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For a perfectly elastic body, Bulk modulus always remains constant and is defined as, $$B=-V_i \frac{\Delta P}{\Delta V} \tag{1}$$ Which means, $$B \left(\frac{V_f -V_i}{V_i}\right)= -(P_f-P_i)$$

But, the definition of bulk modulus in a lot of places is given as, $$B=-V\frac{dP}{dV}\tag{2}$$ In particular, this definition is used to show that for an ideal gas in an isothermal process, we get $B=P$, i.e., the bulk modulus is equal to the pressure for an ideal gas in constant temperature.

Using the definition ($2$) for an elastic body, we get, $$\frac{B}{V}dV = -dP$$ $$\Rightarrow \int_{V_i}^{V_f} \frac{B}{V}dV= -\int_{P_i}^{P_f}dP$$ $$\Rightarrow B\ln\frac{V_f}{V_i}= -(P_f -P_i) $$

Clearly, this doesn't confirm the definition ($1$).

What is a general definition of bulk modulus that we can get ($1$) for an elastic body and ($2$) for an ideal gas? Also, the general definition should not allow us to use definition ($2$) in the case of an elastic body.

If there is no such general definition, why call the quantities in ($1$) and ($2$) with the same name 'Bulk modulus'?

GiorgioP-DoomsdayClockIsAt-89
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Navneet
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2 Answers2

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If $V_f \approx V_i$, then $$\ln\left(\frac{V_f}{V_i}\right) = \ln \left( 1 + \frac{V_f - V_i}{V_i} \right) \approx \frac{V_f - V_i}{V_i}.$$So the two formulas are functionally equivalent so long as the volume does not change significantly during the process.

I suspect that your equation (1) implicitly assumes that $\Delta V = V_f - V_i$ is small compared to $V_i$ (or $V_f$.) The equation (2) is the more general relation.

I should also note that the bulk modulus is not necessarily constant with respect to volume (or pressure), so the integration you have written following (2) is not necessarily correct. For example, for an ideal gas $B = P = NkT/V$, which is very obviously dependent on $V$ or on $P$.

Michael Seifert
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  1. No doubts,- differential forms are always superior over $\Delta$ variants, simply because differential equation form takes into account instant resulting value. So, $$B=-V_0\frac{dP}{dV}\tag{2}$$ describes instant bulk modulus, which is very handy if for some reason bulk modulus of material is function of time, anisotropic or etc, so to say if $B=B(x_1,x_2,x_3,\dots)$. $\Delta$ variant can only get you to the average bulk modulus, no more, no less. Sure, sometimes average is the only thing you care, then it's ok.

  2. You have made an error in integration. $$\int_{V_0}^{V_f} \frac {B}{V_0} dV = B \frac{V_f - V_0}{V_0},$$

in other words,- it's not logarithm of volume ratios, because $V_0$ is not same volume integration variable, but rather initial volume- wiki mentions it, but alas uses wrong notation for initial volume, which may get confused somebody upon integration. So the result actually matches (1).

Agnius Vasiliauskas
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