/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A manometer using oil (density \... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 25

A manometer using oil (density \(0.90 \mathrm{g} / \mathrm{cm}^{3}\) ) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases by \(0.74 \mathrm{cm}\) Hg. (a) By how much does the fluid level rise in the side of the manometer that is open to the atmosphere? (b) What would your answer be if the manometer used mercury instead?

Short Answer

Expert verified
Answer: (a) When oil is used as the manometer fluid, the fluid level rises by approximately 1.206 cm. (b) When mercury is used as the manometer fluid, the fluid level rises by approximately 0.080 cm.

Step by step solution

01

Given data and conversion of units

We are given the manometer fluid(oil) density \(\rho_{oil} = 0.90 \, \mathrm{g/cm^3}\). We know the density of mercury, \(\rho_{Hg} = 13.6 \, \mathrm{g/cm^3}\). The air pressure increased by \(\Delta P = 0.74 \, \mathrm{cm\,Hg}\). To calculate the variation in fluid level, let's first convert cm Hg to a pressure unit (N/m^2 or Pa). We'll use the factor \(1\,\mathrm{cm\,Hg} = 13.5951 \, \mathrm{kg\,m^{-1}\,s^{-2}}\): $$\Delta P = 0.74 \, \mathrm{cm\, Hg} \times 13.5951 \, \mathrm{kg\,m^{-1}\,s^{-2}}$$
02

Calculate the pressure increase in Pa

$$\Delta P \approx 10.0601 \, \mathrm{kg\,m^{-1}\,s^{-2}}$$
03

Calculate the variation in fluid level for oil

To find the variation in fluid level for oil, we can now use the formula \(\Delta P = \rho_{oil} \cdot g \cdot \Delta h_{oil}\), where \(\Delta h_{oil}\) is the change in fluid level we want to find. Rearranging for \(\Delta h_{oil}\), we get: $$\Delta h_{oil} = \frac{\Delta P}{\rho_{oil} \cdot g}$$ To convert the density of oil to \(\mathrm{kg \cdot m^{-3}}\), we have \(\rho_{oil} = 0.9 \mathrm{g\,cm^{-3}} \times 1000 \, \mathrm{kg\,m^{-3}}\). Using \(g \approx 9.81\,\mathrm{m/s^2}\), we can now plug in the values and find the variation in fluid level for oil: $$\Delta h_{oil} = \frac{10.0601\, \mathrm{kg\,m^{-1}\,s^{-2}}}{900 \, \mathrm{kg\,m^{-3}} \cdot 9.81 \, \mathrm{m/s^2}}$$
04

Compute the change in fluid level for oil

$$\Delta h_{oil} \approx 0.001206 \, \mathrm{m} = 1.206 \, \mathrm{cm}$$ The fluid level rises by approximately 1.206 cm in the side of the manometer that is open to the atmosphere when oil is used as the manometer fluid.
05

Calculate the variation in fluid level for mercury

For (b), we follow the same steps (3&4) with \(\rho_{Hg}\) instead of \(\rho_{oil}\): $$\Delta h_{Hg} = \frac{\Delta P}{\rho_{Hg} \cdot g}$$ To convert the density of mercury to \(\mathrm{kg\cdot m^{-3}}\), we have \(\rho_{Hg} = 13.6 \mathrm{g\,cm^{-3}} \times 1000\, \mathrm{kg\,m^{-3}}\). Using the same value of \(g\), we can now plug in the values for mercury and find the change in fluid level: $$\Delta h_{Hg} = \frac{10.0601\, \mathrm{kg\,m^{-1}\,s^{-2}}}{13600 \, \mathrm{kg\,m^{-3}} \cdot 9.81\, \mathrm{m/s^2}}$$
06

Compute the change in fluid level for mercury

$$\Delta h_{Hg} \approx 0.000080 \, \mathrm{m} = 0.080 \, \mathrm{cm}$$ The fluid level rises by approximately 0.080 cm in the side of the manometer that is open to the atmosphere when mercury is used as the manometer fluid.

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