Interpreting a PV diagram
Problem-solving strategy PV diagrams are useful for understanding thermodynamic processes, such as the cycle of a heat engine. It takes some practice to interpret a PV diagram. Start by looking at how pressure and volume change for each individual segment on the PV diagram. Then use any additional information from the problem, such as whether the quantity of gas or heat is constant. Read the text aloud
A process example Thermodynamic process from A to B The diagram shows a process in which a gas goes from A to B and no heat is added. Which of the following is true and why?
  1. The energy of the gas remains constant.
  2. The gas does work on its surroundings and the gas loses energy.
  3. The surroundings do work on the gas and the gas gains energy.
Read the text aloud
Asked: whether the work done is positive, negative, or zero
Given: PV diagram showing pressure decreasing and volume increasing
Relationships: ΔE = Q − PΔV
Solution: ΔV is positive because the diagram shows a volume at B greater than the volume at A. Pressure is also positive; therefore, PΔV > 0 and the gas does work on the surroundings. The energy of the gas must decrease because no heat is added that might offset the work done (i.e., Q = 0).
Answer: Choice b is correct; the gas does work on the surroundings and loses energy in the process.
Thermodynamic cycle example Thermodynamical cycle for gas changing in pressure and volume The diagram on the right shows a cycle in which pressure and volume change for a fixed quantity of gas.
  1. Which part of the process compresses the gas? How do you know?
  2. Does the complete cycle absorb heat energy or give off heat energy? How do you know?
Read the text aloud
Relationships: ΔE = Q − PΔV
Solution:
  1. Compression is a reduction in volume and only process C shows a reduction in volume.
  2. In any complete cycle, the net energy change of the gas is zero. If the net change in work is positive, then heat must be added.
    Process A expands the gas and therefore does work on the surroundings, so PΔV is positive.
    Process C compresses the gas, so the surroundings do work on the gas and PΔV is negative.
    Since the pressure is higher for A than C and the volume change is the same, the net change in work is positive and heat must be added.
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