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The mass of compressed air in a 210 -L steel tank is determined by weighing the tank plus its contents and then subtracting the known weight of the empty tank. The mass of compressed air in the tank is found to be equal to \(3.2 \mathrm{~kg}\). If the tank is located in a room where the temperature is maintained at \(25^{\circ} \mathrm{C}\), estimate the pressure of the air in the tank.

Short Answer

Expert verified
The pressure of the air in the tank is approximately 1310 kPa.

Step by step solution

01

Use Ideal Gas Law Formula

The ideal gas law formula is given as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. Since we need the pressure \( P \), we can rearrange the formula to \( P = \frac{nRT}{V} \).
02

Convert Temperature to Kelvin

The temperature is given in Celsius but needs to be in Kelvin for the formula. Convert by adding 273.15 to the Celsius temperature: \( 25^{\circ}C + 273.15 = 298.15 K \).
03

Calculate Number of Moles \( n \)

Using the formula \( n = \frac{m}{M} \), where \( m \) is the mass of the air (3.2 kg) and \( M \) is the molar mass of air (approximately 28.97 g/mol or 0.02897 kg/mol), calculate \( n \). Convert mass to kg for consistency: \( n = \frac{3.2}{0.02897} \approx 110.47 \text{ moles} \).
04

Substitute Values into Ideal Gas Law Formula

Substitute the known values into the ideal gas law formula \( P = \frac{nRT}{V} \): - \( n = 110.47 \) moles - \( R = 8.314 \) \( \text{J/mol} \cdot \text{K} \), the universal gas constant - \( T = 298.15 \) K - Convert volume to cubic meters: \( 210 \text{ L} = 0.210 \text{ m}^3 \)Substitute these into the formula:\[ P = \frac{110.47 \times 8.314 \times 298.15}{0.210} \approx 1.31 \times 10^6 \text{ Pa} \] or approximately 1310 kPa.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Calculation
Pressure calculation is a crucial part of understanding gases and their interactions in confined spaces, like a steel tank. In this context, we're using the Ideal Gas Law which is expressed as:\[ PV = nRT \]Here, \( P \) stands for pressure, \( V \) for volume, \( n \) for moles of gas, \( R \) for the gas constant, and \( T \) for temperature in Kelvin. To find the pressure \( P \), the equation can be rearranged:\[ P = \frac{nRT}{V} \]This formula shows that pressure is directly proportional to the number of moles and the temperature, while inversely proportional to the volume. For example, if you have more gas (more moles), or it is hotter (higher temperature), the pressure increases, assuming the volume stays constant. This concept helps predict how gases will behave under different conditions, essential for fields like chemistry, physics, and engineering.
Temperature Conversion
When working with gases, temperature must be in the Kelvin scale to correctly use the Ideal Gas Law. Kelvin is the SI unit for temperature and starts at absolute zero, making it essential for calculations involving thermodynamic equations.To convert Celsius to Kelvin:\[ T(\text{K}) = T(\degree C) + 273.15 \]In our example, the room temperature is \( 25^{\circ}C \). To convert it:\[ 25 + 273.15 = 298.15 \text{ K} \]The Kelvin scale ensures that all thermodynamic calculations reflect the physical world accurately. It avoids negative values for temperature, which aren't viable in these equations, and relates directly to the energy within a system.
Mole Calculation
Calculating the number of moles in a gas sample is foundational for determining pressure using the Ideal Gas Law. The number of moles \( n \) is calculated by dividing the mass \( m \) of the gas by its molar mass \( M \):\[ n = \frac{m}{M} \]For this exercise, 3.2 kg of air is used, with air having a molar mass of approximately 0.02897 kg/mol. The calculation is:\[ n = \frac{3.2 \text{ kg}}{0.02897 \text{ kg/mol}} \approx 110.47 \text{ moles} \]This shows how to transition from a mass measurement to a molar quantity, necessary for applying the Ideal Gas Law. Understanding moles - which represent a specific number of particles (Avogadro's number) - connects the macroscopic world to the molecular scale, illustrating chemical interactions.
Gas Constant
The gas constant \( R \) is a vital component of the Ideal Gas Law, with the value 8.314 \( \, \text{J/mol} \cdot \text{K} \). It serves as a bridge between the energy scales and the amount of substance. The gas constant allows calculations involving temperature, volume, and pressure to yield meaningful results across various systems. It's universal, meaning it's applicable in any ideal gas scenario, regardless of specific gas quantities or conditions.Some key points about \( R \):
  • Unit consistency is crucial, combining Joules for energy, moles for the amount of substance, and Kelvin for temperature.
  • The constant links microscopic properties to macroscopic observables, facilitating accurate predictions about system behavior.
  • Understanding \( R \) helps unify concepts in physics and chemistry, showing how energy, matter, and temperature interact in gaseous forms.
Utilizing the gas constant effectively in calculations strengthens our grasp of chemical principles and their real-world applications.

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Most popular questions from this chapter

A 450-L storage tank is completely filled with water at \(25^{\circ} \mathrm{C}\). (a) If the top of the storage tank is left open to the atmosphere and the water in the tank is heated to \(80^{\circ} \mathrm{C},\) what volume of water will spill out of the tank? (b) If the water is cooled back down to \(25^{\circ} \mathrm{C}\), by what percentage will the weight of water in the tank be reduced from its original weight? Neglect the expansion of the tank when the water is heated.

A 1.5-mm-diameter capillary tube is inserted in a liquid, and it is observed that the liquid rises \(15 \mathrm{~mm}\) in the tube and has a contact angle of \(15^{\circ}\) with the surface of the glass tube. If a hydrometer indicates that the liquid has a specific gravity of \(0.8,\) what is the surface tension of the liquid? Would you expect that this same surface tension would be found if the experiment was done using a tube material other than glass? Explain.

A \(0.1-\mathrm{m}^{3}\) tank is filled with air at a temperature and pressure of \(20^{\circ} \mathrm{C}\) and \(400 \mathrm{kPa}\), respectively. What is the weight of air in the tank?

A \(2.0-\mathrm{m}^{3}\) volume of pure oxygen at a temperature and pressure of \(20^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\), respectively, is expanded to a volume of \(4.0 \mathrm{~m}^{3} .\) (a) What is the pressure in the expanded volume if the expansion process is isentropic? What are the initial and final densities of the gas? (b) What is the pressure in the expanded volume if the expansion process is isothermal? What are the initial and final densities of the gas? (c) For the expansion process described in part (b), what amount of heat must be added to the gas?

Convert the following quantities to SI units: 15 gallons per minute, 99 miles per hour, 20 feet per second, 150 cubic feet per minute, 1540 gallons, 28 acres, and 600 horsepower.

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