页数：51页文件大小：3.46 MB来源： Atkins’PhysChem物理化学ppt课件
本课件讲第一定律的方法论：用状态函数与全微分的数学性质建立物理量之间的联系，推导内能与焓随温度、体积、压力的变化关系，并讨论焦耳实验、焦耳-汤姆逊效应、转化温度与气体液化。
Brief review Δ U 0 - p ex Δ VnRTlnw 0 q V = Δ Uqp = Δ H - wqnCp Δ TnCp Δ T = Δ U + VdpnCp Δ T = Δ U + pdV 0 Δ HnCV Δ TnCV Δ TnCV Δ T 0 Δ U q = 0 d V = 0 d p = 0 d T = 0
Brief review
In this chapter we will discuss:
State functions and exact differentials 3.1 state and path functions 3.2 exact and in exact differentials Thermodynamic consequences 3.3 Changes in internal energy 3.4 The temperature dependence of the enthalpy 3.5 The relation between CVandCp
内能的变化
To use the mathematical properties of state functions to draw far - reaching conclusions about the relations between physical properties and establish connections that maybe completely unexpected. To combine measurements of different properties to obtain the value of a property we require.
State functions and exact differentials
Like in mountain climbing The altitude is independence of path, but the climbing distance depends on the path. 3.1 State and path functions
全微分与非全微分
Heating a system from state i to state f “ dU ” is an exact differential An exact differential is an infinitesimal quantity that, when integrated, gives a result that is independent of the path between the initial and final states.
3.2 Exact and in exact differentials
全微分与非全微分
The total energy as heat is the sum of all stages This equation is different with Δ U. First, we cannot write Δ q = q f – q i. Second, we must specified the path of integration. So “ d q ” is an in exact differential. Usually, dqis written as δ q to emphasize that it is in exact.
Also work is an path - dependent function, and d w often written as δ w. 3.2 Exact and in exact differentials
How to use the properties of state functions to establish unexpected relation between the physical properties of a system. Thermodynamic consequences
(a) General considerations 3.3 changes in internal energy … U = f (V, T) V V + d V, U U’
(a) General considerations 3.3 changes in internal energy … T T + d T, U U’
V V + d V, T T + dTU U’ (a) General considerations … U = f (V, T)
(a) General considerations
(a) General considerations … ( U / V) T means the variation of the internal energy of a substanc
(a) General considerations If dU > 0 as the V of the sample expands at constant temperature, that
焦耳实验
James Joule tried to measure the π T by observing the change in temperature of a gas when it expands into a vacuum. When the stop cock was opened, he observed no change in temperature. No temp change: q = 0 Free expansion: w = 0 Hence: U = 0 (b) The Joule experiment
Actually, the heat capacity of his crude apparatus was so large, that the temperature change caused by this expansion of gas was simply too small to measure - so small deviations of real gases were not detected... (b) The Joule experiment
内能与温度压力的关系
How the internal energy varies with temperature when the pressure of the system is kept constant. If we divide both sides of the eqn by d T at const. p. (c) changes in internal energy at constant p We have The expansion coefficient, , of a substance is defined as: Th
H = f (p, T) 3.4 The temperature dependence of the enthalpy or (a) Changes in enthalpy at constant volume:
To divide the eqn bydT at constant volume: (a) Changes in enthalpy at constant volume where T: the isothermal compressibility : Joule - Thomson coefficie
(a) Changes in enthalpy at constant volume
(b) The isothermal compressibility For a perfect gas The higher the pressure of the gas, the lower its co
(c) The Joule - Thomson effect … At the constant enthalpy, orisenthalpic process, the gas expands throu
(c) The Joule - Thomson effect … p 1 p 2 T 1 T 2 Adiabatic wall 多孔材料Throttle process 节流膨胀
(c) The Joule - Thomson effect … The work is done when gas passes through the barrier. At high p side,
(c) The Joule - Thomson effect … w = w i + w f = piVi – pfVf = U f – U i Therefore, U f + pfVf = U i +
(c) The Joule - Thomson effect … μ is measured through μ T, which is called isothermal Joule - Thomson
(c) The Joule - Thomson effect … To measure μ T, the gas is pumped continuously at a steady pressure th
(c) The Joule - Thomson effect
(c) The Joule - Thomson effect … The energy provided by the heater is monitored. The heat can be identi
(c) The Joule - Thomson effect … Real gases have nonzero Joule - Thomson coefficients. Depending on The
(c) The Joule - Thomson effect … In Joule - Thomson experiment d p < 0 If μ < 0, hencedT > 0 The gas he
(c) The Joule - Thomson effect … A gas typically has two in version temperatures, one at high temperatu
(c) The Joule - Thomson effect For a perfect gas, μ = 0, Hence, Δ T = 0 after Joule - Thomson expansi
3.5 the relation between CVandCp … C V ≠ CpFor the work needed to change the V of the system. There are t
According to the definition: Internal pressure Expansion coefficient We have: 3.5 the relation between CVandCp
3.5 the relation between CVandCp … For a perfect gas, P V = nRT
例：在外包绝热材料的带活塞气缸中，原放有101325 P a， 25 ℃ 的一定量某理想气体，将其压缩至192517.5 P a 时，气体温度升至79 ℃。试求每压缩1 mol 该气体时所作的功和系统的焓变。已知该气体的摩尔定容热容近似为25.3 J K - 1 mol - 1。
例： 273 K， 1.0 MP a， 0.01 m 3 的He (g) 经(1) 绝热可逆过程； (2) 绝热且外压恒定在100 kPa 的过程膨胀到末态压力为100 kPa。分别求此二过程的q， w， U 和 H？解：
He (g) n = 4.403 molT 1 = 273 K p 1 = 1.0 × 106 P a V 1 = 0.01 m 3 He (g) n = 4.403 molT 2 = ? P 2 = 1.0 × 105 P a V 2 = ? (1) q = 0, 可逆(2) q = 0 p ex = p 2
(1)
(2) 不是可逆过程， ∴ 不能用绝热过程方程式
Homework Page 68, exercises 2.42 a Page 69, problems 2.16, 3.8, 3.14
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