Homework Assignment # 3 solutions and discussion1. Contouring exercise: On the back of this page is a simplified map of sea level pressure as reported by weather stations across North America at 10 p.m. Eastern Time on 27 January 1999. (A real weather map would contain many more stations and other data in addition to pressure). Draw isobars (lines of constant pressure) on the figure, from 1012 mb up to 1028 mb, contouring every 4 mb (i.e. 1012, 1016, 1020, 1024 and 1028 mb). Label high pressure and low pressure centers with "H" and "L", respectively. Hints: Use a pencil first, then smooth the contours as you trace them in ink. The 1020 mb isobar is probably the easiest contour line to start your contouring with. Remember the guidelines listed on the back of the sheet handed out in quiz section. Bonus: Winds blow counterclockwise around low pressure centers and clockwise around high pressure centers,parallel to the isobars (we will explain this phenomenon in class later). The strength of the wind is proportional to the pressure gradient, which is represented on your drawing by how closely the isobars are spaced. Using this information, draw a few sample wind arrows on your figure; make the length of the arrow proportional to the wind speed.See the solution in class or posted outside the TA office. Also make sure you understand all the rules for contouring! Especially: label your contours; do not extend your contours too far from where you have data; and make sure all the values on one side of a isobar are greater than its value, while values to the other side of the contour are less.2. a) I have 6 kg of dry air in a rigid box. The volume of the box is 5 m3. Using the formula for density, =mass/volume, calculate the density of the air in the box.density = mass / volume = 6 kg / 5 m3 = 1.2 kg/ m3 b) Now I add some air (1.8 kg) to the box, without changing the volume. Calculate the new air density in the box.density = (6.0 kg + 1.8 kg) / 5 m3 = 7.8 kg / 5 m3 = 1.56 kg/ m3 c) Now consider a balloon containing the same mass and with the same volume as the box in part (a). Initially, the density inside the balloon is the same as inside the box. Now we raise the balloon several hundred meters. What happens to the balloon? Why?If the air pressure outside the balloon is less than the air pressure inside the balloon, the walls of the balloon will experience a net outward force and the balloon will quickly expand. If the balloon expands, the air density inside the balloon will decrease because the mass of air inside the balloon does not change but the volume of the balloon increases (=mass/volume). We also know that the air inside the balloon will cool during the expansion process because expansion of the balloon's walls requires work (read: energy), which comes from the internal kinetic energy of the molecules of air inside the balloon. So if the density of the air inside the balloon decreases, and the temperature of the air inside the balloon also decreases, we know from the ideal gas law that the pressure inside the balloon must also decrease as it expands (p=RT).Air pressure always decreases with height in the atmosphere. So as the balloon rises in the atmosphere, it experiences less pressure from the outside and the balloon must expand and cool until the pressure inside the balloon equals the pressure outside the balloon. At anygiven height, the pressure inside the balloon must equal the pressure outside the balloon or the balloon will quickly expand or contract until this condition is met. Note that the reason the balloon expands is because pressure decreases with height, NOT because density decreases with height. The density inside the balloon and outside the balloon may be very different, but as long as the pressure inside the balloon and the pressure outside the balloon are the same, the balloon will remain the same size.d) If the volume of the balloon increases to 5.8 m3 in part (c), but the mass of air inside the balloon stays the same, what is the new density of air inside the balloon? Do you think this is greater than, less than, or equal to the density of air outside the balloon at this height? How about the pressure inside the balloon versus outside the balloon?density = 6 kg / 5.8 m3 = 1.03 kg/ m3 We have already noted that the pressure inside the balloon must be the same as the pressureoutside the balloon. Figuring out the air density inside the balloon versus the air density outside the balloon is a little trickier, and you actually do not know for sure which is higherunless you also know the temperature both inside the balloon and outside the balloon. But here is a good argument for why the density is probably greater inside the balloon: recall that the environmental lapse rate in the troposphere is about -6.5C/km on average, but the dry adiabatic lapse rate is about -10C/km. So if the balloon rises, say, 1 km, the air inside the balloon will be 10C colder than it was at the surface (assuming that the air inside the balloon is relatively dry), while the temperature outside the balloon is probably only about 6.5C colder than the surface air. Since the pressure inside and outside the balloon is the same, then the colder air (the air inside the balloon) will be less dense than the warmer air (the air outside the balloon). You can work this out for yourself using the ideal gas law for both gases:p(inside)=(inside)*R*T(inside)p(outside)=(outside)*R*T(outside)since p(inside)=p(outside), (inside)*R*T(inside)=(outside)*R*T(outside) (inside)=(outside)*T(outside)/T(inside)note that if T(outside) is greater than T(inside), then (inside) will be larger than (outside). You certainly did not have to do this calculation to get credit for this problem, but this is a good illustration of how density, pressure, and temperature are related.e) Do you think the air inside the balloon at this height is colder or warmer than the air inside the box on the ground in part (a)? How do you know?From the argument in part (c), we know that the balloon must have cooled as it rose and expanded. Again, the work of expanding the walls of the balloon requires energy, which comes from the internal kinetic energy (temperature) of the air inside the balloon. Since thebox did not rise, it is at the same temperature as it started. So the temperature of the air inside the balloon is colder than the temperature of the air in the box. Make sure that you make the