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Exercises · 8.12

Q.Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm=1.013×105 Pa1\ \text{atm} = 1.013 \times 10^{5}\ \text{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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Bulk modulus measures how hard it is to compress a substance. For water, a huge pressure increase (100 atm) causes only a tiny volume decrease (0.5 L out of 100 L), giving a bulk modulus of about 2.03×109 Pa2.03 \times 10^{9}\ \text{Pa}. Air at constant temperature has a bulk modulus equal to its pressure, about 1.01×105 Pa1.01 \times 10^{5}\ \text{Pa}. The ratio is roughly 2×1042 \times 10^{4} — water is about 20,000 times harder to compress than air because molecules in a liquid are already tightly packed, while gas molecules are far apart and easily squeezed.


Concept and Intuition

Bulk modulus (BB) is the ratio of the volumetric stress (pressure change ΔP\Delta P) to the volumetric strain (fractional change in volume −ΔV/V0-\Delta V / V_0). The negative sign ensures BB is positive — when pressure increases, volume decreases.

B=−ΔPΔV/V0B = -\frac{\Delta P}{\Delta V / V_0}

Think of it as the "stiffness" of a material against compression. For a gas, molecules are far apart — you can push them closer easily, so BB is small. For a liquid like water, molecules are already nearly touching; pushing them even a little closer requires enormous force, so BB is huge. That's why water feels "incompressible" in everyday life, though it does compress slightly under extreme pressure.


Step-by-Step Calculation

1. Identify the given data

  • Initial volume: V0=100.0 litreV_0 = 100.0\ \text{litre}
  • Final volume: Vf=100.5 litreV_f = 100.5\ \text{litre}
  • Pressure increase: ΔP=100.0 atm\Delta P = 100.0\ \text{atm}
Watch out

The final volume is larger than the initial volume? That would mean expansion, not compression. Check carefully: the problem says "Pressure increase = 100.0 atm" and "Final volume = 100.5 litre". This is a trick — if pressure increases, volume must decrease. The given final volume is likely a misprint or meant to be 99.5 litre. We'll proceed with the correct physics: volume decreases by 0.5 litre.

So the actual volume change is:

ΔV=Vf−V0=−0.5 litre\Delta V = V_f - V_0 = -0.5\ \text{litre}

2. Convert units to SI

Bulk modulus is usually expressed in pascals (Pa). Convert pressure and volume:

  • Pressure: 1 atm=1.013×105 Pa1\ \text{atm} = 1.013 \times 10^{5}\ \text{Pa}, so

ΔP=100.0×1.013×105=1.013×107 Pa\Delta P = 100.0 \times 1.013 \times 10^{5} = 1.013 \times 10^{7}\ \text{Pa}

  • Volume: 1 litre=10−3 m31\ \text{litre} = 10^{-3}\ \text{m}^3, so

V0=100.0×10−3=0.100 m3V_0 = 100.0 \times 10^{-3} = 0.100\ \text{m}^3

ΔV=−0.5×10−3=−5.0×10−4 m3\Delta V = -0.5 \times 10^{-3} = -5.0 \times 10^{-4}\ \text{m}^3

3. Compute volumetric strain

Volumetric strain is the fractional change in volume:

ΔVV0=−5.0×10−40.100=−5.0×10−3\frac{\Delta V}{V_0} = \frac{-5.0 \times 10^{-4}}{0.100} = -5.0 \times 10^{-3}

This is a 0.5% decrease in volume — tiny for such a large pressure.

4. Apply the bulk modulus formula

B=−ΔPΔV/V0=−1.013×107−5.0×10−3=1.013×1075.0×10−3B = -\frac{\Delta P}{\Delta V / V_0} = -\frac{1.013 \times 10^{7}}{-5.0 \times 10^{-3}} = \frac{1.013 \times 10^{7}}{5.0 \times 10^{-3}}

B=2.026×109 PaB = 2.026 \times 10^{9}\ \text{Pa}

Bwater≈2.03×109 PaB_{\text{water}} \approx 2.03 \times 10^{9}\ \text{Pa}

5. Bulk modulus of air at constant temperature

For an ideal gas at constant temperature (isothermal process), Boyle's law applies: PV=constantPV = \text{constant}. Differentiating gives PdV+VdP=0P dV + V dP = 0, so:

dP−dV/V=P\frac{dP}{-dV/V} = P

Thus the isothermal bulk modulus of an ideal gas equals its pressure.

At normal atmospheric pressure:

Bair=Patm=1.013×105 PaB_{\text{air}} = P_{\text{atm}} = 1.013 \times 10^{5}\ \text{Pa} …

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