Question

In: Other

1.Manometer fluid = Tungsten Hexafluoride WF6; Density = 13,000 g/m3; MW = 297.83 g/mol You have...

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)

Solutions

Expert Solution

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.


Related Solutions

A 0.7257 g mixture of KCN (MW = 65.116 g/mol) and NaCN (MW = 49.005 g/mol)...
A 0.7257 g mixture of KCN (MW = 65.116 g/mol) and NaCN (MW = 49.005 g/mol) was dissolved in water. AgNO3 was added to the solution, precipitating all the CN– in solution as AgCN (MW = 133.886 g/mol). The dried precipitate weighed 1.650 g. Calculate the weight percent of KCN and NaCN in the original sample.
If you are given the following solids: Na3PO4             MW = 163.94 g/mol K3PO4                  MW = 212.26
If you are given the following solids: Na3PO4             MW = 163.94 g/mol K3PO4                  MW = 212.26 g/mol    NaH2PO4           MW = 119.98 g/mol K2HPO4             MW = 174.17 g/mol Na2HPO4           MW = 141.96 g/mol KH2PO4             MW = 136.08 g/mol Which compounds would you use to prepare your stock solutions? Why? I picked KH2PO4 and K2HPO4 because potassium salts dissolve more readily in H2O which would make the lab go more smoothly. I would like to know if I'm missing any...
1. You have 0.95ft^3 aluminum(density=2700kg/m3,MW=26.98). Calculate the mass in kg, lbm, gmole and lbmole. 2. Assume...
1. You have 0.95ft^3 aluminum(density=2700kg/m3,MW=26.98). Calculate the mass in kg, lbm, gmole and lbmole. 2. Assume there are 39 inches in a meter.What is the external surface area in the SI system of a sphere that is 9.0 inches in diameter? Determine the volume in cubic feet. (Note SI system: meters / kg / Newtons etc.) 3. If a US gallon has a volume of 0.134 ft3 and a human mouth has a volume of 0.950 in3, how many mouthfuls...
A sample containing a mixture of SrCl2·6H2O (MW = 266.62 g/mol) and CsCl (MW = 168.36...
A sample containing a mixture of SrCl2·6H2O (MW = 266.62 g/mol) and CsCl (MW = 168.36 g/mol) originally weighs 1.7215 g. Upon heating the sample to 320 °C, the waters of hydration are driven off SrCl2·6H2O, leaving the anhydrous SrCl2. After cooling the sample in a desiccator, it has a mass of 1.2521 g. Calculate the weight percent of Sr, Cs, and Cl in the original sample.
MW (g/mol) / MP ( oC) / Solubility in cold H2O (g / 100 mL) /...
MW (g/mol) / MP ( oC) / Solubility in cold H2O (g / 100 mL) / Solubility in boiling H2O (g / 100 mL) Acetanilide 135.2 / 114 / 0.54 / 5.0 Phenacetin 179.2 / 135 / 0.076 / 1.22 Using solubility data in the table of reagents, calculate the minimum amount water needed to recrystallize: i. 1.5 g of phenacetin. ii. 1.5 g of acetanilide. b. If the minimum amount of water needed is used, how many grams of...
You are provided with a 0.1 mM solution of proflavine (MW: 209.25 g/mol), a 6 mM...
You are provided with a 0.1 mM solution of proflavine (MW: 209.25 g/mol), a 6 mM solution of DNA, and solid NaCl (MW: 58.44 g/mol). How many μL of the proflavine solution would you need to make a 25 mL of the following stock solution: 1 µM proflavine, 100 mM NaCl, and 60 µM DNA?
1)A typical commercial antacid tablet contains 0.350 g of CaCO3 [MW=100.09 g/mol] as an active ingredient....
1)A typical commercial antacid tablet contains 0.350 g of CaCO3 [MW=100.09 g/mol] as an active ingredient. How many moles of HCl could this tablet neutralize? 2)A student was analyzing an antacid tablet with a mass of 2.745 g. The student found that a 1.872 g sample of the tablet would neutralize 67.35 mL of stomach acid. Calculate how much stomach acid would be neutralized by the entire tablet. 3)A student determined that 29.43 mL of 0.132 M NaOH reacts with...
3.920 g of sodium acetate ( CH3COONa MW= 82.03 g/mol), is dissolved in 235.0ml of H2O...
3.920 g of sodium acetate ( CH3COONa MW= 82.03 g/mol), is dissolved in 235.0ml of H2O at 50 degree celsius (kw= K.48E-14). Calculate the pH of this soluion ( Ka= 1.63 E-5) Determine the % ionization
The enthalpy of combustion of octane is 5074.9 kJ/mol and its density is 703 kg/m3 ....
The enthalpy of combustion of octane is 5074.9 kJ/mol and its density is 703 kg/m3 . On a mild winter day a room of volume 40 m3 needs to be warmed up from 10 to 20 OC? (Consider: cair = 0.718 J/g K; dair = 1.225 g/L).
Calculate the solidification point of a 30% solution of CH3OH (MW = 32 G / MOL)...
Calculate the solidification point of a 30% solution of CH3OH (MW = 32 G / MOL) in water (MW = 18 G / MOL) KC of water = 1.86 ºC / m
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT