Step 1: Understanding the Concept:
The surface-area-to-volume ratio (SA:V) is a measure of the amount of surface area a three-dimensional object has relative to its volume. This question asks how this ratio changes as the size (characteristic dimension) of the object changes.
Step 2: Key Formula or Approach:
Let's analyze a simple shape, a cube of side length L.
The surface area (SA) is the sum of the areas of the 6 faces: \( SA = 6 \times L^2 \)
The volume (V) is: \( V = L^3 \)
The ratio is: \( \frac{SA}{V} = \frac{6L^2}{L^3} = \frac{6}{L} \)
For a sphere of radius r (characteristic dimension):
Surface area (SA) is: \( SA = 4\pi r^2 \)
Volume (V) is: \( V = \frac{4}{3}\pi r^3 \)
The ratio is: \( \frac{SA}{V} = \frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3}{r} \)
Step 3: Detailed Explanation:
In both cases (and for any shape), the surface area is proportional to the square of the characteristic dimension (\(L^2\) or \(r^2\)), while the volume is proportional to the cube of the characteristic dimension (\(L^3\) or \(r^3\)). The ratio is therefore inversely proportional to the characteristic dimension (\(1/L\) or \(1/r\)).
This means that as the characteristic dimension (L or r) decreases, the surface-area-to-volume ratio increases. This is a critical property of nanomaterials, leading to high reactivity and unique surface-dominated effects.
Conversely, as the dimension increases, the ratio decreases. Thus, options A and D are incorrect.
Step 4: Final Answer:
The surface area to volume ratio increases with a decrease in the characteristic dimension of materials.