To calculate isothermal bulk modulus, I have two methods.

Method 1:








Method 2:

I love calculus


On differentiating,


$\dfrac{dP}{dV}=-\dfrac{P}{V}$ .......(1)


Using equation (1)



Using method (1) I got $P_2$ as the answer whereas in method (2) I got $P_1$ as the answer. I know $P_1\neq P_2$. Which one is correct then?


(2) gives the answer $P$, the pressure at any instant, and not $P_1$. Method(1) is wrong. $B=-\frac{dP}{dV}V$ includes $\frac{dP}{dV}$, which is the instantaneous rate of change of pressure with respect to volume and B varies with P(or V) in this process. What you are doing is calculating the average change over a significant change in pressure and volume. Instantaneous change can be approximated by average change only if the change is very small.

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    $\begingroup$ If I tell you that the ideal gas was initially at pressure $P_1$ and ask you what is the isothermal bulk modulus, what will you answer $\endgroup$ – user203191 Aug 27 '18 at 9:58
  • $\begingroup$ Method (1) is not "wrong" if $P_2$ is very close to $P_1$ - it is just explicitly writing down the limit, rather than using calculus. If the change is very small, then in the limit $P_1$ = $P_2$ = $P$. The mistake in the OP's first method is assuming that $K$ is constant when $P$ changes. Of course $K$ is independent of pressure (to a good approximation) for solids and liquids, but not for gases. $\endgroup$ – alephzero Aug 27 '18 at 12:18
  • $\begingroup$ @LoopBack No matter what the initial pressure is the bulk modulus of an ideal gas for an isothermal process at a particular instant is given by the pressure at that instant. If the change is very small then method (1) is not completely wrong and bulk modulus is quite close to $P_1$, so you can safely say that it's $P_1$. $\endgroup$ – Asit Srivastava Aug 27 '18 at 13:07

Isothermal bulk modulus is defined as volume times negative partial derivative of pressure with respect to volume at constant temperature: 

$K = -V \dfrac{\partial P}{\partial V}$

An ideal gas satisfies the following equation of state: 

$PV = nRT$ 

So the pressure for an ideal gas is given by: 

$P = \dfrac{nRT}{V} $

The partial derivative of pressure w.r.t. volume at constant temperature is: 

$\dfrac{\partial P}{\partial V} = \dfrac{∂\dfrac{nRT}{V}}{∂P}$

$= nRT \dfrac{d\dfrac{1}{V}}{dV} $

$= - nRT/V² $

$= - \dfrac{\dfrac{nRT}{V}} {V}$

$= - \dfrac{P}{V} $


$K = P$


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