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PV diagrams show how a gas changes pressure and volume during a thermodynamic process. This cheat sheet helps students connect graph shape, process type, and work done by or on a gas. It is useful for solving problems involving engines, compression, expansion, and the first law of thermodynamics.

Students need these relationships to interpret diagrams quickly and avoid sign errors.

Key Facts

  • Thermodynamic work done by a gas is W=ViVfPdVW = \int_{V_i}^{V_f} P\,dV, which equals the signed area under a curve on a PV diagram.
  • For an isobaric process with constant pressure, the work is W=PΔV=P(VfVi)W = P\Delta V = P(V_f - V_i).
  • For an isochoric process with constant volume, the work is W=0W = 0 because ΔV=0\Delta V = 0.
  • For an ideal gas undergoing an isothermal process, the work is W=nRTln(VfVi)W = nRT\ln\left(\frac{V_f}{V_i}\right).
  • For an adiabatic ideal gas process, PVγ=constantPV^{\gamma} = \text{constant} and W=PiViPfVfγ1W = \frac{P_iV_i - P_fV_f}{\gamma - 1}.
  • Expansion gives ΔV>0\Delta V > 0 and usually W>0W > 0, meaning the gas does work on the surroundings.
  • Compression gives ΔV<0\Delta V < 0 and usually W<0W < 0, meaning work is done on the gas.
  • For a complete cycle on a PV diagram, the net work is the enclosed area, with clockwise cycles giving Wnet>0W_{\text{net}} > 0 and counterclockwise cycles giving Wnet<0W_{\text{net}} < 0.

Vocabulary

PV Diagram
A graph of pressure PP versus volume VV that represents the thermodynamic state and process of a gas.
Thermodynamic Work
Energy transferred when a gas changes volume under pressure, calculated by W=PdVW = \int P\,dV.
Isobaric Process
A thermodynamic process that occurs at constant pressure, so work is W=PΔVW = P\Delta V.
Isochoric Process
A thermodynamic process that occurs at constant volume, so the gas does no work and W=0W = 0.
Isothermal Process
A thermodynamic process that occurs at constant temperature, with ideal gas work W=nRTln(VfVi)W = nRT\ln\left(\frac{V_f}{V_i}\right).
Adiabatic Process
A thermodynamic process with no heat transfer, so Q=0Q = 0 and for an ideal gas PVγ=constantPV^{\gamma} = \text{constant}.

Common Mistakes to Avoid

  • Using W=PΔVW = P\Delta V for every process is wrong because that formula only applies when pressure is constant.
  • Ignoring the sign of work is wrong because expansion has W>0W > 0 for work done by the gas, while compression has W<0W < 0.
  • Calculating the area under an isochoric line as nonzero is wrong because a vertical line has ΔV=0\Delta V = 0, so W=0W = 0.
  • Confusing work done by the gas with work done on the gas is wrong because these values have opposite signs, so Won gas=Wby gasW_{\text{on gas}} = -W_{\text{by gas}}.
  • Assuming all curved PV paths have the same work between two endpoints is wrong because work depends on the path, not only the initial and final states.

Practice Questions

  1. 1 A gas expands isobarically at P=2.0×105PaP = 2.0 \times 10^5\,\text{Pa} from Vi=0.030m3V_i = 0.030\,\text{m}^3 to Vf=0.080m3V_f = 0.080\,\text{m}^3. Find the work done by the gas.
  2. 2 An ideal gas undergoes an isothermal expansion with n=1.0moln = 1.0\,\text{mol}, T=300KT = 300\,\text{K}, Vi=2.0LV_i = 2.0\,\text{L}, and Vf=6.0LV_f = 6.0\,\text{L}. Calculate WW using R=8.31J/(mol\cdotK)R = 8.31\,\text{J/(mol\cdot K)}.
  3. 3 A thermodynamic cycle encloses an area of 450J450\,\text{J} on a PV diagram and runs counterclockwise. What is WnetW_{\text{net}} for work done by the gas?
  4. 4 Two processes connect the same initial and final states on a PV diagram, but one path stays at higher pressure for most of the expansion. Explain which process does more work and why.