Question

One-half kilogram of propane initially at 4 bar, \(30^{\circ} \mathrm{C}\) undergoes a process to 14 bar, \(100^{\circ} \mathrm{C}\) while being rapidly compressed in a piston-cylinder assembly. Heat transfer with the surroundings at \(20^{\circ} \mathrm{C}\) occurs through a thin wall. The net work is measured as \(-72.5 \mathrm{~kJ}\). Kinetic and potential energy effects can be ignored. Determine whether it is possible for the work measurement to be correct.

Short answer

Evaluate the heat transfer feasibility by comparing calculated \(Q\).

Step-by-step solution

- Identify the given data

The given data is:- Initial pressure, \(P_1\) = 4 bar- Initial temperature, \(T_1\) = 30^{\circ}C- Final pressure, \(P_2\) = 14 bar- Final temperature, \(T_2\) = 100^{\circ}C- Mass of propane, \(m\) = 0.5 kg- Heat transfer with surroundings, \(T_{\text{sur}}\) = 20^{\circ} C- Net work, \(W = -72.5 \mathrm{kJ}\)

- Understand the First Law of Thermodynamics

The First Law of Thermodynamics for a closed system is given by: \[ Q - W = \Delta U \]where \(Q\) is heat transfer, \(W\) is work done, and \(\Delta U\) is the change in internal energy.

- Find specific internal energies from tables

Look up the specific internal energy values for propane at the given states from standard thermodynamic tables:- Specific internal energy at state 1, \(u_1\). (at \(30^{\circ}C\)- Specific internal energy at state 2, \(u_2\). (at \(100^{\circ}C\)These values are typically found in tables or software like REFPROP or an engineering textbook.

- Calculate \Delta U

The change in internal energy \(\Delta U\) is: \[ \Delta U = m \(u_2 - u_1\) \]}, {

- Apply the First Law

Rearrange the First Law equation to solve for \(Q\):: \[ Q = W + \Delta U \]Insert the values for \(W\) and \ \Delta U \ to determine the heat transfer.

- Evaluate Heat Transfer Feasibility

Given the surroundings temperature of \(20^{\circ}C\), compare the calculated \(Q\) with possible realistic heat transfer rates to determine the plausibility of the measured work value.

Key concepts

First Law of Thermodynamics

The First Law of Thermodynamics is crucial in understanding energy transfer in thermodynamic processes. It asserts that energy cannot be created or destroyed, only transferred or converted.
This is mathematically expressed as: \( Q - W = \Delta U \)
where:

• \( Q \): Heat added to the system
• \( W \): Work done by the system
• \( \Delta U \): Change in internal energy of the system

This principle helps us account for all energy transitions, ensuring we can solve for unknowns like heat transfer or work done during a thermodynamic process.
For instance, in the given problem, knowing the work done and the change in internal energy allows us to solve for the heat exchanged.

Internal Energy

Internal Energy (\( U \)) refers to the total energy contained within a system, which includes kinetic energy of molecules and potential energy due to molecular interactions.
It is a state function, meaning it depends only on the state of the system and not on how it reaches that state. Internal energy changes with temperature and phase changes.
For a given mass (\( m \)) of a substance, the change in internal energy (\( \Delta U \)) is: \( \Delta U = m (u_2 - u_1) \)
Here, \( u_1 \) and \( u_2 \) are the specific internal energies at the initial and final states, respectively. Thermodynamic tables or software like REFPROP can provide these values for accurate calculations. In the exercise, determining \( u_1 \) and \( u_2 \) from the tables is critical to solving for \( \Delta U \).

Heat Transfer

Heat transfer (\( Q \)) is the energy exchanged between a system and its surroundings due to a temperature difference.
It flows from the hotter region to the cooler one, affecting the system's internal energy.
The direction of heat flow can be positive (into the system) or negative (out of the system). In the context of the First Law of Thermodynamics, \( Q \) can be directly calculated if the work done (\( W \)) and the change in internal energy (\( \Delta U \)) are known: \( Q = W + \Delta U \)
This equation helps figure out if the measured work aligns with possible heat transfer and change in internal energy values. In practice, comparing given constraints like the surrounding temperature helps assess the feasibility of computed results.

Work Measurement

Work (\( W \)) in thermodynamics is the energy transferred when a force is applied over a distance, or when a system undergoes a volume change under pressure.
In a piston-cylinder assembly, work can be positive (work done by the system) or negative (work done on the system).
The net work done on the system can be measured and used in the First Law of Thermodynamics: \( Q - W = \Delta U \)
In our problem, the net work is given as \( -72.5 \text{kJ} \). Such measurements are essential for understanding the energy dynamics of the process and the feasibility of the system's state changes.

Thermodynamic Tables

Thermodynamic tables provide crucial data like specific volumes, internal energy, enthalpy, and entropy at various states.
These values are essential for solving thermodynamic problems accurately.
For substances like propane, tables can include specific internal energy values (\( u \)) at different temperatures and pressures.
In the given exercise, specific internal energies \( (u_1 \) and \( u_2 \)) are indispensable for calculating \( \Delta U \). Accurate table values enable precise application of thermodynamic principles, ensuring the reliability of the solution.
Always cross-reference your values with standardized tables or reliable software to avoid errors in your calculations.