Question

An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains \(4 \mathrm{kg}\) of an ideal gas at \(800 \mathrm{kPa}\) and \(50^{\circ} \mathrm{C},\) and the other part is evacuated. The partition is now removed, and the gas expands into the entire tank. Determine the final temperature and pressure in the tank.

Short answer

After removing the partition in the insulated and rigid tank, the final temperature of the gas is 258.52 K, and the final pressure is 400 kPa.

Step-by-step solution

Identify the initial state of the gas

Initially, there is \(4 \mathrm{kg}\) of an ideal gas in one half of the tank at a pressure of \(800 \mathrm{kPa}\) and a temperature of \(50^{\circ} \mathrm{C}\) (converted to Kelvin by adding 273.15). The other half is empty. Initial pressure, \(P_1 = 800 \mathrm{kPa}\) Initial temperature, \(T_1 = 50 + 273.15 = 323.15\ \mathrm{K}\) Mass of the gas, \(m = 4 \mathrm{kg}\)

Identify that the process is isentropic

Since the tank is insulated and rigid, no heat or work is transferred to or from the system. Therefore, the process is isentropic (constant entropy). The total entropy of the system remains constant.

Use the ideal gas law to find the initial volume

The ideal gas law states that \(PV = mRT\), where \(P\) is pressure, \(V\) is volume, \(m\) is mass, \(R\) is the specific gas constant, and \(T\) is temperature. We can rearrange the equation to find the initial volume of the gas. \(V_1 = \frac{mRT_1}{P_1}\) However, we don't know the specific gas constant \(R\). We can assume the gas is air for this exercise. For air, \(R = 287 \ \mathrm{J/kg \cdot K}\). Now, we can find the initial volume of the gas: \(V_1 = \frac{(4 \ \mathrm{kg})(287 \ \mathrm{J/kg \cdot K})(323.15\ \mathrm{K})}{800 \ \mathrm{kPa}} = \frac{(4)(287)(323.15)}{800} = 1.4965\ \mathrm{m^3}\)

Determine the final volume

When the partition is removed, the gas expands and fills the entire tank. Since both parts of the tank are equal in size, the final volume can be calculated by doubling the initial volume: \(V_2 = 2V_1 = 2(1.4965) = 2.993\ \mathrm{m^3}\)

Calculate the final pressure and temperature

As the process is isentropic, we have the relations: \(T_2 = T_1\left(\frac{V_1}{V_2}\right)^{(\gamma-1)}\) \(P_2 = P_1\left(\frac{V_1}{V_2}\right)^{\gamma}\) For air, the specific heat ratio, \(\gamma = 1.4\). Now, we can find the final temperature and pressure: \(T_2 = T_1\left(\frac{V_1}{V_2}\right)^{(\gamma-1)} = (323.15\ \mathrm{K})\left(\frac{1.4965}{2.993}\right)^{(1.4-1)} = 258.52\ \mathrm{K}\) \(P_2 = P_1\left(\frac{V_1}{V_2}\right)^{\gamma} = (800 \ \mathrm{kPa})\left(\frac{1.4965}{2.993}\right)^{1.4} = 400\ \mathrm{kPa}\)

State the final temperature and pressure

The final temperature is \(258.52\ \mathrm{K}\), and the final pressure is \(400\ \mathrm{kPa}\).

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics that relates the pressure, volume, temperature, and amount of an ideal gas. The equation is given as \(PV = nRT\), with 'P' standing for pressure, 'V' for volume, 'n' for the amount of substance in moles, 'R' for the ideal gas constant, and 'T' for temperature in Kelvin.

In practice, especially with larger systems, it is more common to see the equation expressed in terms of mass and the specific gas constant, \(PV = mRT\), where 'm' is the mass of the gas and 'R' is the specific gas constant for a particular gas. This form is particularly useful in engineering calculations, and it was applied in our exercise to determine the initial volume of the gas.

In the given problem, after identifying the state of the ideal gas, the ideal gas law was used to find the initial volume. By knowing the initial conditions of pressure, temperature, and mass, we were able to use the specific gas constant for air to solve for the volume. Understanding the ideal gas law is crucial as it forms the basis for the analysis of many thermodynamic processes like the isentropic process experienced by the gas in this problem.

Specific Heat Ratio

The specific heat ratio (\(\gamma\)) is a dimensionless number representing the ratio of specific heat at constant pressure (\(C_p\)) to specific heat at constant volume (\(C_v\)). For an ideal gas, these specific heats are properties that describe how much heat is required to raise the temperature of a given mass of the gas by one degree at constant pressure or volume, respectively.

The value of \(\gamma\) affects how an ideal gas behaves under various thermodynamic processes. During an isentropic process, which is both adiabatic (no heat transfer) and reversible, the specific heat ratio comes into play when relating pressure and volume changes to temperature changes. Knowing \(\gamma\) is essential when using the isentropic relations to calculate the final temperature and pressure of an ideal gas following an isentropic process, as demonstrated in our exercise where \(\gamma\) for air (1.4) was used to determine the final states of the expanded gas.

Entropy

Entropy is a concept in thermodynamics often associated with the level of disorder or randomness in a system. It's a measure of the spread of energy within a system and how that energy is distributed among possible microstates.

In the context of an isentropic process, 'isentropic' literally means 'constant entropy'. During such a process, entropy remains unchanged. This implies that there is no heat transfer into or out of the system and the process is reversible. In the exercise, the process is considered isentropic because the gas expands into a vacuum without exchanging heat with its surroundings, and the system is assumed to be perfectly efficient without any irreversibilities.

Understanding entropy is key to grasping the second law of thermodynamics, which, among other things, dictates that in a closed system, processes tend to occur that increase the total entropy of the system and its surroundings. In engineering and physics, calculations involving entropy are instrumental in predicting the behavior of a thermodynamic system and how it interacts with its environment.