(a) Define pressure - Leaving Cert Physics - Question 12 - 2006

Pressure is a scalar quantity. Unlike vector quantities, which have both magnitude and direction, pressure only has magnitude and acts uniformly in all directions at a point within a fluid. It does not have a specific orientation, which is why it is classified as a scalar.

Boyle's law states that for a fixed mass of an ideal gas at constant temperature, the pressure of the gas is inversely proportional to its volume. This can be mathematically represented as:

$P \propto \frac{1}{V}$

or, in a more conventional form:

$PV = k$

where $k$ is a constant when temperature remains unchanged.

To calculate the pressure at the bottom of the lake, we utilize the relationship that the pressure difference is equal to the density of the fluid times gravity times the height:

$P_{bottom} = P_{top} + h \cdot \rho \, g$

Where $P_{top}$ is the atmospheric pressure (1.01 × 10^6 Pa), and the pressure at the bottom is three times that of the top:

Hence, $P_{bottom} = 3.03 × 10^6 \,\text{Pa}$

To find the depth, rearranging the formula for pressure gives:

$h = \frac{P_{bottom} - P_{top}}{\rho \, g}$

Plugging in the values:

$h = \frac{2.02 × 10^6 \,\text{Pa}}{(1.0 × 10^3 \,\text{kg m}^{-3})(9.8 \,\text{m s}^{-2})} = \frac{2.02 × 10^6}{9800} \approx 206.61 \,\text{m}$

Thus, the depth of the lake is approximately 206.61 meters.