Experiments · Experiment 2

The First Law: Work, Heat and the Ratio of Heat Capacities

Isothermal and adiabatic changes in an ideal gas, and three determinations of gamma.

Objectives

On completing this experiment you should be able to:

  • Measure an isotherm and confirm Boyle's law, allowing for the dead volume of the apparatus.
  • Determine the amount of gas confined and the dead volume from a single linear plot.
  • Measure a reversible adiabat and obtain gamma from the gradient of log P against log V.
  • Determine gamma by the Clement-Desormes method and investigate how the result is biased by the time the tap is left open.
  • Determine gamma by Ruchhardt's oscillating-piston method.
  • Compare the measured values with the equipartition predictions and account for the behaviour of carbon dioxide.
  • Compute the work done, the heat exchanged and the change in internal energy for each process and verify the first law.

Theory

The first law states that the internal energy is a function of state: dU=dQdWdU = dQ - dW, with dW=PdVdW = P\,dV the work done by the system.

For an ideal gas U depends only on temperature, so an isothermal change has dU=0dU = 0 and Q=W=Q = W = nRT ln(V2/V1)\ln(V2/V1). A reversible adiabatic change has Q=0Q = 0, so n CVC_V dT = -P dV, which together with PV = nRT integrates to P VγV^\gamma = constant with gamma = CpCV\frac{C_p}{C_V}.

Because CpC_p - CVC_V = nR for an ideal gas, a measurement of gamma is equivalent to a measurement of CVC_V. Equipartition predicts CV=(32)RC_V = \left(\frac{3}{2}\right)R for a monatomic gas, (52)R\left(\frac{5}{2}\right)R for a diatomic gas whose vibration is frozen out, and larger values whenever low-lying vibrational modes are excited - which is exactly why carbon dioxide, with a bending mode at only 960 K, has gamma near 1.29 rather than 1.4.

Key relations

  • dU=δQPdV\mathrm{d}U = \delta Q - P\,\mathrm{d}V

    First law.

  • P(V+V0)=nRTP(V + V_0) = nRT

    Boyle's law corrected for the dead volume V0V_0.

  • PVγ=constantPV^{\gamma} = \text{constant}

    Reversible adiabat.

  • γ=ln(P1/P0)ln(P1/P2)h1h1h2\gamma = \frac{\ln(P_1/P_0)}{\ln(P_1/P_2)} \simeq \frac{h_1}{h_1-h_2}

    Clement-Desormes.

  • τ=2πmVγA2P\tau = 2\pi\sqrt{\frac{mV}{\gamma A^2 P}}

    Ruchhardt's oscillation period.

Procedure by part

Part A - Isotherm and the dead volume

Set the piston to a series of indicated volumes at constant temperature and record the pressure.

  1. Select the gas and set the thermostat. Allow two minutes for equilibration (the apparatus does this automatically).
  2. Record the pressure at eight or more indicated volumes spanning the full travel.
  3. Plot 1P\frac{1}{P} against the indicated volume V. The result is a straight line of gradient 1nRT\frac{1}{nRT} and intercept V0nRT\frac{V_0}{nRT}.
  4. Obtain the amount of gas n and the dead volume V0V_0 from the gradient and the intercept.
  5. Compute the work done by the gas in expanding between your first and last volumes, and hence the heat absorbed.

Part B - Reversible adiabat

With the cylinder insulated, compress the gas slowly enough to remain reversible but fast enough to remain adiabatic, and record P, V and T.

  1. Select the gas. The apparatus starts each run from the same reference state.
  2. Set the final indicated volume and take the reading.
  3. Repeat for at least eight volumes.
  4. Plot ln P against ln(V + V0)V_0) using the dead volume from Part A. The gradient is -gamma.
  5. Verify that T V(γ1)V^(\gamma-1) is also constant, and compute the work done using W=P1V1P2V2γW = \frac{P_1 V_1 - P_2 V_2}{\gamma -} 1).

Part C - Clement-Desormes

Pressurise the carboy, open the tap for a chosen time, close it, and wait for the gas to return to room temperature.

  1. Select the gas and the initial excess pressure h1 on the oil manometer.
  2. Choose how long the tap is left open, then take the reading. The apparatus reports the final excess pressure h2 after thermal re-equilibration.
  3. Compute gamma = h1h1\frac{h1}{h1} - h2) and also the exact form ln(P1/P0)ln(P1/P2)\frac{\ln(P1/P0)}{\ln(P1/P2)}.
  4. Repeat over a wide range of tap-open times and plot gamma against the open time. Identify the plateau and explain the behaviour at both ends.
  5. Repeat at the best open time for each gas.

Part D - Ruchhardt's method

Drop a precision ball into the vertical tube and time its oscillations on the gas column beneath.

  1. Select the gas and the ball. The apparatus reports the time for twenty complete oscillations.
  2. Repeat at least five times for each combination and take the mean.
  3. Compute the equilibrium gas pressure P=Patm+mg/AP = P_{atm} + mg/A, then gamma = 4 π2\pi^2 m V / (τ2A2)(\tau^2 A^2) P).
  4. Compare the three determinations of gamma for each gas and comment on their relative precision.

Questions to answer in your report

  1. Plot your Clement-Desormes value of gamma against the tap-open time. Explain the shape of the curve and justify the open time you finally adopted. [8]
  2. Argon and nitrogen have gamma close to 53\frac{5}{3} and 75\frac{7}{5} respectively, but carbon dioxide gives about 1.29. Account for all three values quantitatively using equipartition and the vibrational temperatures of the molecules. [8]
  3. For your isothermal expansion between the first and last volumes of Part A, compute W, Q and dU. Do the same for the adiabatic compression of Part B. Show explicitly that the first law is satisfied in both cases. [7]
  4. Why does Ruchhardt's method systematically underestimate gamma, and how large is the effect in your data? [5]