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 0.125 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 0.379g, what will be the pressure difference (in earth atmospheres) between the top and bottom of the oil column?

Answer 1

The pressure difference (in Earth's atmosphere) between the top and bottom of the oil column on Mars is 0.045 atm.

Step-by-step explanation:

(a) To find the density of the oil, we can use the formula for pressure difference in a fluid column: ΔP = ρgh, where ΔP is the pressure difference, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.

We know that the pressure difference is 0.125 atm, the height of the column is 1.50 m, and the acceleration due to gravity on Earth is 9.81 m/s². Plugging in these values, we get:

0.125 atm = ρ(9.81 m/s²)(1.50 m)

Solving for ρ, we get:

ρ = 0.00803 g/cm³

Therefore, the density of the oil must be 0.00803 g/cm³.

(b) If the vehicle is taken to Mars, where the acceleration due to gravity is 0.379g, we can use the same formula to find the pressure difference:

ΔP = ρgh

We know that the height of the column is still 1.50 m, but the acceleration due to gravity is now 0.379g. Plugging in these values, we get:

ΔP = (0.00803 g/cm³)(9.81 m/s²)(0.379)(150 cm)

Solving for ΔP, we get:

ΔP = 0.045 atm

Therefore, the pressure difference (in Earth's atmosphere) between the top and bottom of the oil column on Mars is 0.045 atm.

Answer 2

Step-by-step explanation:

(a) To find the density of the oil, we can use the formula for pressure difference in a fluid column:

ΔP = ρgh

where ΔP is the pressure difference, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.

Plugging in the given values, we have:

0.125 atm = ρgh = ρ(9.81 m/s^2)(1.50 m)

Solving for ρ, we get:

ρ = 0.125 atm / (9.81 m/s^2 x 1.50 m) ≈ 0.00847 g/cm^3

Therefore, the density of the oil must be approximately 0.00847 g/cm^3.

(b) On Mars, the acceleration due to gravity is 0.379 times that of Earth, or g_Mars = 0.379g_Earth. The pressure difference between the top and bottom of the oil column will be:

ΔP_Mars = ρgh_Mars = ρg_Earth(0.379)(1.50 m)

Using the density we found in part (a), we have:

ΔP_Mars = (0.00847 g/cm^3)(9.81 m/s^2)(0.379)(1.50 m) / (1 atm/101325 Pa)

ΔP_Mars ≈ 0.019 atm

So, the pressure difference between the top and bottom of the oil column on Mars will be approximately 0.019 atm, or about 0.15 times the pressure difference on Earth.