Question

A scuba diver's \(2-\mathrm{ft}^{3}\) air tank is to be filled with air from a compressed air line at 120 psia and \(85^{\circ} \mathrm{F}\). Initially, the air in this tank is at 20 psia and \(60^{\circ} \mathrm{F}\). Presuming that the tank is well insulated, determine the temperature and mass in the tank when it is filled to 120 psia.

Short answer

Answer: The final temperature in the tank is approximately \(215^\circ\text{F}\) and the mass in the tank when filled to 120 psia is approximately \(0.387\,\text{lbm}\).

Step-by-step solution

Identify given and unknown quantities

We are given: - Initial volume of the tank: \(V_1 = 2\,\text{ft}^3\) - Initial pressure of the tank: \(P_1 = 20\,\text{psia}\) - Initial temperature of the tank: \(T_1 = 60\,^\circ\text{F}\) - Pressure of the compressed air line: \(P_2 = 120\,\text{psia}\) - Temperature of the compressed air line: \(T_2 = 85\,^\circ\text{F}\) We need to find: - Final temperature in the tank: \(T_f\) - Mass in the tank when filled to 120 psia: \(m_f\)

Convert temperatures to absolute scale

In order to use the ideal gas law, we need to convert the temperatures from Fahrenheit to an absolute scale, such as Rankine (°R) or Kelvin (K). Let's convert to Rankine using the following formula: $$T_\text{R} = T_\text{F} + 459.67$$ Initial temperature in Rankine: $$T_{1,\text{R}} = 60 + 459.67 = 519.67\,\text{°R}$$ Temperature of the compressed air line in Rankine: $$T_{2,\text{R}} = 85 + 459.67 = 544.67\,\text{°R}$$

Use the ideal gas law to find the initial mass

Assuming the air behaves as an ideal gas, we can use the ideal gas law to find the initial mass in the tank: $$PV = mRT$$ where P is the pressure, V is the volume, m is the mass, R is the specific gas constant for air \((R = 53.35\,\text{ft}\cdot\text{lbf}/\text{lbm} \cdot \text{°R})\), and T is the temperature. First, find the initial mass: $$m_1 = \frac{P_1V_1}{RT_{1,\text{R}}} = \frac{20\text{ psia} \cdot 2\,\text{ft}^3}{53.35\,\text{ft}\cdot\text{lbf}/\text{lbm}\cdot\text{°R} \cdot 519.67\,\text{°R}} \approx 0.035\,\text{lbm}$$

Use mass conservation to find the final mass

The air added to the tank from the compressed air line has the same properties as the compressed air line itself. Therefore, we can express the mass-conservation equation as follows: $$m_f = m_1 + m_2$$ The mass of the air added to the tank \(m_2\) can be calculated using the ideal gas law: $$m_2 = \frac{(P_2 - P_1)V_1}{RT_{2,\text{R}}} = \frac{(120-20)\text{ psia} \cdot 2\,\text{ft}^3}{53.35\,\text{ft}\cdot\text{lbf}/\text{lbm}\cdot\text{°R} \cdot 544.67\,\text{°R}}\approx 0.352\,\text{lbm}$$ Now, find the final mass: $$m_f = 0.035\,\text{lbm} + 0.352\,\text{lbm} = 0.387\,\text{lbm}$$

Find the final temperature in the tank

Use the ideal gas law again to find the final temperature \(T_f\): $$T_f = \frac{P_2V_1}{m_fR} = \frac{120\text{ psia}\cdot 2\,\text{ft}^3}{0.387\,\text{lbm} \cdot 53.35\,\text{ft}\cdot\text{lbf}/\text{lbm}\cdot\text{°R}} \approx 674.7\,\text{°R}$$ Now, convert the final temperature back to Fahrenheit: $$T_f = T_\text{R} - 459.67 = 674.7 - 459.67 = 215^\circ\text{F}$$

Summary

The final temperature in the tank is approximately \(215^\circ\text{F}\) and the mass in the tank when filled to 120 psia is approximately \(0.387\,\text{lbm}\).

Key concepts

Temperature Conversion

Understanding temperature conversion is essential when solving problems involving thermodynamics and the ideal gas law. In our daily life, we primarily use the Celsius and Fahrenheit scales for measuring temperature. However, in thermodynamics, temperatures are often required to be in absolute terms, meaning on the Kelvin or Rankine scale.

For example, converting Fahrenheit to Rankine is done by adding 459.67, as in the problem where the initial temperature of the air in the tank is converted to Rankine before using the ideal gas law. This conversion is crucial because it represents an absolute temperature scale where zero indicates no thermal energy, in contrast with Fahrenheit where temperatures could be below zero without signifying the absence of thermal energy.

Always remember to use the correct scale when applying equations in thermodynamics, as using the incorrect temperature scale would result in erroneous calculations and defy the laws of physics.

Why Rankine for Thermodynamics?

The usage of Rankine in thermodynamics is particularly common in the United States and in engineering practices tied to the Imperial system. It's directly related to the Fahrenheit scale and is convenient for those accustomed to thinking in degrees Fahrenheit but require an absolute temperature scale for their calculations.

Mass Conservation

Mass conservation is a fundamental principle in both physics and chemistry, asserting that the mass in a closed system must remain constant over time. This concept is integral to solving problems using the ideal gas law as it allows us to understand how the mass of a gas behaves during compression and expansion processes, provided there are no leaks or mass transfers into or out of the closed system in question.

In the exercise, the mass conservation principle is used to find the final mass of the gas in the tank. By following this law, the sum of the initial mass of the air in the tank and the mass of the air added from the compressed air line gives us the total mass in the tank after it is filled to 120 psia. This approach is a clear application of mass conservation, which tells us that the mass of an isolated system will not change as long as nothing is added or removed.

Real-life Implications

Mass conservation has real-life implications far beyond textbook problems. It is an essential factor in engineering, environmental science, and any system where mass flow and balance are relevant. It aids in predicting how systems will behave and helps in designing processes that rely on the accurate control of material amounts.

Specific Gas Constant

The specific gas constant, often denoted with the symbol R, is a key parameter in the ideal gas law equation. It is specific to the chemical composition of a gas, which means that each gas has its own unique specific gas constant. For air, the specific gas constant used in engineering and physics is typically 53.35 ft·lbf/lbm·°R using the Imperial system, or 287.05 J/kg·K using the SI system.

The specific gas constant is derived from the universal gas constant when divided by the molar mass of the gas. In the ideal gas law equation (\(PV = mRT\)), it relates the pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) of a gas to its mass (\(m\)), ensuring that units are consistent and correct. This constant makes it possible to solve for variables such as mass, temperature, and volume in thermodynamic systems.

Knowing the specific gas constant for air allowed us to calculate the mass of the air before and after compression in the scuba tank problem. Without this constant, we would be unable to complete the calculations necessary to determine the final conditions in the tank.