Question

Show that the adiabatic bulk modulus, defined as \(B=-V(d p / d V)\), for an ideal gas is equal to \(\gamma p\).

Short answer

The adiabatic bulk modulus of an ideal gas is equal to \(B = \gamma P\), where \(\gamma\) is the adiabatic index, and \(P\) is the pressure.

Step-by-step solution

Recall the ideal gas law and adiabatic process relation

We begin by recalling the ideal gas law: \(PV = nRT\) where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature. For an adiabatic process, we have the relation: \(PV^{\gamma} = constant\) where \(\gamma = \frac{C_p}{C_v}\) is the adiabatic index.

Differentiate the adiabatic process relation with respect to the volume

Now, we differentiate the adiabatic process relation with respect to the volume, keeping the constant term in mind: \(\frac{d}{dV}(PV^{\gamma}) = 0\) Next, we apply the product rule to obtain an expression for \(\frac{dP}{dV}\): \(P\gamma V^{\gamma - 1} + V^{\gamma}\frac{dP}{dV} = 0\) Now, we solve for \(\frac{dP}{dV}\): \(\frac{dP}{dV} = -\frac{P\gamma}{V} \cdot V^{\gamma - 1}\)

Calculate the adiabatic bulk modulus

Now, we substitute our expression for \(\frac{dP}{dV}\) into the definition of the adiabatic bulk modulus: \(B = -V\left(-\frac{P\gamma}{V} \cdot V^{\gamma - 1}\right)\) We can simplify this expression: \(B = P\gamma V^{\gamma - 1}\)

Use the adiabatic process relation to replace \(V^{\gamma -1}\)

We rewrite the adiabatic process relation as: \(V^{\gamma - 1} = \frac{constant}{P}\) Substitute this expression back into our expression for \(B\): \(B = P\gamma \cdot \frac{constant}{P}\) Finally, simplify to obtain the desired result: \(B = \gamma P\) Hence, we have shown that the adiabatic bulk modulus of an ideal gas is equal to \(\gamma P\).