Question

The air in an insulated, rigid compressed-air tank whose volume is \(0.5 \mathrm{m}^{3}\) is initially at \(4000 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) Enough air is now released from the tank to reduce the pressure to 2000 kPa. Following this release, what is the temperature of the remaining air in the tank?

Short answer

Answer: The temperature of the remaining air in the tank after the release is approximately -126.57°C.

Step-by-step solution

Analyze the Initial Conditions

Given: Initial Pressure: \(P_1 = 4000 \mathrm{kPa}\) Initial Temperature: \(T_1 = 20^{\circ} \mathrm{C}\)

Convert to Absolute Temperature

Convert the temperature from Celsius to Kelvin: \(T_1 (K) = T_1 (^{\circ} \mathrm{C}) + 273.15\) \(T_1 (K) = 20 + 273.15\) \(T_1 = 293.15 \mathrm{K}\)

Determine the Final Pressure

Given: Final Pressure: \(P_2 = 2000 \mathrm{kPa}\)

Employ the Ideal Gas Law

Using the Ideal Gas Law, we have the equation: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) Since the volume remains constant (\(V_1 = V_2\)), we can simplify the equation to: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)

Solve for the Final Temperature

To find the final temperature, \(T_2\), we can rearrange the equation: \(T_2 = \frac{P_2T_1}{P_1}\) Plugging in the given values: \(T_2 = \frac{2000 \mathrm{kPa} \times 293.15 \mathrm{K}}{4000 \mathrm{kPa}}\) \(T_2 = 146.575 \mathrm{K}\)

Convert to Celsius

To convert the final temperature from Kelvin to Celsius, we use the following equation: \(T_2 (^{\circ} \mathrm{C}) = T_2 (\mathrm{K}) - 273.15\) \(T_2 (^{\circ} \mathrm{C}) = 146.575 - 273.15\) \(T_2 = -126.575 ^{\circ} \mathrm{C}\) The temperature of the remaining air in the tank after the release is approximately \(-126.57 ^{\circ} \mathrm{C}\).

Key concepts

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In practice, it outlines how thermal energy is converted to and from other forms of energy and how it affects matter. The ideal gas law is a fundamental aspect of thermodynamics that describes how gases behave under various conditions of pressure, volume, and temperature.

Within the context of the given exercise, we see thermodynamics at play through the behavior of the air within a rigid insulated tank as it cools down after a decrease in pressure. Since the tank is insulated and rigid, we assume that no heat enters or leaves the system (an adiabatic process), and the volume of air remains constant. The cooling of the air is a result of the energy required to expand the gas and perform work against atmospheric pressure outside the tank, which is aligned with the first law of thermodynamics, concerning the conservation of energy.

Pressure-Volume-Temperature Relationship

The pressure-volume-temperature (PVT) relationship in gases is crucial in understanding how gases will respond to changes in conditions. For an ideal gas, these relationships are defined by the ideal gas law, \(PV = nRT\), where \(P\) represents pressure, \(V\) stands for volume, \(T\) indicates temperature, \(n\) signifies the amount of substance in moles, and \(R\) is the gas constant.

In the exercise, we're focused on a system where the volume is constant (\(V_1 = V_2\)), meaning we do not need to account for changes in volume when solving for the new temperature post-pressure change. The relationship simplifies to \(P_1/T_1 = P_2/T_2\), reflecting that at constant volume, pressure is directly proportional to temperature. This is a specific case of the combined gas law and is derived when only temperature and pressure change.

Gas Laws

The gas laws describe how gases behave under various conditions. Among these laws, Boyle's Law illustrates the inverse relationship between pressure and volume, Charles's Law demonstrates the direct relationship between volume and temperature, and Gay-Lussac’s Law presents the direct relationship between pressure and temperature, all considering a fixed amount of gas. The Ideal Gas Law unifies these individual gas laws and also includes Avogadro's principle, which states that equal volumes of ideal gases at the same temperature and pressure contain the same number of particles.

In our exercise, we primarily utilize the portion of the Ideal Gas Law that corresponds to Gay-Lussac’s Law since we're interested in the change in temperature as a result of a change in pressure while maintaining a constant volume. As we improve our understanding of these laws, we can predict the outcome of various thermodynamic processes efficiently and accurately.

Temperature Conversion

Temperature conversion between scales is a basic but vital skill in thermodynamics and various scientific disciplines. The Celsius and Kelvin scales are commonly used in scientific contexts, with Kelvin being the SI unit for temperature. The Kelvin scale is an absolute scale where 0 K is absolute zero, the point at which atomic motion ceases.

To convert from Celsius to Kelvin, we add \(273.15\) to the Celsius temperature, as seen in the exercise. Vice versa, to convert from Kelvin to Celsius, we subtract \(273.15\) from the Kelvin temperature. This conversion is crucial for accurately applying the Ideal Gas Law, which requires temperature to be in Kelvin to ensure consistency with other units in the law's equation. In the given task, temperature conversion is necessary to use the Ideal Gas Law and complete the final step of determining the air's temperature in Celsius after the pressure change.