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1.Manometer fluid = Tungsten Hexafluoride WF6; Density = 13,000 g/m3; MW = 297.83 g/mol You have sent a probe to an unexplored planet and need to figure out what the gravity is on the planet before you can go there. The probe is equipped with a manometer that was calibrated on earth. Pressurized tanks are attached to each side with Tank A being a constant pressure. On earth, the Tank B was set to a pressure of 40 Pa which gave a height difference of 21 cm in the monometer fluid with the tank B side fluid being lower. The manometer was sent to the planet and where tank B pressure was set to 3.7 Pa. This gave a height difference of 6.1 cm with the fluid being higher on the tank B side. What is the gravity constant of the new planet? (Note: density of fluid will not change, and the fluid is immiscible with the other gases)
This problem can be solved by comparing the manometer equations in two different situations i.e at the earth and at an unexplored planet.
Manometer fluid density = 13000 g/m³ = 13000*10-3 kg/m³
Considering the case in earth we develop an equation as follows:
Pa = a constant., Pb1= 40 Pa , h1= 21cm = 21*10-2 m , g = 9.8 m/s² , rho = density of manometer fluid = 13000 *10-3 kg/m³
Since level of fluid in tank B is lower it means that Pb1> Pa.
Therefore Pb1 - Pa = h1* (rho) * g
40- Pa = 21*10-2 * 9.8 * 13000*10-3 let it be equation 1
Next on on considering the case of unexplored planet the equation is as follows:
Pa = a constant., Pb2 = 3.7 Pa, h2 = 6.1cm = 6.1* 10-2 m , let gravity constant in this planet be g p which is to be found out.
rho remains the same for both planet.
Pa - Pb2 = h2* (rho)* g p
Pa - 3.7= 6.1*10-2 * g p * 13000*10-3 let it be equation 2
On addition of both equation 1&2 Pa get cancelled
We obtain 40 -3.7 = ( 13000*10-3 *9.8* 21*10-2) +
( 13000*10-3 * g p * 6.1* 10-2)
Hence g p can be found out by rearranging the above equation
And g p is found to be 12.037 m/s²
Therefore gravity constant of new planet is found to be 12.037 m/s².