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The isothermal bulk modulus of elasticity of a gas is \[1.5 \times {10^5}N{m^{ - 2}}\] . Its adiabatic bulk modulus of elasticity will be (if \[\gamma = 1.4\] )
A. \[3 \times {10^5}N{m^{ - 2}}\]
B. \[2.1 \times {10^5}N{m^{ - 2}}\]
C. \[1.5 \times {10^5}N{m^{ - 2}}\]
D. \[\infty \]

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Last updated date: 13th Jun 2024
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Answer
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Hint: We know that the equation for the adiabatic bulk modulus is \[{B_{adiabatic}} = \gamma p\] and the equation for the isothermal bulk modulus is \[{B_{isothermal}} = p\] . Now divide these equations to get the following relation \[{B_{adiabatic}} = \gamma \times {B_{isothermal}}\] and then calculate the value of \[{B_{adiabatic}}\] .

Complete answer:
Before moving on to the mathematical calculations let’s first discuss isothermal bulk modulus and adiabatic bulk modulus.
Adiabatic process - Without moving heat or mass between a thermodynamic system and its environment, an adiabatic process happens. Unlike an isothermal cycle, an adiabatic process only transfers energy to the atmosphere as a task.
Isothermal process - An isothermal process is a system shift in which the temperature stays constant. This usually happens when a system is in contact with an outside thermal reservoir and the system adjusts slowly enough to allow the system to continue to adapt through heat exchange to the temperature of the reservoir.
Isothermal bulk modulus – It is the ratio of change in pressure \[\left( {\Delta p} \right)\] to the change in the volume \[\left( {\dfrac{{\Delta V}}{V}} \right)\] at gas temperature \[T\] .
\[{B_{isothermal}} = - \dfrac{{\Delta p}}{{\left( {\dfrac{{\Delta V}}{V}} \right)}}\]
But as in an isothermal process, the volume does not change
\[{B_{isothermal}} = p\] (Equation 1)
Adiabatic process – It is the ratio of change in pressure to the change in volume when the process is adiabatic.
\[{B_{adiabatic}} = - \dfrac{{\Delta p}}{{\left( {\dfrac{{\Delta V}}{V}} \right)}} = \gamma p\] (Equation 2)
Here, \[\gamma = \] The ratio of specific heat
Dividing equation 2 by equation 1, we get
\[\dfrac{{{B_{adiabatic}}}}{{{B_{isothermal}}}} = \dfrac{{\gamma p}}{p}\]
\[\dfrac{{{B_{adiabatic}}}}{{{B_{isothermal}}}} = \gamma \]
\[{B_{adiabatic}} = \gamma \times {B_{isothermal}}\]
\[{B_{adiabatic}} = 1.4 \times 1.5 \times {10^5} = 2.1 \times {10^5}N{m^{ - 2}}\]

So, the correct answer is “Option B”.

Note:
The bulk modulus is based on Hooke’s law. According to Hooke’s law, the strain in a material is directly proportional to the stress applied.
\[\dfrac{{Strain}}{{Stress}} = B\]
Here, \[B = \] Bulk’s modulus
The above equation is just for the volumetric extension of material, we have different equations for the areal and linear extension of a material.