Question

An insulated vertical piston-cylinder device initially contains \(0.11 \mathrm{m}^{3}\) of air at \(150 \mathrm{kPa}\) and \(22^{\circ} \mathrm{C}\). At this state, a linear spring touches the piston but exerts no force on it. The cylinder is connected by a valve to a line that supplies air at \(700 \mathrm{kPa}\) and \(22^{\circ} \mathrm{C}\). The valve is opened, and air from the high-pressure line is allowed to enter the cylinder. The valve is turned off when the pressure inside the cylinder reaches 600 kPa. If the enclosed volume inside the cylinder doubles during this process, determine ( \(a\) ) the mass of air that entered the cylinder, and \((b)\) the final temperature of the air inside the cylinder.

Short answer

Answer: (a) 0.3458 kg, (b) 590.3 K

Step-by-step solution

Determine the initial mass of air

Using the Ideal Gas Law, \(PV = m R T\), where \(P\) is the pressure, \(V\) is the volume, \(m\) is the mass of the gas, \(R\) is the gas constant, and \(T\) is the temperature. For air, \(R = 0.287 \, \mathrm{kJ/kg \cdot K}\). The initial conditions are \(P_1 = 150 \, \mathrm{kPa}\), \(V_1 = 0.11 \, \mathrm{m^3}\), and \(T_1 = 22 + 273.15 = 295.15 \, \mathrm{K}\). We need to find the mass \(m_1\) of the air: \(m_1 = \frac{P_1 V_1}{R T_1} = \frac{(150)(0.11)}{(0.287)(295.15)} = 0.1734 \, \mathrm{kg}\)

Find the mass of air that entered the cylinder

When the pressure inside the cylinder reaches \(600 \, \mathrm{kPa}\), the volume doubles (\(V_2 = 2 V_1 = 0.22 \, \mathrm{m^3}\)). At this point, let's denote the mass of the air that entered the cylinder as \(m_{in}\). The total mass of the air inside the cylinder will be \(m_1 + m_{in}\). Since we know the final pressure (\(P_2 = 600 \, \mathrm{kPa}\)), final volume (\(V_2 = 0.22 \, \mathrm{m^3}\)), and that the temperature of the air entering the cylinder is \(T_{in} = 22 + 273.15 = 295.15 \, \mathrm{K}\), we can use the Ideal Gas Law to find \(m_{in}\): \(P_2 V_2 = (m_1 + m_{in}) R T_{in}\) We can solve for \(m_{in}\): \(m_{in} = \frac{P_2 V_2}{R T_{in}} - m_1 = \frac{(600)(0.22)}{(0.287)(295.15)} - 0.1734 = 0.3458 \, \mathrm{kg}\) So, the mass of air that entered the cylinder is \(0.3458 \, \mathrm{kg}\).

Determine the final temperature of the air inside the cylinder

We assume the process during which the air enters the cylinder is polytropic with a polytropic exponent, \(n\). To calculate the final temperature, we use the polytropic process equation: \(P_1 V_1^n = P_2 V_2^n\) We can rearrange the equation and find the polytropic exponent, \(n\): \(n = \frac{\ln (P_2/P_1)}{\ln (V_2/V_1)} = \frac{\ln (600/150)}{\ln (0.22/0.11)} = 1.4\) Finally, we can find the final temperature, \(T_2\): \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) \(T_2 = \frac{P_2 V_2 T_1}{P_1 V_1} = \frac{(600)(0.22)(295.15)}{(150)(0.11)} = 590.3 \, \mathrm{K}\) The final temperature of the air inside the cylinder is \(590.3 \, \mathrm{K}\). So, the mass of air that entered the cylinder is \((a) \, 0.3458 \, \mathrm{kg}\) and the final temperature of the air inside the cylinder is \((b) \, 590.3 \, \mathrm{K}\).

Key concepts

Piston-Cylinder Device

A piston-cylinder device is a common piece of laboratory and industrial equipment used in the field of thermodynamics. It is essentially a cylinder with a moveable piston inside that can compress or expand gases within. The pressure, volume, and temperature of the contained gas can change depending on external conditions or internal reactions.

For educational purposes, it's a perfect tool for demonstrating the principles of the ideal gas law, as it neatly encapsulates a controlled amount of gas in a variable volume controlled by the piston. During a process, as gas is added or removed, or as the gas is heated or cooled, the piston moves to maintain mechanical equilibrium with the pressure of the gas.

Polytropic Process

In thermodynamics, a polytropic process is a type of reversible process that involves heat exchange. The process is characterized by the equation \( P V^n = \text{constant} \), where \( P \) is the pressure of the gas, \( V \) is the volume, and \( n \) is the polytropic index. The value of \( n \) can vary depending on the specific heat capacities of the gas and determines the nature of the process.

For instance, \( n=0 \) corresponds to an isobaric process (constant pressure), \( n=1 \) to an isothermal process (constant temperature), and \( n=\infty \) to an isochoric process (constant volume). When dealing with real-world scenarios, the value of \( n \) often falls between these ideal processes, reflecting the heat transfer and work done during the process.

Thermodynamic Temperature

Thermodynamic temperature is one of the fundamental concepts in the study of thermodynamics. It represents the absolute measure of temperature and is one of the principal parameters of the state of a system. The thermodynamic temperature scale is defined in Kelvin (K), where 0 K, or absolute zero, is theoretically the lowest possible temperature, at which point no thermal energy is present in a system.

The temperature of a substance directly influences its pressure and volume if the amount of substance and its chemical composition remain constant, as described in the ideal gas law \( PV = nRT \), where \( R \) is the universal gas constant and \( T \) is the thermodynamic temperature. In our exercise, the temperature is initially given in degrees Celsius and must be converted to Kelvin by adding 273.15 before applying the ideal gas law.

Mass and Volume Relationships

Understanding mass and volume relationships is crucial when examining the behavior of gases. According to the ideal gas law, \( PV = mRT \), where \( m \) is the mass of the gas, these two properties are inversely related, e.g., as the volume of a gas increases, the density (mass per unit volume) decreases, assuming that temperature and pressure remain constant.

The ideal gas law allows us to calculate the mass of a gas when its pressure (\( P \)), volume (\( V \)), and temperature (\( T \) in Kelvin) are known, as long as we know the specific gas constant \( R \) for the gas in question. In our exercise, finding the mass of air that entered the cylinder is a practical application of the relationship between mass and volume, taking into account that air also has mass and occupies space. Making accurate calculations in these scenarios is important for various scientific and engineering applications, such as designing HVAC systems or predicting the behavior of an air-filled balloon as it rises in the atmosphere.