91Ó°ÊÓ

Thermodynamics is a branch of physics that studies the relationships between heat, work, temperature, and energy. In this exercise, we focus specifically on the concept of work done during gas compression. When a gas is compressed, energy is transferred in the form of work.
In thermodynamics, several laws help us understand these energy exchanges, notably the first law which is also known as the law of energy conservation. It states that energy cannot be created or destroyed, only transformed from one form to another. This fundamental understanding is crucial when calculating the work done on or by a system such as a gas during compression or expansion.

In this problem, the pressure and volume of the gas are linearly related. This means that as one variable changes, the other changes in a directly proportional manner. This linear relationship can be mathematically described using the equation

\text{ \(p = aV + b\)},

where -a is the slope of the line ( it tells us how much the pressure changes for a given change in volume). - b is the y-intercept ( it tells us what the pressure would be if the volume were zero). Knowing this relationship allows us to solve for the constants a and b using the given boundary conditions (initial and final pressure and volume values).

To find - a and b, we set up two equations based on the given data points and solve them simultaneously. For instance, in this problem, we had

\( 100 = a(0.3) + b\) (initial condition)

and \( 300 = a(0.1) + b\)
(final condition) . By solving these equations, we fine that \( a = -1000\) (kPa/m^3) and \( b = 400\) (kPa). This linear relationship is crucial in setting up the problem for the subsequent steps where we integrate the work equation.

The work done by or on a gas during a process can often be determined using an integration of the pressure-volume relationship. In this case of linear compression, we utilize a simplified form of the work equation:
\[ W = \frac{1}{2} (p_1 + p_2)(V_1 - V_2) \] This formula is specifically useful when the relationship between pressure and volume is linear. It essentially averages the initial and final pressures and multiplies by the change in volume.

Here’s the step-by-step breakdown of how we applied it:

• First, substitute the given pressures (\(p_1 = 100\) kPa and \(p_2 = 300\) kPa
• Next, use the as volumes \(V_1 = 0.3 \text{ m}^3 and \) V_2 = 0.1 \text{m}^3\(
• Now, calculate the changes: \)\frac{1}{2} (100 + 300)(0.3 - 0.1)\(
• This becomes: \)\frac{1}{2} (400)(0.2) \text{kPa} \text{m}^3\(
• Which further simplifies to: \)\frac{1}{2} (80) \text{kJ}\(
• Finally, we find the work done, \)W = 40\( \)\text{kJ}$

This equation allows us to easily compute the work involved in gas processes with linear pressure-volume relationships, providing clear insight into energy transfer within thermodynamic systems.