q₂ + W
    • For a Carnot (ideal) refrigerator: Q₁/Q₂ = T₁/T₂

    Significance:

    • Sets theoretical efficiency limits for cooling systems
    • Explains why refrigerators and air conditioners require electricity
    • Establishes that perpetual cooling machines are impossible
    • Provides the foundation for calculating the minimum energy requirements of cooling processes
    • Demonstrates the fundamental thermodynamic asymmetry in nature
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