Thermodynamics: Energy and Systems

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Questions and Answers

A rigid container holds an ideal gas. If heat is added to the gas, which statement is correct regarding the process?

  • The process is isothermal, and the pressure increases.
  • The process is isochoric, and the pressure increases. (correct)
  • The process is adiabatic, and the temperature remains constant.
  • The process is isobaric, and the volume increases.

During an adiabatic compression of an ideal gas, what happens to the gas's temperature and internal energy?

  • Temperature decreases, internal energy remains constant.
  • Temperature increases, internal energy increases. (correct)
  • Temperature remains constant, internal energy increases.
  • Temperature increases, internal energy decreases.

In a thermodynamic process, 500 J of heat is added to a system, and the system does 300 J of work. What is the change in internal energy of the system?

  • -200 J
  • -800 J
  • 800 J
  • 200 J (correct)

Which of the following statements accurately describes an isolated system?

<p>Neither energy nor matter can enter or leave. (A)</p> Signup and view all the answers

For an isothermal process, which of the following statements is always true?

<p>The change in internal energy is zero. (B)</p> Signup and view all the answers

An ideal gas expands isothermally and its volume doubles. If the initial pressure was 2 atm, what is the final pressure?

<p>1 atm (A)</p> Signup and view all the answers

A gas undergoes an isobaric process at a pressure of $2 × 10^5$ Pa. If the volume increases from $1 m^3$ to $3 m^3$, how much work is done by the gas?

<p>$4 × 10^5 J$ (B)</p> Signup and view all the answers

For a cyclic process, what must be true about the change in internal energy (ΔU) and the net heat flow (Q)?

<p>ΔU = 0 and Q = W (D)</p> Signup and view all the answers

Which thermodynamic process is characterized by the relationship $P_1V_1^γ = P_2V_2^γ$, where γ is the heat capacity ratio?

<p>Adiabatic (D)</p> Signup and view all the answers

A monoatomic ideal gas undergoes an adiabatic expansion. If the initial temperature is 300 K and the volume doubles, what is the final temperature? (Assume γ = 5/3)

<p>189 K (D)</p> Signup and view all the answers

Flashcards

Closed System

Energy can enter/leave, but not matter.

Isolated System

Neither energy nor matter can enter or leave.

Open System

Both matter and energy can enter or leave.

Isochoric Process

Constant volume, work done is zero.

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Isobaric Process

Constant pressure.

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Isothermal Process

Constant temperature, internal energy is zero.

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Adiabatic Process

No heat exchange.

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Cyclic Process

Returns a system to its initial state.

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Gamma (γ)

Ratio of molar heat capacities (Cp / Cv).

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Ideal Gas Law

PV = nRT

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Study Notes

Thermodynamics

  • Thermodynamics explores the relationship among heat, mechanical work, and internal energy.
  • The first law of thermodynamics: the change in internal energy (ΔU) equals the heat added (Q) minus the work done (W), expressed as ΔU = Q - W.

Internal Energy (U)

  • Internal energy (U) is directly proportional to temperature.
  • Heat flowing into the system results in a positive Q, increasing internal energy.
  • Heat flowing out of the system results in a negative Q, decreasing internal energy.

Work (W)

  • Work done by the system is positive, which decreases the internal energy.
  • Work done on the system is negative, increasing the internal energy.

Open, Closed, and Isolated Systems

  • Open System: Allows the exchange of both matter and energy with the surroundings.
  • Closed System: Allows energy exchange but not matter exchange.
  • Isolated System: Prevents exchange of both energy and matter, maintaining constant internal energy (ΔU = 0).

Thermodynamic Processes

  • Isochoric Process: Constant volume (ΔV = 0), resulting in zero work done (W = 0).
  • Isobaric Process: Constant pressure (ΔP = 0).
  • Isothermal Process: Constant temperature (ΔT = 0), leading to no change in internal energy (ΔU = 0).
  • Adiabatic Process: No heat exchange occurs (Q = 0).

Isochoric Process Equations

  • W = 0 because volume remains constant.
  • ΔU = Q, indicating the change in internal energy equals heat flow.
  • Q = n * CV * ΔT, used for heat flow calculation.
    • n = number of moles.
    • CV = molar heat capacity at constant volume for an ideal gas.
    • ΔT = change in temperature, measured in Kelvin or Celsius.
    • P1 / T1 = P2 / T2 is a notable equation.

Isobaric Process Equations

  • W = P * ΔV, determines work using pressure and volume change.
  • W = n * R * ΔT, alternative calculation for work.
  • Q = n * CP * ΔT, is used to calculate heat flow.
    • CP = molar heat capacity at constant pressure.
  • ΔU = Q - W, calculates the change in internal energy.
  • ΔU = n * CV * ΔT, provides another method to calculate the change in internal energy.
  • V1 / T1 = V2 / T2 is a useful equation.

Isothermal Process Equations

  • ΔU = 0, the change in internal energy is always zero during the process.
  • Q = W, heat flow equals the work done.
  • W = n * R * T * ln(Vf / Vi), work calculation using volume.
    • Vf = final volume.
    • Vi = initial volume.
  • W = n * R * T * ln(Pi / Pf), work calculation using pressure.
    • Pi = initial pressure.
    • Pf = final pressure.
  • P1 * V1 = P2 * V2, this is a key equation.

Adiabatic Process Equations

  • Q = 0, signifying no heat flow.
  • ΔU = -W, change in internal energy equals negative work done.
  • ΔU = n * CV * ΔT calculates change in internal energy.
  • W = -n * CV * ΔT calculates work done.
  • CV = (3/2) * R for monoatomic gas

Molar Heat Capacity

  • CP = (5/2) * R, describes molar heat capacity at constant pressure for a monoatomic ideal gas.
  • For diatomic gases, CV ≈ (5/2) * R and CP ≈ (7/2) * R.
  • For polyatomic gases, CV ≈ (7/2) * R and CP ≈ (9/2) * R, but these values are approximate.

Ideal Gas Law

  • PV = nRT (Ideal Gas Law)
  • Value of R = 8.3145 J/(mol·K)
  • Value of R = 0.08206 Lâ‹…atm/(mol·K)

Conversions

  • 1 Pascal * 1 cubic meter = 1 Joule
  • 1 liter * 1 atm = 101.3 Joules
  • 1 atm = 1.013 * 10^5 Pascals
  • 1 cubic meter = 1,000 liters

PV Diagrams

  • The area under a PV diagram indicates the work done on or by the gas.

Cyclic Process

  • Returns a system to its initial state.
  • Total work is the area inside the cycle on a PV diagram.
  • ΔU = 0, therefore Q = W.
  • Efficiency (η) = (Output Work / Heat Input) * 100%

Not State Function

  • Work is not a state function

Additional Equations for Isobaric Process

  • Q = P * CP * ΔV / R (Heat equation)
  • P * ΔV = NR * ΔT (Ideal Gas Law)

Gamma Ratio

  • γ (gamma) = CP / CV (ratio of molar heat capacities)
  • CV = R / (γ - 1)
  • CP = R * γ / (γ - 1)

Adiabatic Expansion/Compression

  • P1 * V1^γ = P2 * V2^γ
  • T1 * V1^(γ-1) = T2 * V2^(γ-1)

Isobaric Process

  • Pressure remains constant during the process.
  • Volume increases as temperature increases.
  • Governed by Charles's Law: V1 / T1 = V2 / T2.

Cyclic Process

  • The work done by the gas is equivalent to the enclosed area within the cycle on a PV diagram.
  • In a rectangular cycle, work = change in pressure × change in volume, involving both isobaric and isochoric processes.
  • With changing pressure at a constant rate, work done equals average pressure × volume change.
  • Average pressure formula: (Initial Pressure + Final Pressure) / 2.
  • Work formula: [(Initial Pressure + Final Pressure) / 2] × Change in Volume.
  • Clockwise direction indicates positive work.
  • Counterclockwise direction indicates negative work.
  • Net heat flow is equal to work done (Q = W).
  • Change in internal energy (ΔU) equals zero.
  • Net heat flow is the difference between heat entering (Qh) and heat leaving (Ql).
  • Qh is the total heat absorbed.
  • Ql is the total energy released.
  • Work is the difference between total heat absorbed and total energy released (W = Qh - Ql).
  • Efficiency Calculation: Work performed / Total heat input (W / Qh), often expressed as a percentage.

Isothermal Process

  • Temperature is constant (ΔT = 0).
  • Displayed as an isothermal graph on a PV diagram.
  • Work equation: W = nRT * ln(V_final / V_initial)
  • Work equation: W = nRT * ln(P_final / P_initial)
  • The Change in internal energy (ΔU) is zero.
  • ΔU = nCvΔT, where ΔU is zero because ΔT is zero.
  • Since ΔU = Q - W = 0, Q = W.
  • Volume and pressure are inversely proportional.
  • This process follows Boyle's Law: P1V1 = P2V2

Adiabatic Process

  • There is virtually no heat exchange with the surroundings (Q ≈ 0).
  • Change in internal energy matches work done: ΔU = W
  • ΔU = nCvΔT applies to all processes.
  • Therefore, W = -nCvΔT
  • Cv = R / (γ - 1), with R as 8.3145 J/(mol·K)
  • Cp = R * γ / (γ - 1)
  • γ represents the ratio of Cp to Cv.
  • P1V1^γ = P2V2^γ
  • T1V1^(γ-1) = T2V2^(γ-1)
  • W = (-Cv / R) * (P_final * V_final - P_initial * V_initial)
  • Consequently, W = -1 / (γ - 1) * (P_final * V_final - P_initial * V_initial)

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