Phase Transitions & Clausius-Clapeyron Eq.: Pressure, Temp., and Entropy in 3He & Lead - P, Assignments of Thermal Physics

Download Phase Transitions & Clausius-Clapeyron Eq.: Pressure, Temp., and Entropy in 3He & Lead - P and more Assignments Thermal Physics in PDF only on Docsity! 1    Fall 2008 Phys 461 Solution Set 7 Schroeder 5.34 (a) At a fixed temperature, the phase that has a smaller volume will be more stable as you increase the pressure. From the phase diagram of 3He, the solid phase is more stable at high pressures, therefore, the sold phase has a smaller volume. But the phase boundary around 0.3 K has a negative slope. According to the Clausius- Clapeyron relation, V S dT dP ∆ ∆ = . Therefore, if volume of the solid phase is smaller than the volume of the liquid phase at 0.3 K, then entropy of the solid phase must be large than the entropy of the liquid phase at this temperature. (b) The third law of thermodynamics says that the entropy of either phase must go to zero as 0→T . Therefore, the difference in entropy S∆ between the two phases, and hence the slope of the phase boundary, must go to zero as 0→T . (c) In an adiabatic, quasistatic compression, the entropy should remain unchanged. However, in the case of 3He, conversion to a solid at very low temperatures tends to increase the entropy of the system. Therefore, to keep the entropy a constant, the temperature of the system must drop. This method of cooling 3He makes it possible to reach temperatures as low as 1 mK. Schroeder 5.35 The Clausius-Clapeyron equation gives the relationship between pressure and temperature that describes the phase boundary between two phases, in terms of the change in entropy S∆ and change in volume V∆ between the two phases, as: VT L V S dT dp ∆ = ∆ ∆ = where L is the latent heat per mole. Consider the phase boundary between the gas phase and the liquid phase, where PRTVVV liquidgas /≈−=∆ for 1 mole, assuming that liquidgas VV >> and that the gas can be treated as an ideal gas. We further assume that L is a constant over some temperature and pressure range, to get: )/exp(constantconstant)ln(22 RTLpRT LPdT RT L P dPP RT L dT dp −×=⇒⎟ ⎠ ⎞ ⎜ ⎝ ⎛−×=⇒=⇒= 2    Adkins 10.1 At 1 atm, the melting temperature Tm for lead is 327°C = 600K. From the Clausius-Clapeyron relation L VPTT VT L dT dp ∆∆ =∆⇒ ∆ = For 1 kg of lead, L = 24.5 kJ and the difference in volume between the liquid phase and the solid phase of lead is give be 3434 3 4 1003.010101.1 1 10065.1 1 mmmV −×≈⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × − × =∆ Therefore, for p∆ = 99 atm ≈ 99×105 N/m2, the change in the melting temperature is given by 7.327)100(72.0 105.24 1003.06001099 3 34 2 5 ≈=⇒= × ×× ××=∆ − atmpTK J mK m NT m °C Adkins 10.2 We showed in class, and also in problem 5.35 from Schroeder, that the vapor pressure between liquid and gas, or solid and gas, can be approximated as )/exp(0 RTLPP −×≈ We can take the logarithm on both sides to get RTLPP /)mmHg/ln()mmHg/ln( 0 −= where the pressures P and Po are divided by 1 mm Hg to make the quantity inside the logarithm dimensionless. Compare the above equation with the vapor pressure relation for solid magnesium for temperatures between 700 and 739K: 6.19107.1)mmHg/ln( 4 + × −= T KP Therefore L/R = 1.7×104 K⇒ kJ/mol 141K107.1J/mol/K 3.8 4 ≈××≈L