Bulk Modulus of Gases

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The volume of a gas changes when pressure applied on it is varied. The bulk modulus of a gas is defined as the ratio of volumetric stress to the volumetric strain i.e, \begin{align} B=-\frac{\Delta p}{(\Delta V/V)}\nonumber \end{align} where $\Delta p$ is a change in pressure and $\Delta V$ is change in volume. The negative sign indicates that the gas compresses (volume decreases) when applied pressure is increased. The unit of bulk modulus is Pascal.

Bulk Modulus of Gases
Compression of a gas due to increases in applied pressure. The bulk modulus of a gas is $B=-V\mathrm{d}p/\mathrm{d}V.$

The pressure on a gas can be changed through an isothermal process (constant temperature) or an adiabatic process (constant entropy). A gas has different bulk moduli for isothermal and adiabatic processes.

The isothermal bulk modulus of an ideal gas is the ratio of change in pressure ($\Delta p$) to the fractional change in volume ($\Delta V/V$) at constant gas temperature $T$. Differentiate the ideal gas equation $pV=nRT$ to get $p\Delta V+ V\Delta p=0$ i.e., $\Delta V/V=-\Delta p/p$. Note that $\Delta T=0$ because the temperature is constant. Substitute $\Delta V/V$ in the defining equation to get the isothermal bulk modulus \begin{align} B_\mathrm{isothermal}=-\frac{\Delta p}{(\Delta V/V)}=p. \nonumber \end{align} The isothermal bulk modulus of an ideal gas is equal to its pressure.

The adiabatic bulk modulus of an ideal gas is the ratio of change in pressure to the fractional change in volume when these changes are carried out through an adiabatic process (no heat exchange with surrounding). The equation of state of an ideal gas for an adiabatic process is $pV^\gamma=\mathrm{const}$, where $\gamma$ is the ratio of specific heats. Differentiate $pV^\gamma=\mathrm{const}$ to get $p\gamma V^{\gamma-1}\Delta V+ \Delta p V^\gamma=0$ i.e., $\Delta V/V=-\Delta p/\gamma p$. Substitute in the defining equation to get the adiabatic bulk modulus \begin{align} B_\mathrm{adiabatic}=-\frac{\Delta p}{(\Delta V/V)}=\gamma p. \nonumber \end{align} Note that $B_\mathrm{adiabatic}=\gamma B_\mathrm{isothermal}>B_\mathrm{isothermal}$. The bulk modulus of a gas in an isobaric process is zero.

The compressibility of a gas is the reciprocal of its bulk modulus i.e., $K=1/B$.

The speed of sound in a gas is related to its bulk modulus by $ v=\sqrt{B/\rho} $, where $\rho$ is the density of gas.

Solved Problems from IIT JEE

Problem from IIT JEE 1998

A given quantity of an ideal gas is at pressure $p$ and absolute temperature $T$. The isothermal bulk modulus of the gas is,

  1. 2p/3
  2. p
  3. 3p/2
  4. 2p

Solution: Isothermal bulk modulus is ratio of volumetric stress to volumetric strain at constant temperature i.e., \begin{align} B=-\frac{\mathrm{d}p}{\mathrm{d}V/V}=\left. -V\frac{\mathrm{d}p}{\mathrm{d}V}\right|_{T}. \end{align} Differentiate the ideal gas equation, $pV=nRT$, at constant $T$ to get, \begin{align} \left.\frac{\mathrm{d}p}{\mathrm{d}V}\right|_{T}=-\frac{p}{V}. \end{align} Substitute ${\mathrm{d}p}/{\mathrm{d}V}$ from second equation into first equation to get $B=p$.

Problem from IIT JEE 2005

The pressure of a medium is changed from $1.01\times{10}^{5}$ Pa to $1.165\times{10}^{5}$ Pa and change in volume is 10% keeping temperature constant at 20℃. The bulk modulus of the medium is

  1. $204.8\times{10}^{5}$ Pa
  2. $102.4\times{10}^{5}$ Pa
  3. $51.2\times{10}^{5}$ Pa
  4. $1.55\times{10}^{5}$ Pa

Solution: Substitute the values in the formula for bulk modulus to get \begin{align} B&=-\frac{\Delta p}{\Delta V/V}\nonumber\\ &=-\frac{1.01\times{10}^{5}-1.165\times{10}^{5}}{0.1}\nonumber\\ &=1.55\times{10}^{5} \mathrm{Pa}.\nonumber \end{align}

Problem from IIT JEE 1995

A gaseous mixture enclosed in a vessel of volume $V$ consists of one mole of gas A with $\gamma=C_p/C_V=5/3$ and another gas B with $\gamma=7/5$ at a certain temperature $T$. The molecular weights of the gases A and B are $4$ and $32$, respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation $pV^{19/13}=\mathrm{constant}$, in adiabatic processes.

  1. Find the number of moles of the gas B in the gaseous mixture.
  2. Compute the speed of sound in the gaseous mixture at 300 K.
  3. If $T$ is raised by 1 K from 300 K, find the percentage change in the speed of sound in the gaseous mixture.
  4. The mixture is compressed adiabatically to $1/5\,\mathrm{th}$ of its initial volume $V$. Find the change in its adiabatic compressibility in terms of the given quantities.

Solution: Eliminate $C_p$ from the relations $C_p-C_v=R$ and $\gamma=C_p/C_v$ to get, $C_v={R}/{(\gamma-1)}$. The internal energies of the gases A and B are $U_\text{A}=n_\text{A} C_\text{$v$,A} T$ and $U_\text{B}=n_\text{B} C_\text{$v$,B} T$ where $C_\text{$v$,A}={R}/{(\gamma_A-1)}$ and $C_\text{$v$,B}={R}/{(\gamma_B-1)}$. The internal energy of the mixture is \begin{align} U=U_\text{A}+U_\text{B}=\left[\frac{n_A}{\gamma_A-1}+\frac{n_B}{\gamma_B-1}\right] R T.\nonumber \end{align} The molar specific heat of the mixture, at the constant volume is \begin{align} \label{bsa:eqn:1} C_\text{$v$,m}&=\frac{1}{n_A+n_B}\frac{\mathrm{d}U}{\mathrm{d}T} \nonumber\\ &=\frac{1}{n_A+n_B}\left[\frac{n_A}{\gamma_A-1}+\frac{n_B}{\gamma_B-1}\right] R \nonumber\\ &=\frac{R}{\gamma_m-1}. \end{align} The adiabatic equation for the mixture is, $pV^{19/13}=\text{constant}$, which gives $\gamma_m=19/13$. Substitute $n_A=1$, $\gamma_A=5/3$, $\gamma_B=7/5$, and $\gamma_m=19/13$ in above equation to get $n_B=2$.

Molecular mass of the mixture is \begin{align} M_m & =\frac{M_\text{total}}{n_{m}}=\frac{n_AM_A+n_BM_B}{n_A+n_B} \nonumber\\ &=\frac{1(4)+2(32)}{1+2}=22.67 \,\mathrm{g/mol}. \nonumber \end{align} The speed of sound in the mixture is \begin{align} v&=\sqrt{\frac{\gamma_m RT}{M_m}}=401\,\mathrm{m/s}. \nonumber \end{align} Differentiate above equation with respect to $T$ and simplify to get \begin{align} \frac{\Delta v}{v}=\frac{1}{2}\frac{\Delta T}{T}=\frac{1}{2\times300}=0.167\%. \nonumber \end{align} The bulk modulus is defined as \begin{align} \label{bsa:eqn:3} B=-\frac{\mathrm{d}p}{\mathrm{d}V/V}=-V\frac{\mathrm{d}p}{\mathrm{d}V}. \end{align}

Differentiate $pV^{\gamma_m}=\text{constant}$, to get ${\mathrm{d}p}/{\mathrm{d}V}=-{\gamma_m p}/{V}$ for an adiabatic process. Substitute it in above equation to get $B=\gamma_m p$. Thus, adiabatic compressibility is \begin{align} \beta={1}/{B}={1}/{(\gamma_m p)}.\nonumber \end{align} Let $\beta={1}/{(\gamma_m p)}$ and $\beta^\prime={1}/{(\gamma_m p^\prime)}$ be the compressibilities when the volumes are $V$ and $V^\prime={V}/{5}$. The adiabatic equation, $pV^{\gamma_m}=p^\prime {V^\prime}^{\gamma_m}$, gives $p^\prime=5^{\gamma_m}\, p$. Also, $p={n_mRT}/{V}$. Substitute the values to get \begin{align} \Delta\beta&=\beta^\prime-\beta=\frac{1}{\gamma_m p}\left(5^{-\gamma_m}-1\right) \nonumber\\ &=-8.27\times{10}^{-5}V.\nonumber \end{align}

Questions on Bulk Modulus

Question 1: The ratio of isothermal bulk modulus to the adiabatic bulk modulus for a monatomic gas is given by

A. 5/3
B. 0.6
C. 1.4
D. 1.0

Question 2: The temperature-volume (T-V) diagram for $n$ moles of an ideal gas undergoing an adiabatic process is shown in the figure. The adiabatic bulk modulus of the gas at the point P is

T-V diagram for adiabatic process
A. $5nRT_0/3V_0$
B. $nR(2+T_0/V_0)$
C. $nR(1+T_0/V_0)$
D. $2nRT_0/V_0$

Related Topics

  1. Isothermal and Adiabatic Processes
  2. Speed of Sound in Gases

References and External Links

  1. IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
  2. Concepts of Physics Part 2 by HC Verma (Link to Amazon)
  3. Bulk Modulus (Hyperphysics)
  4. Bulk Modulus of Gases (Cambridge)
  5. Adiabatic Bulk Modulus (pdf, MIT)