College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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1.15 please help me compute the variations of the temperature with respect to height Ty.
**1.14 Problem Statement:**

Show that a homogeneous atmosphere (density independent of height) has a finite height that depends only on the temperature at the lower boundary. Compute the height of a homogeneous atmosphere with surface temperature \( T_0 = 273 \, \text{K} \) and surface pressure \( 1000 \, \text{hPa} \). (Use the ideal gas law and hydrostatic balance.)

**Solution Derivation:**

Starting with the differential equation for hydrostatic balance:

\[
\frac{dp}{dz} = -\rho g
\]

By integrating from 0 to \( h \):

\[
\int_{p_0}^{0} dp = -\int_{0}^{h} \rho g \, dz
\]

This simplifies to:

\[
-p_0 = -\rho g h \implies p_0 = \rho g h
\]

Using the ideal gas law:

\[
p_0 = \rho R T_0
\]

By substitution:

\[
\rho = \frac{p_0}{R T_0} = \frac{1000 \times 10^2}{287 \times 273}
\]

Find the height \( h \):

\[
h = \frac{R T_0}{g} = 7.99 \, \text{km}
\]

**1.15 Problem Statement:**

For the conditions of the previous problem, compute the variation of the temperature with respect to height.

**Note:** The equations integrate hydrostatic balance with the ideal gas law to derive the relationship between atmospheric pressure, density, temperature, and height. Calculations involve using constants such as the specific gas constant for dry air, \( R = 287 \, \text{J/(kg·K)} \), and the acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \).
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Transcribed Image Text:**1.14 Problem Statement:** Show that a homogeneous atmosphere (density independent of height) has a finite height that depends only on the temperature at the lower boundary. Compute the height of a homogeneous atmosphere with surface temperature \( T_0 = 273 \, \text{K} \) and surface pressure \( 1000 \, \text{hPa} \). (Use the ideal gas law and hydrostatic balance.) **Solution Derivation:** Starting with the differential equation for hydrostatic balance: \[ \frac{dp}{dz} = -\rho g \] By integrating from 0 to \( h \): \[ \int_{p_0}^{0} dp = -\int_{0}^{h} \rho g \, dz \] This simplifies to: \[ -p_0 = -\rho g h \implies p_0 = \rho g h \] Using the ideal gas law: \[ p_0 = \rho R T_0 \] By substitution: \[ \rho = \frac{p_0}{R T_0} = \frac{1000 \times 10^2}{287 \times 273} \] Find the height \( h \): \[ h = \frac{R T_0}{g} = 7.99 \, \text{km} \] **1.15 Problem Statement:** For the conditions of the previous problem, compute the variation of the temperature with respect to height. **Note:** The equations integrate hydrostatic balance with the ideal gas law to derive the relationship between atmospheric pressure, density, temperature, and height. Calculations involve using constants such as the specific gas constant for dry air, \( R = 287 \, \text{J/(kg·K)} \), and the acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \).
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