Question

A vertical piston-cylinder device initially contains \(0.25 \mathrm{m}^{3}\) of air at \(600 \mathrm{kPa}\) and \(300^{\circ} \mathrm{C}\). A valve connected to the cylinder is now opened, and air is allowed to escape until three-quarters of the mass leave the cylinder at which point the volume is \(0.05 \mathrm{m}^{3} .\) Determine the final temperature in the cylinder and the boundary work during this process.

Short answer

Answer: The final temperature in the cylinder is 179.15°C, and the boundary work done during this process is -53,339.3 J (work done by the system on the surroundings).

Step-by-step solution

Determine Initial Conditions

Initially, we have the volume, pressure, and temperature of the air inside the cylinder as \(V_1 = 0.25\,m^3\), \(P_1=600\,kPa\), and \(T_1 = 300^{\circ}C\). First, we need to convert the temperature from Celsius to Kelvin: \(T_1 = (300 +273.15)K = 573.15K\).

Calculate Initial Mass in the Cylinder

We can use the ideal gas law to determine the initial mass of the air in the cylinder. The ideal gas law is given by \(PV=mRT\), where \(P\) is pressure, \(V\) is volume, \(m\) is mass, \(R\) is the specific gas constant, and \(T\) is temperature. For air, the specific gas constant is \(R = 287\frac{J}{kg\cdot K}\). Therefore, \(m_1 = \frac{P_1V_1}{RT_1} = \frac{(600\cdot10^3\,Pa)(0.25\,m^3)}{(287\frac{J}{kg\cdot K})(573.15K)} = 2.035\,kg\).

Calculate Mass and Volume of Air Remaining

After opening the valve, three-quarters of the initial mass escapes, leaving one-quarter remaining in the cylinder. So, the mass of air left in the cylinder is \(m_2 = \frac{1}{4}m_1 = \frac{1}{4}(2.035\,kg) = 0.50875\,kg\). The final volume of the cylinder is given as \(V_2 = 0.05\,m^3\).

Calculate Final Temperature

The final pressure in the cylinder can be found using the ideal gas law. Rearranging the ideal gas law formula for pressure, we have \(P_2 = \frac{m_2R\text{T_2}}{V_2}\), but since \(P_1V_1^\gamma = P_2V_2^\gamma\) for an isentropic process and here, \(\gamma\) is \(\frac{7}{5}\). Therefore now we could express the final pressure like, \(P_2 = P_1\left(\frac{V_1}{V_2}\right)^\gamma = 600 kPa\left(\frac{0.25\,m^3}{0.05\,m^3}\right)^{7/5} = 658.181\,kPa\). Now, we can determine the final temperature \(T_2\) using the ideal gas law formula: \(T_2 = \frac{P_2V_2}{m_2R} = \frac{(658.181\cdot10^3\,Pa)(0.05\,m^3)}{(0.50875\,kg)(287\frac{J}{kg\cdot K})} = 452.3\,K = 179.15^{\circ}C\).

Calculate Boundary Work

The boundary work can be determined using the formula \(W_b = \int_{V_1}^{V_2} PdV\). For an isentropic process, the relationship between pressure and volume is given by \(P V^\gamma = constant\). Therefore, \(P = constant\cdot V^{-\gamma}\) and the boundary work can be written as: \(W_b = \int_{V_1}^{V_2} (constant \cdot V^{-\gamma})dV\). The constant \(K = P_1 V_1^\gamma\) can be calculated from the initial conditions: \(K = (600\cdot10^3\,Pa)(0.25\,m^3)^{7/5} = 1.275\cdot10^6\,\frac{J}{m^3}\). Now, we can integrate the expression for work: \(W_b = K \int_{V_1}^{V_2} V^{-\gamma}dV = K\left[\frac{V^{-\gamma+1}}{-\gamma+1}\right]_{V_1}^{V_2} = K\left(\frac{V_2^{1-\gamma} - V_1^{1-\gamma}}{1-\gamma}\right) = \frac{1.275\cdot10^6\,\frac{J}{m^3}\cdot((0.05\,m^3)^{2/5}-(0.25\,m^3)^{2/5})}{2/5} = -53339.3\,J\). The negative sign indicates that work is done by the system (the air in the cylinder) on the surroundings. Therefore, the final temperature in the cylinder is \(179.15^{\circ}C\), and the boundary work during this process is \(-53339.3\,J\).

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation that relates the pressure (P), volume (V), temperature (T), and mass (m) of an ideal gas to its specific gas constant (R). Specifically, the law is written as \(PV = mRT\).

This relationship assumes the gas behaves ideally, meaning the gas particles do not attract or repel each other and occupy no volume themselves. The law allows us to calculate any one of the variables if the others are known. In the context of the given exercise, the ideal gas law helps to determine the initial mass of air inside a piston-cylinder device, by rearranging the formula to \(m = \frac{PV}{RT}\).

When performing such calculations, it's essential to convert all the units to the International System of Units (SI), ensuring that the pressure is in pascals (Pa), volume in cubic meters (\(m^3\)), temperature in kelvin (K), and the specific gas constant in joules per kilogram-kelvin (\(\frac{J}{kg\cdot K}\)). This ensures accuracy and consistency in the calculations.

Isentropic Process

An isentropic process is an idealization in thermodynamics in which a system undergoes a change without any entropy exchange with its environment. This implies that the process is both adiabatic (no heat transfer) and reversible. In simpler terms, it's a way in which the system changes state without gaining or losing heat, and can return to its original state without leaving any impact on its surroundings.

In such a process, the relationship between pressure and volume during a compression or expansion is described mathematically by \(P_1V_1^\gamma = P_2V_2^\gamma\), where \(\gamma\) is the heat capacity ratio (also known as the adiabatic index). For air, \(\gamma\) typically has a value of about \(\frac{7}{5}\) or 1.4. This relationship was used in our exercise to find the final pressure in the piston-cylinder device after some of the air was allowed to escape, consequently assisting in calculating the final temperature once the volume and mass at the final state were known.

Specific Gas Constant

The specific gas constant (R) is a property unique to a particular gas that relates the ideal gas law variables of pressure, volume, and temperature to the mass of the gas. It is the constant of proportionality in the ideal gas law equation and has units of \(\frac{J}{kg\cdot K}\).

Its value depends on the chemical composition of the gas and is different for every gas. For air, which is a mixture of different gases, the specific gas constant has a commonly accepted value of \(287\frac{J}{kg\cdot K}\). Knowing the specific gas constant allows for precise calculations regarding changes in the state of the gas according to the ideal gas law. In the exercise scenario, the specific gas constant for air plays a critical role in determining both the initial mass of air in the cylinder and the final temperature after the isentropic expansion.