2. We study how barometric pressure varies with altitude. Let P(2) be a function that gives the barometric pressure at altitude z (in meters), with = = 0 corresponding to sea level. From Newton's law it is not too difficult to show (you can do it for fun!) that dP dz = -p(=)9. where p(z) is the density of air at altitude z. Note that the pressure at sea level is roughly Po = P(0) = 101.325 kPa, and the gravitational acceleration is g ~ 9.8 m/$2. (a) First, assume that p(2) = po is a constant. Solve the differential equation to find P(=). According to this model, assuming that the density of air is about po = 1.225 kg/mo, what is the pressure on Mount Robson (= = 3954 m)? On Mount Everest (= = 8848 m)? Is that realistic? (b) Let us improve our model by assuming instead that air is compressible, in which case the density varies with altitude. From the ideal gas law, we find that M P(2 ) = RT P(=), where M ~ 0.02897 kg/mol is the molar mass of Earth's air (which we assume to be constant), R ~ 8.314 kg . m'/(s' . mol . K) is the gas constant, and T is the temperature. We assume that the temperature is constant: T = 288 K. Solve the differential equation to find P(=). According to this model, what is the pressure on Mount Robson (= = 3954 m)? On Mount Everest (z = 8848 m)? Is this more realistic than the first model? (c) But... is temperature really constant with altitude? Certainly not. Let us now assume that T decreases linearly with altitude: T(2) = To - Le for some constants To and L (L is called the lapse rate). We end up with the differential equation dP gMP(z) dz R(To - Lz)Solve the differential equation to find P(2). Assuming that To = 288 K and L = 0.0065 K/m, what is the pressure on Mount Robson (2 = 3954 m)? On Mount Everest (2 = 8848 m)? Is that model realistic