As a thunderstorm passes over an area, typically the air temperature will decrea
ID: 802838 • Letter: A
Question
As a thunderstorm passes over an area, typically the air temperature will decrease and the surface pressure will increase. The increase in pressure is a hydrostatic response to the changing temperature (and density) of the air. On 6 May 2005 a wx station in Boulder, Colorado reported a decrease in temperature from 74 degrees F to 59 degrees F and an increase in pressure from 826 hPa to 827.3 hPa as a thunderstorm passed over the station. Assume that the cooling of the air was uniform over some depth of the atmosphere. Calculate the depth of the air using the hypsometric equation that would need to be cooled by the observed amount to produce the observed change in pressure.
Hypsometric: RT/g(ln)(P1/P2)=change in z
Explanation / Answer
answer- when a thunderstorm passing over an area it is characterized by decrease in temperature and increse in pressure. thunderstorm occurs for a short period of interval of time. this increse in pressure during thunderstorm events it is hydrostatic respone to the changing temperature. temperature decrese and density also increse . cooling of air is uniform over somedepth.
T1 = 74 , T2 = 59 and i will take final temperature as temperature at the time of thunderstorm generation. temperature unit should be in Kelvin So T= 59 farenheit = 288.15K
Pressure P 1 = 82 hpa and pressure P 2 = 827.3
hypsometric equation- it shows the varation in the geopotential height with respect to pressure depend on temperature
Zt = Z1 - Z2 = RT/g * log (P1/P2)
where R is universtal gas contant for dry air , T is temperature in kelvin and P is pressure
Z = 8.314*288.15 / 9.8 log (826.3/827.3)
So depth of the air will be 242 meter which will need to be cooled by the observed amount to produce change in pressure.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.