The tank below is filled with air at 25°C under absolute pressure of 60 kPa. The mass of air in the tank is nearest (a) 1.76x106 kg 4 m (b) 1.43 kN (c) 4.96 kg (d) 112 kg (e) 19.8 kg 1.5 m OT CO"
The tank below is filled with air at 25°C under absolute pressure of 60 kPa. The mass of air in the tank is nearest (a) 1.76x106 kg 4 m (b) 1.43 kN (c) 4.96 kg (d) 112 kg (e) 19.8 kg 1.5 m OT CO"
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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![### Problem Statement:
The tank below is filled with air at 25°C under absolute pressure of 60 kPa. The mass of air in the tank is nearest to:
(a) 1.76×10⁶ kg
(b) 1.43 kN
(c) 4.96 kg
(d) 112 kg
(e) 19.8 kg
### Diagram Description:
There is a cylindrical tank depicted in the diagram. The cylinder has hemispherical ends. The dimensions of the tank are provided as follows:
- The total length of the cylinder (including the hemispherical ends) is 4 meters.
- The diameter of the cylindrical portion (same as the diameter of the hemispherical ends) is 1.5 meters.
### Explanation for Calculation:
To determine the mass of the air in the tank, one would typically use the Ideal Gas Law, which is stated as:
\[ PV = mRT \]
where:
- \( P \) is the absolute pressure in the tank (60 kPa)
- \( V \) is the volume of the tank
- \( m \) is the mass of the air
- \( R \) is the specific gas constant for air
- \( T \) is the absolute temperature in the tank (which should be converted to Kelvin)
The volume \( V \) of the tank is calculated by the sum of the volume of the cylindrical part and the volume of the two hemispherical ends.
For the cylindrical part:
\[ V_{cylinder} = \pi r^2 h \]
For the hemispherical ends (combined to form a sphere):
\[ V_{sphere} = \frac{4}{3} \pi r^3 \]
Summing these volumes will provide the total volume of the tank. Substitute this into the Ideal Gas Law to solve for \( m \), the mass of air.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F502330eb-9fa1-41eb-8716-374edceb5782%2Ff4514cba-f7b7-4b65-a869-8ddb562fe935%2F4bhicy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
The tank below is filled with air at 25°C under absolute pressure of 60 kPa. The mass of air in the tank is nearest to:
(a) 1.76×10⁶ kg
(b) 1.43 kN
(c) 4.96 kg
(d) 112 kg
(e) 19.8 kg
### Diagram Description:
There is a cylindrical tank depicted in the diagram. The cylinder has hemispherical ends. The dimensions of the tank are provided as follows:
- The total length of the cylinder (including the hemispherical ends) is 4 meters.
- The diameter of the cylindrical portion (same as the diameter of the hemispherical ends) is 1.5 meters.
### Explanation for Calculation:
To determine the mass of the air in the tank, one would typically use the Ideal Gas Law, which is stated as:
\[ PV = mRT \]
where:
- \( P \) is the absolute pressure in the tank (60 kPa)
- \( V \) is the volume of the tank
- \( m \) is the mass of the air
- \( R \) is the specific gas constant for air
- \( T \) is the absolute temperature in the tank (which should be converted to Kelvin)
The volume \( V \) of the tank is calculated by the sum of the volume of the cylindrical part and the volume of the two hemispherical ends.
For the cylindrical part:
\[ V_{cylinder} = \pi r^2 h \]
For the hemispherical ends (combined to form a sphere):
\[ V_{sphere} = \frac{4}{3} \pi r^3 \]
Summing these volumes will provide the total volume of the tank. Substitute this into the Ideal Gas Law to solve for \( m \), the mass of air.
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