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Given: P1 = 100 kPa, V1 = 0.082 m³, V₂ = 0.0054 m³, gamma = 1.4.

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2 parts
1.1

−33.24 kJ-33.24 \text{ kJ}

1.2

44.02 kJ44.02 \text{ kJ}

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Part 1.1· 2 steps

The work done during adiabatic compression

The process from point 1 to 2 is adiabatic compression (PVγ=CPV^\gamma = C).
Given: P1=100 kPaP_1 = 100 \text{ kPa}, V1=0.082 m3V_1 = 0.082 \text{ m}^3, V2=0.0054 m3V_2 = 0.0054 \text{ m}^3, γ=1.4\gamma = 1.4.

  1. Calculate the pressure at point 2 (P2P_2) using the adiabatic relation.

    P1V1γ=P2V2γP_1V_1^\gamma = P_2V_2^\gamma P2=P1(V1V2)γP_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma P2=100 kPa(0.082 m30.0054 m3)1.4P_2 = 100 \text{ kPa} \left(\frac{0.082 \text{ m}^3}{0.0054 \text{ m}^3}\right)^{1.4} P2=100 kPa(15.185)1.4P_2 = 100 \text{ kPa} (15.185)^{1.4} P2=100 kPa×39.810P_2 = 100 \text{ kPa} \times 39.810 P2=3981 kPaP_2 = 3981 \text{ kPa}

  2. Calculate the work done during the adiabatic compression.

    The work done for a reversible adiabatic process is given by:

    W1−2=P2V2−P1V11−γW_{1-2} = \frac{P_2V_2 - P_1V_1}{1-\gamma} W1−2=(3981 kPa)(0.0054 m3)−(100 kPa)(0.082 m3)1−1.4W_{1-2} = \frac{(3981 \text{ kPa})(0.0054 \text{ m}^3) - (100 \text{ kPa})(0.082 \text{ m}^3)}{1-1.4} W1−2=21.4974 kJ−8.2 kJ−0.4W_{1-2} = \frac{21.4974 \text{ kJ} - 8.2 \text{ kJ}}{-0.4} W1−2=13.2974 kJ−0.4W_{1-2} = \frac{13.2974 \text{ kJ}}{-0.4} W1−2=−33.2435 kJW_{1-2} = -33.2435 \text{ kJ}

    The negative sign indicates work is done on the gas.

    −33.24 kJ\boxed{-33.24 \text{ kJ}}

Part 1.2

44.02 kJ44.02 \text{ kJ}

The work done during the constant pressure process

The process from point 3 to 4 is a constant pressure process.
Given: P3=6200 kPaP_3 = 6200 \text{ kPa}, V3=0.0054 m3V_3 = 0.0054 \text{ m}^3, V4=0.0125 m3V_4 = 0.0125 \text{ m}^3.

  1. Calculate the work done for the constant pressure process.

    W3−4=P3(V4−V3)W_{3-4} = P_3(V_4 - V_3) W3−4=6200 kPa(0.0125 m3−0.0054 m3)W_{3-4} = 6200 \text{ kPa} (0.0125 \text{ m}^3 - 0.0054 \text{ m}^3) W3−4=6200 kPa(0.0071 m3)W_{3-4} = 6200 \text{ kPa} (0.0071 \text{ m}^3) W3−4=44.02 kJW_{3-4} = 44.02 \text{ kJ} 44.02 kJ\boxed{44.02 \text{ kJ}}

Part 1.3· 2 steps

The volume at point 3 and point 5 in litres

From the diagram: V3=0.0054 m3V_3 = 0.0054 \text{ m}^3 and V5=0.082 m3V_5 = 0.082 \text{ m}^3.
To convert from cubic meters to litres, multiply by 1000 (1 m3=1000 L1 \text{ m}^3 = 1000 \text{ L}).

  1. Convert V3V_3 to litres.

    V3=0.0054 m3×1000 L/m3=5.4 LV_3 = 0.0054 \text{ m}^3 \times 1000 \text{ L/m}^3 = 5.4 \text{ L}

  2. Convert V5V_5 to litres.

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