-Z/Zo 3 The density of air as a function of altitude above sea level is modeled as p(z) = poe` where po = 1.2 kg/m³, and zo = 16, 855 m. A spherical balloon with radius r = 0.2 m is filled with Helium gas (MH₂ = 4 g/mol) to STP (P = 1 atm, T = 273 K). What is the maximum height the balloon can float up to assuming constant pressure and temperature within the balloon? Assume that the density of air is constant across the volume of the balloon for a specific height, and that the balloon's rubber has negligible mo
-Z/Zo 3 The density of air as a function of altitude above sea level is modeled as p(z) = poe` where po = 1.2 kg/m³, and zo = 16, 855 m. A spherical balloon with radius r = 0.2 m is filled with Helium gas (MH₂ = 4 g/mol) to STP (P = 1 atm, T = 273 K). What is the maximum height the balloon can float up to assuming constant pressure and temperature within the balloon? Assume that the density of air is constant across the volume of the balloon for a specific height, and that the balloon's rubber has negligible mo
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![**Density of Air as a Function of Altitude**
The density of air as a function of altitude above sea level is modeled as:
\[ \rho(z) = \rho_0 e^{-z/z_0} \]
where:
- \( \rho_0 = 1.2 \, \text{kg/m}^3 \)
- \( z_0 = 16,855 \, \text{m} \)
A spherical balloon with a radius of \( r = 0.2 \, \text{m} \) is filled with Helium gas (\( M_{\text{He}} = 4 \, \text{g/mol} \)) to STP (\( P = 1 \, \text{atm} \), \( T = 273 \, \text{K} \)).
**Problem Statement:**
What is the maximum height the balloon can float up to, assuming constant pressure and temperature within the balloon? Assume that the density of air is constant across the volume of the balloon for a specific height, and that the balloon’s rubber has negligible mass.
**Step-by-Step Solution:**
1. **Calculate Volume of the Balloon:**
The volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
Substituting \( r = 0.2 \, \text{m} \):
\[
V = \frac{4}{3} \pi (0.2)^3 \approx 0.0335 \, \text{m}^3
\]
2. **Determine the Mass of Helium in the Balloon:**
Using the Ideal Gas Law \( PV = nRT \), where \( R \) is the gas constant (\( 8.314 \, \text{J/(mol} \cdot \text{K)} \)):
\[
n = \frac{PV}{RT}
\]
Given \( P = 1 \, \text{atm} = 101325 \, \text{Pa} \) and \( T = 273 \, \text{K} \):
\[
n = \frac{101325 \times 0.0335}{8.314 \times 273} \approx 1.49](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F651b911d-2c53-4f54-b4b6-29cb6ac234d2%2F87e96e0f-a199-4ace-8352-675df4670efa%2F5awbtnq_processed.png&w=3840&q=75)
Transcribed Image Text:**Density of Air as a Function of Altitude**
The density of air as a function of altitude above sea level is modeled as:
\[ \rho(z) = \rho_0 e^{-z/z_0} \]
where:
- \( \rho_0 = 1.2 \, \text{kg/m}^3 \)
- \( z_0 = 16,855 \, \text{m} \)
A spherical balloon with a radius of \( r = 0.2 \, \text{m} \) is filled with Helium gas (\( M_{\text{He}} = 4 \, \text{g/mol} \)) to STP (\( P = 1 \, \text{atm} \), \( T = 273 \, \text{K} \)).
**Problem Statement:**
What is the maximum height the balloon can float up to, assuming constant pressure and temperature within the balloon? Assume that the density of air is constant across the volume of the balloon for a specific height, and that the balloon’s rubber has negligible mass.
**Step-by-Step Solution:**
1. **Calculate Volume of the Balloon:**
The volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
Substituting \( r = 0.2 \, \text{m} \):
\[
V = \frac{4}{3} \pi (0.2)^3 \approx 0.0335 \, \text{m}^3
\]
2. **Determine the Mass of Helium in the Balloon:**
Using the Ideal Gas Law \( PV = nRT \), where \( R \) is the gas constant (\( 8.314 \, \text{J/(mol} \cdot \text{K)} \)):
\[
n = \frac{PV}{RT}
\]
Given \( P = 1 \, \text{atm} = 101325 \, \text{Pa} \) and \( T = 273 \, \text{K} \):
\[
n = \frac{101325 \times 0.0335}{8.314 \times 273} \approx 1.49
Transcribed Image Text:### Thermal Properties of Various Materials
Below is a table summarizing the thermal properties of several materials. This information is crucial in understanding how different substances respond to changes in temperature, including their heat capacities, latent heats, and densities.
| Material | α (°C⁻¹) | c (J/(kg K)) | C (J/(mol K)) | Lf (J/kg) | Lv (J/kg) | Density (kg/m³) |
|-----------------|------------------|----------------|------------------|----------------|-----------------|--------------------|
| Gold | - | 129 | 25 | - | - | 19,300 |
| Copper | 1.65×10⁻⁵ | - | - | - | - | 8,960 |
| Aluminum | 2.3×10⁻⁵ | - | - | - | - | 2,700 |
| Stainless Steel | 1.75×10⁻⁵ | - | - | - | - | 7,500 |
| Water | - | 4190 | 75 | 3.33×10⁵ | 22.6×10⁵ | 1,000 |
| Ethyl Alcohol | - | 2400 | 110 | 1.09×10⁵ | 8.79×10⁵ | 790 |
| Ice | - | 2090 | 37.6 | - | - | 917 |
### Constants
- **1 atm = 1.01×10⁵ Pa**
- **Nᴀ = 6.022×10²³ particles/mol**
- **R = 8.3145 J/(mol K)**
- **kB = 1.38×10⁻²³ J/K**
- **σ = 5.67×10⁻⁸ W/(m²K⁴)**
### Explanation of Columns:
- **α (°C⁻¹)**: The coefficient of linear thermal expansion, showing how much the material expands per degree Celsius increase in temperature.
- **c (J/(kg K))**: The
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