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Chapter 10: Problem 21

Atmospheric pressure on Venus is about 90 times that on Earth. A steel sphere with a bulk modulus of 160 GPa has a volume of \(1.00 \mathrm{cm}^{3}\) on Earth. If it were put in a pressure chamber and the pressure were increased to that of Venus (9.12 MPa), how would its volume change?

Short Answer

Expert verified
Answer: The approximate volume of the steel sphere on Venus is 0.99994363 cm鲁.

Step by step solution

01

Convert GPa to MPa

To maintain consistency between pressure units, we will convert the bulk modulus of steel from GPa to MPa by multiplying by 1000. 1 GPa = 1,000 MPa So, the bulk modulus B = 160 GPa 脳 1000 = 160,000 MPa
02

Calculate the change in pressure

We know that the atmospheric pressure on Venus is 90 times that of Earth. Since we are given the pressure on Venus (9.12 MPa), we can find Earth's pressure and calculate the change in pressure. Earth's pressure = Venus' pressure / 90 Earth's pressure = 9.12 MPa / 90 = 0.101 MPa Now, we can calculate the change in pressure (\(\Delta P\)) which is the difference between pressure on Venus and pressure on Earth. \(\Delta P\) = Pressure on Venus - Pressure on Earth \(\Delta P\) = 9.12 MPa - 0.101 MPa = 9.019 MPa
03

Calculate the change in volume

Using the formula for the change in volume due to pressure, we can now calculate the change in volume of the steel sphere (\(\Delta V\)). \(\Delta V = \frac{-V_0 \Delta P}{B}\) \(\Delta V = \frac{-(1.00 \mathrm{cm}^3)(9.019 \mathrm{MPa})}{160,000 \mathrm{MPa}}\) \(\Delta V \approx -0.00005637 \mathrm{cm}^3\)
04

Calculate the final volume of the sphere on Venus

Now we can find the final volume (\(V_f\)) of the sphere by adding the change in volume to the initial volume. \(V_f = V_0 + \Delta V\) \(V_f = 1.00 \mathrm{cm}^3 - 0.00005637 \mathrm{cm}^3\) \(V_f \approx 0.99994363 \mathrm{cm}^3\) The volume of the steel sphere on Venus would be approximately \(0.99994363 \mathrm{cm}^3\).

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Most popular questions from this chapter

A gibbon, hanging onto a horizontal tree branch with one arm, swings with a small amplitude. The gibbon's CM is 0.40 m from the branch and its rotational inertia divided by its mass is \(I / m=0.25 \mathrm{m}^{2} .\) Estimate the frequency of oscillation.
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