A gas contained in a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where p1 =1bar, V1 = 1m3, U1 =400kJ and p * 2 = 10 bar, V2 = 0.1 m3, U2 = 450 kJ: Process A: Constant-volume process from state 1 to a pressure of 10 bar, followed by a constantpressure process to state 2. Process B: Process from 1 to 2 during which the pressure-volume relation is pV = constant. Kinetic and potential effects can be ignored. For each of the processes A and B, (a) sketch the process on p-V coordinates, (b) evaluate the work, in kJ, and (c) evaluate the heat transfer, in kJ.

The correct answer and explanation is :

Alright, let’s work through this carefully.

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First, given data:

• State 1:
  ( p_1 = 1\, \text{bar} = 100\, \text{kPa} )
  ( V_1 = 1\, \text{m}^3 )
  ( U_1 = 400\, \text{kJ} )
• State 2:
  ( p_2 = 10\, \text{bar} = 1000\, \text{kPa} )
  ( V_2 = 0.1\, \text{m}^3 )
  ( U_2 = 450\, \text{kJ} )

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(a) Sketch on p-V Diagram:

• Process A:
• Step 1: Constant volume compression (vertical line up on p-V diagram) from 1 bar to 10 bar.
• Step 2: Constant pressure compression (horizontal line left) from ( V_1 = 1\, \text{m}^3 ) to ( V_2 = 0.1\, \text{m}^3 ) at ( p = 10\, \text{bar} ).
• Process B:
• ( pV = \text{constant} ) (a hyperbolic curve, called an isothermal-like shape but NOT necessarily isothermal).

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(b) Work done (W)

Process A:

• Step 1 (constant volume):
  ( W = 0 ) (because ( dV = 0 ))
• Step 2 (constant pressure):
  [
  W = p (V_2 – V_1)
  ]
  Substituting values:
  [
  W = 1000\, \text{kPa} \times (0.1 – 1) \, \text{m}^3 = 1000 \times (-0.9) = -900\, \text{kJ}
  ]

So, Work for Process A = (-900) kJ

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Process B:

Since ( pV = \text{constant} ), it is a polytropic process with ( n = 1 ).

Work formula for ( n = 1 ) process:
[
W = p_1 V_1 \ln\left( \frac{V_2}{V_1} \right)
]

Substituting:
[
W = (100\, \text{kPa})(1\, \text{m}^3) \ln\left( \frac{0.1}{1} \right)
]
[
= 100 \times \ln(0.1)
]
[
= 100 \times (-2.3026)
]
[
= -230.26\, \text{kJ}
]

So, Work for Process B = (-230.26) kJ

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(c) Heat transfer (Q)

From First Law of Thermodynamics:
[
Q = \Delta U + W
]
where:
[
\Delta U = U_2 – U_1 = 450 – 400 = 50\, \text{kJ}
]

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Process A:

[
Q_A = \Delta U + W_A
]
[
Q_A = 50 + (-900)
]
[
Q_A = -850\, \text{kJ}
]

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Process B:

[
Q_B = \Delta U + W_B
]
[
Q_B = 50 + (-230.26)
]
[
Q_B = -180.26\, \text{kJ}
]

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Summary of Results:

	Work (kJ)	Heat Transfer (kJ)
Process A	-900	-850
Process B	-230.26	-180.26

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Explanation (300+ words):

This problem involves analyzing two different thermodynamic processes that take a gas between the same starting and ending conditions, but using different paths. In Process A, the gas first undergoes a constant volume compression, meaning the piston is locked, and the pressure increases from 1 bar to 10 bar without any volume change. Since the volume doesn’t change, no work is done during this step (( W = 0 )). After reaching 10 bar, the gas is compressed at constant pressure from a volume of 1 m³ to 0.1 m³. For a constant pressure process, the work is simply the pressure multiplied by the volume change. Here, because the volume is decreasing, the work comes out negative, meaning the system is compressed.

In Process B, the gas undergoes a process where ( pV ) remains constant. This is known as a polytropic process with ( n = 1 ). The work done during this process involves integrating ( p \, dV ) while keeping ( pV = \text{constant} ), leading to the logarithmic formula for work. The work is again negative because the volume is decreasing during compression, although it is a smaller magnitude compared to Process A.

Once the work is known for each process, we use the first law of thermodynamics to find the heat transfer: ( Q = \Delta U + W ). The internal energy change ( \Delta U ) is positive, meaning the gas absorbs internal energy, but because the work terms are negative and larger in magnitude, the heat transfer ends up being negative for both processes. A negative heat transfer indicates that the system loses heat to the surroundings.

The results show that Process A requires more compression work (–900 kJ) compared to Process B (–230.26 kJ), and correspondingly, loses more heat. This highlights how the path between states significantly affects work and heat interactions even if the initial and final states are identical.

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