91Ó°ÊÓ

The Pressure-Volume (P-V) relationship is fundamental in understanding the behaviors of gases during expansion or compression. In a gas system, pressure (P) and volume (V) are inversely related when temperature is constant, a concept described by Boyle's Law. However, in our context, we are dealing with a linear change between pressure and volume.
When we have a situation where gas expands or contracts, plotting its pressure against volume on a graph gives us what is known as an indicator diagram. Here, pressure drops from an initial value of \(6 \times 10^{5} \; \text{Nm}^{-2}\) to \(4 \times 10^{5} \; \text{Nm}^{-2}\) as the volume increases from \(1 \; \text{m}^3\) to \(3 \; \text{m}^3\).
In this particular problem, the indicator diagram shows a straight line, signifying a linear relationship between pressure and volume. This suggests that as volume increases, pressure decreases at a constant rate. Therefore, we can use a linear equation to describe this relationship.

In a linear indicator diagram, pressure and volume change along a straight line path on the P-V graph. This is significant because it simplifies analyzing the behavior of the gas system by representing the changes with a linear equation.
To set up such a diagram in our exercise, identify two points:

• Initial state: \((V_1, P_1) = (1, 6 \times 10^{5})\)
• Final state: \((V_2, P_2) = (3, 4 \times 10^{5})\)

Drawing a line between these points gives us the pathway of the gas expansion process. This straight line tells us each increment in volume corresponds to a decrement in pressure.
To find the equation of this line, we calculate the slope \(m\) using the formula:\[ m = \frac{(4 \times 10^{5} - 6 \times 10^{5})}{(3 - 1)} = -10^{5} \]
The line’s equation becomes a useful tool to predict values of pressure at any volume within the range. We use the formula \(P = mV + c\), where \(c\) can be found using known coordinates.

The calculation of work done by the gas involves using integral calculus, specifically the definite integral. In this context, work done by a gas can be found by calculating the area under the line on the P-V diagram.
Using the derived equation for the pressure-volume relationship, \(P = -10^{5}V + 7 \times 10^{5}\), the expression for work done as the gas expands from \(V_1\) to \(V_2\) is:
\[ W = \int_{1}^{3} (-10^{5}V + 7 \times 10^{5}) \, dV \]
Carrying out this integration, we find:\[ W = \left[ -\frac{10^{5}V^2}{2} + 7 \times 10^{5}V \right]_{1}^{3} \] By computing the value, first substitute the upper limit \(V = 3\) and then subtract the value with the lower limit \(V = 1\): \[ W = \left(-\frac{10^{5}(3)^2}{2} + 7 \times 10^{5} \cdot 3\right) - \left(-\frac{10^{5}(1)^2}{2} + 7 \times 10^{5} \cdot 1\right) \] This simplifies to: \[ W = (-45 \times 10^{5} + 21 \times 10^{5}) - (-0.5 \times 10^{5} + 7 \times 10^{5}) = 3 \times 10^{5} \; \text{J} \] This final result represents the work done by the gas as it expands under the conditions given, confirming the choice \(b)\) \(3 \times 10^{5} \mathrm{~J}\).