Solveeit Logo
Login

Question

Question: Compute the bulk modulus of water from the following data: Initial volume=100.0 liter, Pressure incr...

Compute the bulk modulus of water from the following data: Initial volume=100.0 liter, Pressure increases=100.0 atm (1atm=1.013×105Pa1atm=1.013\times {{10}^{5}}Pa), Final volume=100.5 liter. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Explanation

Solution

Water is denser than air so water can be compressed less than air. Convert all the units into SI units. Calculate the difference between the final and initial volume to get increase in volume, apply in the formula to get the bulk modulus.

Formulas used:
Bulk Modulus=p(VV1)=pV1V\dfrac{p}{(\dfrac{V}{{{V}_{1}}})}=\dfrac{p{{V}_{1}}}{V}

Complete step-by-step answer:
The bulk modulus of elasticity is used to describe the mechanical properties of solids. The bulk elastic properties of a material let us know how much it will compress under a given amount of external pressure. It is important to calculate the ratio of the change in pressure to the fractional volume compression. The value is denoted with a symbol of K. It has the dimension of force per unit area. In the English system, it is expressed in the units per square inch. Newton per square meter (N/m2) is in the metric system.
All the data given to us are
V1{V_1}=100.0l =100.0×10−3100.0\times {{10}^{-3}} m3{m^{3}},
V2{V_2} = 100.5l = 100.5\times {{10}^{-3}}$$${m^{3}}$, p=100 atm=100\times 1.013\times {{10}^{5}}PawhereV1isinitialvolume,V2isfinalvolumeandpisanincreaseinpressure.Letusfindtheincreaseinvolume(V)now,Pa where V1 is initial volume, V2 is final volume and p is an increase in pressure. Let us find the increase in volume (V) now, V={{V}{2}}-{{V}{1}}=100.5\times {{10}^{-3}}-100.0\times {{10}^{-3}}=0.5\times {{10}^{-3}}{{m}^{3}}Asweknow,Bulkmodulus= As we know, Bulk modulus=\dfrac{p}{(\dfrac{V}{{{V}{1}}})}=\dfrac{p{{V}{1}}}{V}$$………. (1)

Now let’s substitute the given values in Equation (1)
Bulk modulus = 100×1.013×105×100.0×10−30.5×10−3=2.026×109\dfrac{100\times 1.013\times {{10}^{5}}\times 100.0\times {{10}^{-3}}}{0.5\times {{10}^{-3}}}=2.026\times {{10}^{9}}Pa.
Bulk modulus of air = 1×105Pa1\times {{10}^{5}}Pa
Ratio of bulk modulus of water to air = 2.026×1091×105\dfrac{2.026\times {{10}^{9}}}{1\times {{10}^{5}}}= 2.026×1042.026\times {{10}^{4}}
As air is more compressible than water therefore the ratio is very large.

Note: In this particular question, converting the units into SI units is important to obtain the correct answer. Bulk modulus demonstrates the change in the volume of a system due to the application of tensile stress uniformly over the entire surface of the system. This stress leads to change in pressure.