Question

A:What must be the density of the oil?

B:If the vehicle is taken to Mars, where the acceleration due to gravity is 0.379 g, what will be the pressure difference (in earth atmospheres) between the top and bottom of the oil column?

Answer 1

QUESTION IS NOT COMPLETE IF U HAVE THIS QUESTION

You are designing a machine for a space exploration vehicle. It contains an enclosed column of oil that is 1.50 m tall, and you need the pressure difference between the top and the bottom of this column to be 0.120 atm.
A)What must be the density of the oil?
B)If the vehicle is taken to Mars, where the acceleration due to gravity is 0.379 g, what will be the pressure difference (in earth atmospheres) between the top and bottom of the oil column?

ANSWER IS

Assuming the the fluid (oil) is incompressible (no change in density over the length of the column) and constant temperature you can use the following simple relationship:

dP = den * g * height

where dP is the pressure gradient across the column (in this case dP = 0.12 atm). den is the density of the fluid, g is the gravitational constant and height is...well the height of the column of oil. To solve the first part of your question we will assume that your space vehicle is positioned on the surface of the earth so assume g = 9.8 m/sec^2.

Now solve the equation for "den" and rewrite dP in N/m^2 instead of atm. N/m^2 are also referred to as Pascals (Pa). 1 atm = 101000N/m^2 or 101 kN/m^2.

den = (0.12 atm * 101000 N/m^2/1 atm)/(9.8m/sec^2 * 1.5m) = 824 kg/m^3 <=== density of the oil

Now solving part B...with the space craft on the Martian surface and assuming the same temperature of the oil as on Earth. Using the same equation as above but replacing the gravitational constant with g = 0.379 * 9.8m/sec^2 = 3.71 m/sec^2

Determine the pressure differential in order to maintain a column of oil at a height of h = 1.5 m.

dP = (824kg/m^3)*(3.71m/sec^2)*(1.5m) = 4591 N/m^2

To get this in Earth atmospheres:
dP = 4591 N/m^2 * (1atm/101000N/m^2) = 0.0454 atm