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1.2 The work done during the constant pressure process

Computer Science

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44.02 kJ\text{44.02 kJ}

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Here is the solution for question 1.2:

1.2 The work done during the constant pressure process

The constant pressure process is from state 3 to state 4.
From the P-V diagram:

  • Pressure at state 3 and 4, P3=P4=6200 kPaP_3 = P_4 = 6200 \text{ kPa}
  • Volume at state 3, V3=0.0054 m3V_3 = 0.0054 \text{ m}^3
  • Volume at state 4, V4=0.0125 m3V_4 = 0.0125 \text{ m}^3

The formula for work done during a constant pressure process is:

W=P×(Vfinal−Vinitial)W = P \times (V_{final} - V_{initial})

In this case, W3−4=P3×(V4−V3)W_{3-4} = P_3 \times (V_4 - V_3).

  1. Substitute the values into the formula.

    W3−4=6200 kPa×(0.0125 m3−0.0054 m3)W_{3-4} = 6200 \text{ kPa} \times (0.0125 \text{ m}^3 - 0.0054 \text{ m}^3)
  2. Calculate the change in volume.

    W3−4=6200 kPa×(0.0071 m3)W_{3-4} = 6200 \text{ kPa} \times (0.0071 \text{ m}^3)
  3. Perform the multiplication.

    W3−4=44.02 kJW_{3-4} = 44.02 \text{ kJ}

    The work done during the constant pressure process (3-4) is 44.02 kJ\boxed{\text{44.02 kJ}}.

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