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Chapter 12: Problem 8

When a given quantity of an ideal monoatomic gas is at pressure \(P\) and absolute temperature \(T\), then the adiabatic bulk modulus of the gas will be (A) \(P\) (B) \(\frac{5}{3} P\) (C) \(T\) (D) \(\frac{5 T}{2}\)

Short Answer

Expert verified
The adiabatic bulk modulus of the given ideal monoatomic gas is (B) \(\frac{5}{3} P\), as derived using the ideal gas law, adiabatic process equation, and the definition of bulk modulus.

Step by step solution

01

Ideal Gas Law

According to the ideal gas law, for a given quantity of an ideal gas: \(PV = nRT\) where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
02

Adiabatic Process Equation

For an adiabatic process in an ideal monoatomic gas, the following relationship holds: \(PV^{\gamma} = constant\) where \(\gamma\) is the adiabatic index and has a value of \(\frac{5}{3}\) for a monoatomic gas.
03

Differentiate the Adiabatic Process Equation with Respect to Volume

We differentiate both sides of the adiabatic process equation with respect to volume V, keeping the constant on the right side as zero since it is a constant. \(\frac{d}{dV}(PV^{\gamma}) = 0\) By applying the product rule and the chain rule, we get: \(P\gamma V^{\gamma - 1} + V^{\gamma}\frac{dP}{dV} = 0\)
04

Express the Differentiated Equation in Terms of Pressure

Now we need to express the differentiated equation in terms of pressure P. To do this, we substitute the ideal gas law in the differentiated equation by solving it for V and plugging the value into the differentiated equation. From the ideal gas law, we can write: \(V = \frac{nRT}{P}\) Substitute this expression back into the differentiated equation: \(P\gamma\left(\frac{nRT}{P}\right)^{\gamma - 1} + \left(\frac{nRT}{P}\right)^{\gamma}\frac{dP}{dV} = 0\) Simplify by canceling the common terms and isolating the \(\frac{dP}{dV}\) term: \(\frac{dP}{dV} = -P\gamma\left(\frac{nRT}{P}\right)^{-\gamma}\)
05

Calculate the Adiabatic Bulk Modulus

The adiabatic bulk modulus, B, is defined as: \(B = -V\frac{dP}{dV}\) Substitute the expression of \(\frac{dP}{dV}\) found in the previous step and the expression for V from the ideal gas law: \(B = -\frac{nRT}{P} \times\left(-P\gamma\left(\frac{nRT}{P}\right)^{-\gamma}\right)\) Simplify the equation, and we obtain: \(B = \frac{5}{3} P\) So the correct answer for the adiabatic bulk modulus of the gas is (B) \(\frac{5}{3} P\).

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Most popular questions from this chapter

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