Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 x 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

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To compute the bulk modulus (K) of water, we can use the formula:

K=−ΔPΔVVK=−VΔVΔP

Where:

• ΔPΔP is the change in pressure,
• ΔVΔV is the change in volume, and
• VV is the initial volume.

Given:

• Initial volume V=100.0V=100.0 litres,
• Pressure increase ΔP=100.0ΔP=100.0 atm (which is equivalent to 100.0×1.013×105100.0×1.013×105 Pa),
• Final volume V′=100.5V′=100.5 litres.

We can plug these values into the formula to find the bulk modulus of water.

Kwater=−100.0×1.013×105100.5−100.0100.0Kwater=−100.0100.5−100.0100.0×1.013×105

Kwater=−100.0×1.013×1050.5Kwater=−0.5100.0×1.013×105

Kwater=−2.026×108 PaKwater=−2.026×108 Pa

Now, let's compare this with the bulk modulus of air at constant temperature. The bulk modulus of air is significantly smaller than that of water. Air is compressible, meaning it can easily be squeezed into a smaller volume under pressure, hence its bulk modulus is much lower compared to water.

In simple terms, the ratio of the bulk modulus of water to that of air is large because water is much less compressible compared to air. When you apply pressure to water, it doesn't compress easily, so you need to exert a lot more force to change its volume even slightly. On the other hand, air is highly compressible, so it takes much less force to change its volume. This fundamental difference in compressibility is why the bulk modulus ratio is large.