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What is the actual work done during mechanically reversible, adiabatic compression when the process has an efficiency of 80%?

  • 1248 J (correct)
  • 998 J
  • 665 J
  • 831 J
  • What is the change in heat (Q) during the mechanically reversible cooling process with an irreversible efficiency of 80%?

  • -998 J
  • -250 J
  • 665 J
  • -1829 J (correct)
  • How does the work done in an irreversible process compare to the reversible process for mechanically reversible cooling?

  • It can either be more or less depending on conditions
  • It remains equal to reversible work
  • It is greater than reversible work (correct)
  • It is less than reversible work
  • What is the value of Q for the entire cycle considering the calculated changes in the processes mentioned?

    <p>-883 J</p> Signup and view all the answers

    When performing adiabatic processes, work ( W) must be considered. What was the total work done in the process?

    <p>883 J</p> Signup and view all the answers

    What is the relationship between internal energy change, work, and heat in these thermodynamic processes?

    <p>Q = ΔU - W</p> Signup and view all the answers

    In the case of mechanically reversible processes, how is work (W) affected if the efficiency is less than 100%?

    <p>W increases</p> Signup and view all the answers

    What is the formula used to calculate the work done for the irreversible cooling process detailed in the content?

    <p>W = 665/0.80</p> Signup and view all the answers

    What is the final temperature after the adiabatic compression when the volume changes to 0.008263 m³?

    <p>462.69 K</p> Signup and view all the answers

    In the given adiabatic process, the work done during compression is measured as what value?

    <p>3420 J</p> Signup and view all the answers

    For the constant-volume step in the process, what is the heat transfer value?

    <p>-3420 J</p> Signup and view all the answers

    What is the relationship between Q and W for the process described?

    <p>Q = -W</p> Signup and view all the answers

    What is the heat capacity used in the calculations for the ideal gas?

    <p>20.785 J/K</p> Signup and view all the answers

    In the ideal gas compression process, the specific heat capacity is represented by which notation?

    <p>C_V</p> Signup and view all the answers

    During which process is no work done according to the provided information?

    <p>Constant-volume step</p> Signup and view all the answers

    What does the ideal gas equation suggest about the work done in reversible processes?

    <p>It is directly proportional to the volume change.</p> Signup and view all the answers

    What is the approximate value of γ for diatomic gases?

    <p>1.4</p> Signup and view all the answers

    In thermodynamics, how is the work for an irreversible process primarily calculated?

    <p>By performing a two-step procedure</p> Signup and view all the answers

    Which of the following statements is true regarding property changes in reversible and irreversible processes?

    <p>They depend only on the initial and final states.</p> Signup and view all the answers

    How does the work of an irreversible process compare to that of a reversible process if it produces work?

    <p>The reversible work is larger and gets multiplied by efficiency.</p> Signup and view all the answers

    What is the relationship between equations for property changes and the type of process (reversible or irreversible)?

    <p>They are valid for both types of processes.</p> Signup and view all the answers

    When calculating work for a process that requires work, how does the reversible process value compare to the actual irreversible process?

    <p>It is larger and must be multiplied by efficiency.</p> Signup and view all the answers

    What is the approximate γ value for simple polyatomic gases such as CO2 and NH3?

    <p>1.3</p> Signup and view all the answers

    What common characteristic do the equations for dUig, dHig, ΔUig, and ΔHig share?

    <p>They only apply to ideal gases.</p> Signup and view all the answers

    Study Notes

    Work and Heat in Irreversible Processes

    • For mechanically reversible, adiabatic compression, work ( W_{rev} = 998 , \text{J} ); with 80% efficiency, actual work ( W = \frac{998}{0.80} = 1248 , \text{J} ).
    • First law of thermodynamics gives ( Q = \Delta U_{ig} - W = 998 - 1248 = -250 , \text{J} ).

    Cooling Process

    • Mechanically reversible cooling process requires work ( W_{rev} = 665 , \text{J} ); actual work ( W = \frac{665}{0.80} = 831 , \text{J} ).
    • Change in internal energy ( \Delta U_{ig} = -998 , \text{J} ) leads to heat transfer ( Q = -998 - 831 = -1829 , \text{J} ).

    System Work

    • In a process where work is done by the system, irreversible work in absolute value is less than ( -1495 , \text{J} ).
    • Actual work done ( W = 0.80 \times (-1495) = -1196 , \text{J} ).
    • Since the internal energy change is zero in this step, heat transfer is ( Q = \Delta U_{ig} - W = 0 + 1196 = 1196 , \text{J} ).

    Overall Cycle Summary

    • Total change for the entire cycle shows ( \Delta U_{ig} ) and ( \Delta H_{ig} = 0 ).
    • Cumulative heat transfer is ( Q = -250 - 1829 + 1196 = -883 , \text{J} ).
    • Total work done in the cycle ( W = 1248 + 831 - 1196 = 883 , \text{J} ).

    Characteristics of Real Gases

    • Equations derived for mechanically reversible processes apply similarly to irreversible processes for ideal gases.
    • Property changes ( dU_{ig}, dH_{ig}, \Delta U_{ig}, \Delta H_{ig} ) depend solely on the states' initial and final conditions.

    Calculating Irreversible Work

    • Work for irreversible processes is usually calculated in two steps:
      • First, determine ( W ) for a reversible process accomplishing the same state change.
      • Multiply or divide this value by efficiency to obtain actual work.
    • If the process produces work, actual irreversible ( W ) is greater than reversible; if it requires work, the opposite holds true.

    Additional Notes on Ideal Gases

    • For adiabatic compression, initial step results in a final volume of ( 0.008263 , \text{m}^3 ) with a corresponding temperature of ( 462.69 , \text{K} ).
    • Work for compression using ( C_V ) is ( W = C_V \Delta T = (20.785)(462.69 - 298.15) = 3420 , \text{J} ).
    • In the constant-volume step, no work is done; heat transfer is ( Q = \Delta U_{ig} = 20.785(298.15 - 462.69) = -3420 , \text{J} ).
    • Notably, ( Q = -W ) indicates path-dependency, where heat and work are intricately linked in thermodynamic processes.

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