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Example 15-4 Two Thermodynamic Processes: Isochoric and Isobaric Set Up We’ll use the ideal gas law to determine the final volume of the gas. For the isochoric (constant-volume) process A S B, the gas does no work, so the quantity of heat Q that enters the gas is equal to the internal energy change U of the gas. (The gas is cooled in this process, so Q 6 0 and U 6 0.) We’ll determine U from the temperature change. For the isobaric (constant-pressure) process B S C, the work W that the gas does is given by Equation 15-4; this is positive because the gas expands. The internal energy change U is again given by the temperature change. We’ll then determine Q for this process from the first law of thermodynamics. (a) Use the ideal gas law to find the final volume Vf. A 3.03 × 105 The light blue curve is an isotherm. 1.01 × 105 B 1.20 × 10 C –3 Vf = ? Volume (m3) Process B → C isobaric (constant pressure) Figure 15-9 Isochoric cooling and isobaric expansion In this two-step process for an ideal gas, how much work is done, how much heat flows into the gas, and by how much does the internal energy change? Ideal diatomic gas: 5 U = nRT 2 Ideal gas law: pV = nRT Process A (14-7) system = gas (15-9) C The gas pushes upward and does work on the piston: W > 0. (15-4) Isochoric (constant-volume) process: Heat flows out of the gas: Q < 0. Process B Work done in an isobaric (constant-pressure) process: W = p(Vf 2 Vi) B The volume of the gas is constant: W = 0. First law of thermodynamics: (15-2) U = Q - W U = Q Solve Process A → B isochoric (constant volume) Pressure (Pa) The pV diagram in Figure 15-9 shows two thermodynamic processes that occur in an ideal diatomic gas, for which the internal energy is U = 15>22nRT. The gas is in a cylinder with a moveable piston (see Figure 15-5) and is initially in the state labeled A on the diagram, at pressure pi = 3.03 * 105 Pa and volume Vi = 1.20 * 1023 m3. The piston is first locked in place so that the volume of the gas cannot change, and the cylinder is cooled so that the temperature and pressure of the gas both decrease. When the gas is in state B, the pressure of the gas is pf = 1.01 * 105 Pa = 1.00 atm, the same as the air pressure outside the cylinder. The piston is then unlocked so it is free to move, and the cylinder is slowly heated so that the gas expands at constant pressure. The temperature of the gas increases until in the final state (C in Figure 15-9), when the gas is at the same temperature as in the initial state A. (a) What is the final volume of the gas? (b) Find the values of W, U, and Q for the isochoric process A S B. (c) Find the values of W, U, and Q for the isobaric process B S C. (d) Find the values of W, U, and Q for the net process A S B S C. Heat flows into the gas to make it expand: Q > 0. system = gas The initial pressure and volume of the gas are pi = 3.03 * 105 Pa Vi = 1.20 * 1023 m3 From the ideal gas law, Equation 14-7, piVi = nRTi where Ti is the initial temperature of the gas in state A. The final pressure of the gas is pf = 1.01 * 105 Pa The ideal gas law tells us that pfVf = nRTf where Tf is the final temperature of the gas in state C. The initial and final temperatures are the same (Ti = Tf) and so piVi = pfVf Vf = 13.03 * 105 Pa2 11.20 * 10-3 m3 2 piVi = = 3.60 * 10-3 m3 pf 1.01 * 105 Pa The final pressure is one-third of the initial pressure, and the final volume is three times the initial volume. (b) For the isochoric process A S B the work done is WA S B = 0. Find the change in internal energy U and quantity of heat Q for this process. The change in internal energy U is the difference between the value of U in state B and the value of U in state A: 5 5 UA S B = UB - UA = nRTB - nRTi 2 2 We don’t know the number of moles of gas n, nor do we know the temperatures Ti (in the initial state A, for which the pressure is pi and the volume is Vi) and TB (in state B, for which the pressure is pf and the volume is Vi). But from the ideal gas law pV = nRT, so piVi = nRTi and pfVi = nRTB So the change in internal energy is 5 5 5 UA S B = pfVi - piVi = 1pf - pi 2Vi 2 2 2 5 5 = 11.01 * 10 Pa - 3.03 * 105 Pa2 11.20 * 10-3 m3 2 2 = -606 J (Remember that 1 Pa = 1 N>m2 and 1 N # m = 1 J.) From Equation 15-9 for an isochoric process, QA S B = UA S B = 2606 J Heat flows out of the gas and the internal energy of the gas decreases. (c) Find W, U, and Q for the isobaric process B S C. The work done in an isobaric process is given by Equation 15-4. The constant pressure is pf = 1.01 * 105 Pa, and the volume increases from Vi = 1.20 * 1023 m3 to Vf = 3.60 * 1023 m3: WB S C = pf (Vf 2 Vi) = (1.01 * 105 Pa) (3.60 * 1023 m3 2 1.20 * 1023 m3) = +242 J The gas does positive work as it expands. As in part (b), we calculate the internal energy change with the aid of the ideal gas law: 5 5 nRTf - nRTB 2 2 5 5 5 = pfVf - pfVi = pf 1Vf - Vi 2 2 2 2 UB S C = UC - UB = This is just 5>2 times the above expression for WB S C, so 5 5 WB S C = 1 +242 J2 = +606 J 2 2 From the first law of thermodynamics, the heat that flows into the gas in this process is UB S C = QB S C = UB S C + WB S C = (+606 J) + (+242 J) = +848 J (d) Find the values of W, U, and Q for the combined process A S B S C. The total amount of work done by the gas is WA S B + WB S C = 0 + (+242 J) = +242 J The total change in internal energy of the gas is UA S B + UB S C = (2606 J) + (+606 J) = 0 The total quantity of heat that flows into the gas is QA S B + QB S C = 2606 J + (+848 J) = +242 J Reflect For the combined process, there is zero net change in the internal energy U of the ideal gas. That’s because U depends only on the number of moles of gas and the temperature, which have the same values in the final state C as in the initial state A. In the combined process 242 J of heat flows into the gas, and the gas uses this energy to do 242 J of work.