Question

Two kilograms of Refrigerant \(134 a\), initially at 2 bar and occupying a volume of \(0.12 \mathrm{~m}^{3}\), undergoes a process at constant pressure until the volume has doubled. Kinetic and potential energy effects are negligible. Determine the work and heat transfer for the process, each in \(\mathrm{kJ}\).

Short answer

Work done is 24 kJ and heat transfer is also 24 kJ.

Step-by-step solution

- Determine Initial Specific Volume

Given the initial state with a pressure of 2 bar and a volume of 0.12 m^3, calculate the specific volume using: \(v_1 = \frac{V}{m} = \frac{0.12 \, \mathrm{m^3}}{2 \, \mathrm{kg}} = 0.06 \, \mathrm{m^3/kg}\)

- Determine Final Volume and Specific Volume

Since the volume doubles at constant pressure, the final volume is: \(V_2 = 2 \times 0.12 \, \mathrm{m^3} = 0.24 \, \mathrm{m^3} \) The final specific volume is: \(v_2 = \frac{V_2}{m} = \frac{0.24 \, \mathrm{m^3}}{2 \, \mathrm{kg}} = 0.12 \, \mathrm{m^3/kg}\)

- Use the Ideal Gas Equation

Since the refrigerant can be treated as an ideal gas in this case, use the ideal gas equation for both states. \ P V = mRT\ For initial state: \(2 \, \mathrm{bar} \times 0.12 \, \mathrm{m^3} = 2 \, \mathrm{kg} \times R \times T_1 \) For final state: \(2 \, \mathrm{bar} \times 0.24 \, \mathrm{m^3} = 2 \, \mathrm{kg} \times R \times T_2 \)

- Calculate the Work Done

The work done by the system at constant pressure is \(W = P \Delta V\) \(W = 2 \, \mathrm{bar} \times (0.24 \, \mathrm{m^3} - 0.12 \, \mathrm{m^3}) \times 100 \, \mathrm{J/\mathrm{m^3}} = 24 \, \mathrm{kJ}\)

- Apply the First Law of Thermodynamics

Using the first law of thermodynamics, \(\Delta U = Q - W\) The change in internal energy, \(\Delta U\), for an ideal gas at constant pressure is \(\Delta U = m C_v \Delta T\) where \(C_v\) is the specific heat at constant volume. Since this calculation provides approximations, we'll assume \(\Delta U\) to be nearly zero.

- Solve for Heat Transfer

Rearrange the first law equation to solve for \(Q\): \(Q = \Delta U + W\) Since \Delta U \ is negligible, \(Q = 0 + 24 \, \mathrm{kJ} = 24 \, \mathrm{kJ}\)

Key concepts

Ideal Gas Equation

The ideal gas equation is a fundamental law in thermodynamics. It is represented as \[ PV = nRT \]Here,

\t
• \t\tP is the pressure\t
\t
• \t\tV is the volume \t
\t
• \t\tn is the amount of substance \t
\t
• \t\tR is the universal gas constant \t
\t
• \t\tT is the temperature in Kelvin\t

In our exercise, Refrigerant 134a is treated as an ideal gas, and we use this equation to connect the initial and final states of the process.To find the required parameters at different states, this equation can simplify the process. By knowing the pressure and volume, we can solve for temperature (T1 and T2). This provides a clear path to solving our thermodynamics problems.

Specific Volume Calculation

Specific volume is a key concept in thermodynamics and refers to the volume occupied by a unit mass of a substance. It is calculated as:\[ v = \frac{V}{m} \]Where,

\t
• \t\tV is the total volume \t
\t
• \t\tm is the mass\t

In our given problem, we initially have a pressure (P1) of 2 bar and a volume of 0.12 m^3. This allows us to calculate the initial specific volume:\[ v_1 = \frac{0.12 \, \mathrm{m^3}}{2 \, \mathrm{kg}} = 0.06 \, \mathrm{m^3/kg} \]As the volume doubles during the process, the final volume is 0.24 m^3, leading to a final specific volume:\[ v_2 = \frac{0.24 \, \mathrm{m^3}}{2 \, \mathrm{kg}} = 0.12 \, \mathrm{m^3/kg} \]

First Law of Thermodynamics

The first law of thermodynamics relates the change in internal energy to the heat added to the system and work done by the system. It is given by:\[ \Delta U = Q - W \]In this expression,

\t
• \t\t\( \Delta U \) is the change in internal energy\t
\t
• \t\tQ is the heat transfer into the system\t
\t
• \t\tW is the work done by the system on its surroundings\t

In our example, for an ideal gas assuming negligible kinetic and potential energy effects, we approximate the change in internal energy, \( \Delta U \), to be nearly zero. The relation can then be simplified to calculate the heat transfer:\[ Q = \Delta U + W \]Using this equation, we find the heat transfer of 24 kJ by knowing the work done (24 kJ) and assuming minimal change in internal energy.

Constant Pressure Process

A constant pressure process is crucial in thermodynamic analyses and simplification. When a process occurs at constant pressure, it is often referred to as an isobaric process. This means that the pressure (P) remains unchanged, which simplifies the application of thermodynamic equations.For our problem of Refrigerant 134a, maintaining constant pressure (2 bar) while the volume changes allows for straightforward calculation of work done using \[ W = P \Delta V \]Given that the initial volume, \( V_1 \), is 0.12 m^3 and the final volume, \( V_2 \), is 0.24 m^3, the change in volume, \( \Delta V \), is:\[ \Delta V = V_2 - V_1 = 0.24 \, \mathrm{m^3} - 0.12 \, \mathrm{m^3} = 0.12 \, \mathrm{m^3} \]Substituting into the work equation, we find\[ W = 2 \, \mathrm{bar} \times 0.12 \, \mathrm{m^3} \times 100 \, \mathrm{J/m^3} = 24 \, \mathrm{kJ} \]

Work and Heat Transfer

Understanding work and heat transfer is vital in thermodynamics. Work (W) is the energy transferred when a force moves an object. In thermodynamic processes, it is generally related to the change in volume. The equation is:\[ W = P \Delta V \]Heat transfer (Q) is the energy transferred due to a temperature difference. According to the first law of thermodynamics, it is calculated as:\[ Q = \Delta U + W \]Where \( \Delta U \) is the change in internal energy. In our example where we assume \( \Delta U \) to be zero due to negligible kinetic and potential energy effects, the heat transfer equals the work done by the system:\[ Q = 24 \, \mathrm{kJ} \]