1 PHYS 3141 MIDTERM EXAM SOLUTIONS 1. (15 pts) First law: dU = dQ+dW Ideal gas: U = ?2NRT = NcV T " cV = ?dQ dT ! V = 1N ?@U @T ! V = ?2R # ) dU = NcV dT Heat capacity at constant pressure: cP = 1N ?dQ dT ! P ) dQP=cte = NcPdT Quasistatic work: dW = ¡PdV Introducing dU, dQ and dW in the flrst law: NcV dT = NcPdT ¡PdV , N(cP ¡cV )dT = PdV (1) The equation of state for an ideal gas is: PV = NRT. At constant pressure: PdV = NRdT. Introducing this in equation (1): N(cP ¡cV )dT = NRdT For an arbitrary temperature change, this results in: cP ¡cV = R which is what we wanted to proof. 2. (a) (7 pts) No work is done. There are no external forces involved in the process: The vacuum cannot exert a force on the gas nor the gas can exert a force on the vacuum. It is true that the volume of the gas changes, but ¡PdV can only be identifled with a quasistatic process and the free expansion of a gas is non-quasistatic. (b) (7 pts) No heat transfer occurs. One cannot exchange heat with a vacuum and no heat ows through an adiabatic wall. 2 (c) (7 pts) By the flrst law of Thermodynamics: ¢U = W +Q = 0+0 = 0. Then: U1 = U2. The gas is ideal from now on. (d) (7 pts) For an ideal gas, U = ?2NRT, with ? the microscopic degrees of freedom of the constituent particles. Since U1 = U2 and U = U(T) for an ideal gas, then it follows that T1 = T2. The temperature of an ideal gas does not change in a free expansion process. (e) (7 pts) Equation of state for an ideal gas: PV = NRT. Since the initial and flnal states of the free expansion process are equilibrium states, then: P1V1 = NRT1 and P2V2 = NRT2. Dividing these two equations and using that T1 = T2 (N = cte), we obtain: P2 = P1V1V 2 (f) (7 pts) Deflnition of entropy: ¢S = Z 2 1 ?dQ T ! QS For a quasistatic process, dW = ¡PdV. Using this and the flrst law of Thermodynamics: dQ T = dU T ¡ dW T = NcV T dT + P T dV = NcV T dT + NR V dV where we have used the equation of state for an ideal gas in the last step. This last equation can be integrated from state 1 to state 2 to give: ¢S = S2 ¡S1 = NcV ln T 2 T1 ¶ +NR ln V 2 V1 ¶ (2) Since T1 = T2, the entropy change in the free expansion process is: ¢S = NR ln V 2 V1 ¶ (g) (7 pts) By the flrst law of Thermodynamics, ¢U = W + Q = W + 0 = W, since the process is adiabatic. Given that W 6= 0 ) ¢U 6= 0. The internal energy must change. (h) (7 pts) For an ideal gas, U = U(T). Since the internal energy changes, the temperature of the gas must change. Since the volume in situation 4 is identical to the volume in situation 1, but the temperature must have changed, using the equation of state of an ideal gas (N = cte), the pressure of the gas must have changed. 3 (i) (7 pts) * For the quasistatic process, we could use: dW = ¡PdV and thus: W = ¡ Z 4 3 PdV (Since the process is, in addition, adiabatic, we can use PV = cte to do the integral) * For the non-quasistatic process, we need to use the deflnition of work in terms of the acting external forces: W = Z 4 3 ~Fext ¢d~r where d~r is the inflnitesimal displacement vector. * In both cases, if we know ¢U, since Q = 0, we could also use the flrst law of Thermody- namics: W = ¢U. (j) (7 pts) The answer is provided by equation (2). For the free expansion process: ¢S1!2 = NR ln V 2 V1 ¶ For the compression process: ¢S3!4 = NcV ln T 4 T3 ¶ +NRln V 4 V3 ¶ = NcV ln T 4 T3 ¶ +NRln V 1 V2 ¶ = NcV ln T 4 T3 ¶ ¡¢S1!2 Then: ¢Stotal = ¢S1!2 +¢S3!4 = NcV ln T 4 T3 ¶ Since U4 > U3, given that the work in the compression process is positive, and U = U(T) for an ideal gas: T4 > T3. As a result, ¢Stotal > 0. 3. (15 pts) The processes are quasistatic. Then: dQ = TdS. The temperature is always positive, so heat is absorbed (dQ > 0) when S is increasing (dS > 0) and heat is released (dQ < 0) when S is decreasing (dS < 0). Thus, heat is absorbed (arrow points in) in the quadrants below the horizontal line and heat is released (arrow points out) in the quadrants above the horizontal line. 4 S T