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GEF2200 Spring 2018: Solutions Thermodynam- Ics 1 GEF2200 spring 2018: Solutions thermodynam- ics 1 A.1.T What is the difference between R and R∗? R∗ is the universal gas constant, with value 8.3143JK−1mol−1. R is the gas constant for a specific gas, given by R∗ R = (1) M where M is the molecular weight of the gas (usually given in g/mol). In other words, R takes into account the weight of the gas in question so that mass can be used in the equation of state. For the equation of state: m pV = nR∗T = R∗T = mRT (2) M It is important to note that we usually use mass m in units of [kg], which requires that the units of R is changed accordingly (if given in [J/gK], it must be multiplied by 1000 [g/kg]). A.2.T What is apparent molecular weight, and why do we use it? Apparent molecular weight is the average molecular weight for a mixture of gases. We introduce ∗ it to calculate a gas constant R = R =Md for the mixture , where Md is the apparent molecular weight of i different gases given by Equation (3.10): P m P m P n M M = i = i = i i d P mi n n Mi X ni = M (3) n i In meteorology the most common apparent molecular weight is the one of air. WH06 3.19 Determine the apparent molecular weight of the Venusian atmosphere, assuming that it consists of 95% CO2 and 5% N2 by volume. What is the gas constant for 1 kg of such an atmosphere? (Atomic weights of C, O and N are 12, 16 and 14 respectively.) Concentrations by volume (See exercise A.8.T): V v = N2 (4) N2 V V v = CO2 (5) CO2 V 1 Assuming ideal gas, we have total volume V = VN2 + VCO2 at a given temperature (T ) and pressure (p). This means volume concentration is equal to molecular fraction: V n R?T=p n v = i = i = i (6) i V nR?T=p n The appearent molecular mass is the molecular mass a mixture of n moles of gass and a total mass m appears to have: m P m P n M M = = i = i i ve n n n X ni X = M = v M (7) n i i i Inserting values for CO2 and N2: Mve = 0:95 × (12 + 2 × 16) g/mol +0:05 × (2 × 14) g/mol = 43:2g/mol (8) This is quite higher than for our atmosphere, and gives the gas constant R? 8:3143JK−1mol−1 Rve = = −1 Mve 43:2gmol = 0:1925Jg−1K−1 = 192:5Jkg−1K−1 (9) which is lower than for the Earth’s atmosphere. A.4.T Show that the gas constant for moist air is greater than for dry air. We have that R∗ R = (10) M So proving that the gas constant is greater for moist than dry air is the same as proving that the apparent molecular weight of moist air, Mw, is smaller than that for dry air, Md. For dry air the apparent molecular weight is: P m m M = i = d (11) d P mi n Mi d Adding mass mv of water vapor with molecular weight Mv gives P mi + mv md + mv Mw = = P mi + mv md + mv Mi Mv Md Mv md + mv = Md (12) m + m Md d v Mv 2 Since Mv < Md, we see that Mw < Md (lighter than dry air), which means that Rw > R. If, on the other hand, we had added a gas x with Mx > Md, then Mnew > Md (heavier than dry air) and Rnew < R. A.5.T Why do we introduce virtual temperature? Because R is dependent on humidity, we move this dependency from R to T : RT = RdTv (13) so that we can use the gas constant for dry air in calculations. Virtual temperature is the temperature dry air would have to have the same density at the same pressure as the moist package. Since moist air has lower apparent molecular weight than dry air, this means that viritual temperature must always be higher than absolute temperature. A.6.T Should we use virtual temperature when the gas in question is water vapor? When we look at a single gas we use the gas constant for that gas. For water vapor we use −1 −1 Rv = 461JK kg . We therefore do not use virtual temperature. WH06 3.20 If water vapor comprises 1% of the volume of the air (i.e. if it accounts for 1% of the molecules in air), what is the virtual temperature correction? The virtual temperature depends on the temperature and ratio between the partial pressure of water vapor over the total pressure e=p. Since we assume that the gasses can be approximated as ideal gases, the volume fraction is equal to the molefraction and thus equal to the e=p by the ∗ ∗ e eV nvR T nv equation of state pV = nR T : = = ∗ = . p pV ndR T nd Finally, we can now compute Tv as a function of T , by using e=p = 0:01. T Tv = e 1 − p (1 − ) 1 T =T (14) v 1 − 0:01(1 − 0:622) T v = 1:00379 T If we take T = 288K as an example, then Tv = T · 1:00379 = 289:09K. A.7.T The pressure of water vapor in a sample of air at 20◦C taken at sea level is 18hPa. What is the 3 mole fraction of water vapor? Hint: Make use of Dalton’s law and equation (3.6). We use that the pressure at sea level is approximately 1013hPa. The ideal gas law as stated in (3.6) is pV = nR∗T (15) The partial pressure exerted by water vapor is the pressure it would exert had it alone occupied the same volume as the total gas mixture. We may therefore use (3.6) again on water vapor alone: ∗ eV = nvR T (16) where nv is the number of moles of water vapor contained in the volume V . Combining the two equations, we find that n e 18hPa v = = = 0:0178 (17) n p 1013hPa (From this deduction, we see that the mole fraction of any gas component in a mixture equals the pressure fraction exerted by that component.) So the mole fraction of water vapor in this case is 1.78 %. A.9.T Derive the hydrostatic equation. Why must the atmospheric pressure decrease with height? The hydrostatic equation gives the relationship between gravity and pressure for an air mass at rest, a relationship between a change in pressure dp and a change in height dz (see Figure 1). Since the pressure at a given height is due to the weight of the above air, we expect the pressure to decrease with height, meaning that a decrease in pressure equals an increase in height. We need to consider the forces acting upon the layer, and to do so, we look at the edges of the layer (height z and z + dz. Gravity works downwards, and the force on the top of the layer is Fg(z + dz). Furthermore, the layer acts on the air below with a force Fg(z). Since the layer is at rest, we have from Newton’s thrid law that the layer experience a similar force in the opposite direction (−Fg(z)). For a body with mass m, the gravity is mg = %V g, where V is the volume. Given that the layer has a thickness dz and area A, and therefore a volume dV = Adz, we have (assuming positive axis upwards): Fg = Fg(z + dz) − Fg(z) = −%gA(z + dz − z) = −%gAdz (18) Forces acting upwards: Pressure (force per area). The upward force at the layer top is opposed by a similar force acting on the layer (−Fp(p + dp)). Fp = Fp(p) − Fp(p + dp) = [p − (p + dp)] A = −Adp (19) 4 Figure 1: The forces on a layer of air. Where Fp(p) > Fp(p + dp), that is, dp is negative. Fp is then value of the force acting upwards. To find the equilibrium between gravity and pressure forces, we set Fp + Fg = 0, and cancel A. −%gdz − dp = 0 (20) It is very common to do such calculations per area, leaving out A from the start. When using vector calculation (k-vector): Fgk = −%gA(z + dz)k − (−%gAz)k = −%gAdzk (21) Fpk = [p − (p + dp)] Ak = −Adpk (22) In any case, when the sum of forces is zero, we get dp = −%g (23) dz A.10.T What is the geopotential? Use this quantity together with the hydrostatic equation and the ideal gas law to derive the hypsometric equation. The geopotential at a given location is the energy required to lift 1 kg from sea level and up to that location. Z z φ(z) = gdz0 (24) 0 5 Dividing by the average surface acceleration of gravity, we get the geopotential height. 1 Z z Z = gdz0 (25) g0 0 Using the hydrostatic equation, the equation of state and the definition of virtual temperature: dp p p = −%g = − g = − g (26) dz RT RdTv Using the definition of geopotential height gdz = g0dZ (27) we have p dp = − g0dZ (28) RdTv Which we integrate from p1;Z1 to p2;Z2: R T dp dZ = − d v g0 p Z Z2 Z p2 R T dp dZ = − d v (29) Z1 p1 g0 p Z p2 Rd dp = − Tv (30) g0 p1 p By introducing the weighted mean virtual temperature R p2 T dp p1 v p Tv = (31) R p2 dp p1 p and inserting into Equation (30), we may write Z p2 Rd dp Z2 − Z1 = − Tv (32) g0 p1 p Rd p2 = − Tv ln (33) g0 p1 This means that the thickness is linearly proportional to the mean temperature of the layer.
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