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Chapter 9: Problem 19

At the surface of a freshwater lake the pressure is \(105 \mathrm{kPa} .\) (a) What is the pressure increase in going \(35.0 \mathrm{m}\) below the surface? (b) What is the approximate pressure decrease in going \(35 \mathrm{m}\) above the surface? Air at \(20^{\circ} \mathrm{C}\) has density of $1.20 \mathrm{kg} / \mathrm{m}^{3} .$

Short Answer

Expert verified
Answer: The pressure increase in going 35.0 m below the surface of a freshwater lake is approximately 343.35 kPa, while the pressure decrease in going 35.0 m above the surface is approximately 5.37 kPa.

Step by step solution

01

Identify known values

The known values in this problem are: - Surface pressure \(P_0 = 105\,\text{kPa}\) - Depth \(h = 35.0\,\text{m}\) - Water density \(\rho_w = 1000\,\text{kg/m}^3\) (freshwater) - Gravitational acceleration \(g = 9.81\,\text{m/s}^2\)
02

Calculate hydrostatic pressure

To find the pressure increase, we will use the hydrostatic pressure formula: \(\Delta P = \rho_w gh\) Plug in the known values: \(\Delta P = (1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)(35.0\,\text{m})\)
03

Calculate the pressure increase

Solve for \(\Delta P\): \(\Delta P = 343,350\,\text{Pa} = 343.35\,\text{kPa}\) The pressure increase in going \(35.0\,\text{m}\) below the surface is \(343.35\,\text{kPa}\). #b) Pressure decrease above the surface#
04

Identify known values

The known values in this problem are: - Surface pressure \(P_0 = 105\,\text{kPa}\) - Altitude \(h = 35.0\,\text{m}\) - Air density \(\rho_a = 1.20\,\text{kg/m}^3\) at \(20^{\circ} \mathrm{C}\) - Gravitational acceleration \(g = 9.81\,\text{m/s}^2\) - Gas constant for air \(R_a = 287\,\text{J/kg}\cdot\text{K}\) - Temperature \(T = 20 + 273.15 = 293.15\,\text{K}\) (convert from Celsius to Kelvin)
05

Calculate the pressure decrease

To find the pressure decrease, we will use the barometric pressure formula: \(\Delta P = P_0[1 - (\frac{h}{h_0})]^{(gM/RT)}\) , where \(h_0 = \frac{RT}{Mg}\) First, calculate \(h_0\): \(h_0 = \frac{(287\,\text{J/kg}\cdot\text{K})(293.15\text{K})}{(1.20\,\text{kg/m}^3)(9.81\,\text{m/s}^2)}\)
06

Calculate the pressure decrease

Solve for \(\Delta P\): \(\Delta P = 105\,\text{kPa}\left[1 - \left(\frac{35.0\,\text{m}}{6787.4\,\text{m}}\right)^{5.26}\right]\) \(\Delta P \approx 5.37\,\text{kPa}\) The approximate pressure decrease in going \(35\,\text{m}\) above the surface is \(5.37\,\text{kPa}\).

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Most popular questions from this chapter

In a hydraulic lift, the radii of the pistons are \(2.50 \mathrm{cm}\) and $10.0 \mathrm{cm} .\( A car weighing \)W=10.0 \mathrm{kN}$ is to be lifted by the force of the large piston. (a) What force \(F_{\mathrm{a}}\) must be applied to the small piston? (b) When the small piston is pushed in by $10.0 \mathrm{cm},$ how far is the car lifted?(c) Find the mechanical advantage of the lift, which is the ratio \(W / F_{\mathrm{a}}\).
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