Correct statement about thermodynamics process

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songoku
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Homework Statement
An ideal gas is taken through two cycles shown in Figure a and b. In Figure a, the cycle consists of process A (solid line) and adiabatic process (dash line). In figure b, the cycle consists of process B (solid line) and isothermal process (dash line). Which of the following statements is true?
A) The heat of both processes A and B are released;
B) The heat of both processes A and B are absorbed;
C) The heat of process A is released, while the heat of process B is absorbed;
D) The heat of process A is absorbed, while the heat of process B is released.
Relevant Equations
ΔU = Q + W
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I know process B absorbs heat but I can't determine the heat of process A.

In adiabatic process, Q = 0 but process A is not adiabatic. I only know both W and ΔU will be negative for process A but how to know Q?

Thanks
 
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Chestermiller said:
What is the value of ##\Delta U## when a gas undergoes a thermodynamic cycle, starting and ending in exactly the same thermodynamic state?
ΔU will be zero since there is no change in temperature. But sorry I don't understand the direction of the hint since process A does not start and end in exactly same thermodynamic state.

Thanks
 
Chestermiller said:
For process A, I think they mean the entire cycle, not just the solid line. I think it also includes the dashed line.
I think they mean the solid line. Please see below.
songoku said:
ΔU will be zero since there is no change in temperature. But sorry I don't understand the direction of the hint since process A does not start and end in exactly same thermodynamic state.
For process A, write the first law for the dashed and solid line:

##\Delta U_A^{\text{solid}}=Q_A^{\text{solid}}+W_A^{\text{solid}}##

##\Delta U_A^{\text{dashed}}=Q_A^{\text{dashed}}+W_A^{\text{dashed}}##

You know that ##\Delta U_A^{\text{solid}}=\Delta U_A^{\text{dashed}}##

What else do you know?
How do ##W_A^{\text{solid}}## and ##W_A^{\text{dashed}}## compare?

Repeat along similar lines with process B asking yourself the same questions.
 
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I understand.

Thank you very much for the help and explanation Chestermiller and kuruman