Question

A piston-cylinder device initially contains \(0.07 \mathrm{m}^{3}\) of nitrogen gas at \(130 \mathrm{kPa}\) and \(120^{\circ} \mathrm{C}\). The nitrogen is now expanded polytropically to a state of \(100 \mathrm{kPa}\) and \(100^{\circ} \mathrm{C}\). Determine the boundary work done during this process.

Short answer

Question: Calculate the boundary work done during a polytropic expansion of nitrogen gas with initial volume 0.07 m³, initial pressure 130 kPa, initial temperature 120°C, final pressure 100 kPa, and final temperature 100°C. Answer: The boundary work done during this polytropic process is -0.827 kJ. Since the work is negative, it indicates work done on the system (nitrogen gas).

Step-by-step solution

Convert the temperatures to Kelvin

To convert temperatures from Celsius to Kelvin, we add 273.15 to each value. \(T_1 = 120^{\circ} \mathrm{C} + 273.15 = 393.15 \, \mathrm{K}\) \(T_2 = 100^{\circ} \mathrm{C} + 273.15 = 373.15 \, \mathrm{K}\)

Calculate the initial and final specific volumes

To find the specific volumes (\(v_1\) and \(v_2\)) at each state, we use the ideal gas law: \[Pv = RT\] where \(P\) is pressure, \(v\) is specific volume, \(R\) is the specific gas constant for nitrogen, and \(T\) is the temperature. For nitrogen, \(R = 0.2968 \,\mathrm{kJ/(kg.K)}\). First, we calculate the initial and final specific volumes (\(v_1\) and \(v_2\)) using the ideal gas law. \(v_1 = \frac{R T_1}{P_1} = \frac{0.2968 \,\mathrm{kJ/(kg.K)} \times 393.15 \,\mathrm{K}}{130 \,\mathrm{kPa}} = 0.8938 \,\mathrm{m}^3/\mathrm{kg}\) \(v_2 = \frac{R T_2}{P_2} = \frac{0.2968 \,\mathrm{kJ/(kg.K)} \times 373.15 \,\mathrm{K}}{100 \,\mathrm{kPa}} = 1.1096 \,\mathrm{m}^3/\mathrm{kg}\)

Calculate the polytropic exponent 'n'

We use the polytropic process equation: \[P_1 v_1^n = P_2 v_2^n\] Rearranging the equation to find the value of 'n': \[n = \frac{\ln(P_1/P_2)}{\ln(v_1/v_2)}\] Plugging in the values: \[n = \frac{\ln(130/100)}{\ln(0.8938/1.1096)} = 1.1656\]

Calculate the boundary work done

Now that we have the polytropic exponent 'n', we can find the boundary work formula for a polytropic process: \[W_{b} = \frac{m(R)(T_2 - T_1)}{1 - n}\] We need to find the mass 'm' of the nitrogen gas. From the ideal gas law: \[m = \frac{P_1 V_1}{R T_1}\] \[\Rightarrow m = \frac{130 \,\mathrm{kPa} \times 0.07 \, \mathrm{m}^3}{0.2968 \,\mathrm{kJ/(kg.K)} \times 393.15 \, \mathrm{K}} = 0.0782 \, \mathrm{kg}\] Now we can calculate the boundary work: \[W_{b} = \frac{0.0782 \, \mathrm{kg} \times 0.2968 \, \mathrm{kJ/(kg.K)}(373.15 \, \mathrm{K} - 393.15 \, \mathrm{K})}{1 - 1.1656}\] \[W_{b} = -0.827 \, \mathrm{kJ}\] The boundary work done during this polytropic process is \(-0.827 \, \mathrm{kJ}\). Since the work is negative, it indicates work done on the system (nitrogen gas).

Key concepts

Polytropic Process

A polytropic process is a type of thermodynamic process that follows a specific mathematical relationship between pressure and volume. This relationship can be represented as
\( P V^n = \text{constant} \),
where \( P \) is the pressure, \( V \) is the volume, and \( n \) is the polytropic exponent. The value of \( n \) can vary, defining different types of processes. For example, when \( n = 1 \), it represents an isothermal process, where the temperature remains constant. If \( n = 0 \), it describes an isobaric process, with constant pressure, and if \( n = \text{infinity} \), it's an isochoric process, where the volume remains unchanged.
In a polytropic process, the boundary work performed involves expanding or compressing a gas, which can either absorb or release energy. The sign of the boundary work indicates the direction of energy transfer: positive work means the system has done work on the surroundings, while negative work means work has been done on the system.

Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics that describes the behavior of an ideal gas. Represented as
\( PV = nRT \),
where \( P \) stands for pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the absolute temperature in Kelvins.
The ideal gas law assumes that gases consist of many tiny particles moving in random motion, without any volume of their own and no attraction or repulsion between them. This law is the combination of Boyle's law, Charles's law, and Avogadro's law. It allows us to calculate any one of the four variables (pressure, volume, amount, and temperature) if the other three are known, proving extremely useful in a variety of thermodynamic calculations.

Specific Volume

Specific volume is a property used in thermodynamics to describe the volume occupied by a unit mass of a substance, typically a fluid. It is the inverse of density and is represented by the symbol \( v \), where
\( v = \frac{V}{m} \),
with \( V \) representing the volume and \( m \) representing mass. The specific volume tells us about how much space one kilogram of a substance takes up, and its units are commonly \( \text{m}^3/\text{kg} \).
When analyzing processes involving gases, the specific volume is particularly important as it relates closely to the ideal gas law. By manipulating the ideal gas law, we can express specific volume in terms of pressure and temperature for an ideal gas, showing how these properties change during thermodynamic processes. In practical calculations, specific volume is used to determine other important properties, like the work done in a thermodynamic cycle.

Polytropic Exponent

The polytropic exponent, denoted by \( n \), is a crucial parameter in characterizing a polytropic process. It defines the nature of the thermodynamic process a fluid undergoes when being compressed or expanded. The value of \( n \) depends on the specific heat capacities of the gas, as well as the heat transfer involved in the process. It's also closely related to the adiabatic index, which is the ratio of specific heats at constant pressure and constant volume.
Determining the polytropic exponent is necessary for calculating work and other properties during a polytropic process. It's calculated from the pressure and specific volume at two different states using the formula mentioned in the given exercise. Depending on the value of \( n \), different types of work equations are used to calculate the boundary work in the process.