The pressure [tex]$p$[/tex] of an ideal gas of density [tex]$p$[/tex] is given by the equation
[tex]$p=\frac{1}{3} p\left(c^2\right)$[/tex]
where [tex]$\left\langle c^2\right\rangle$[/tex] is the mean-square-speed (i.e. it is a quantity measured as [speed][tex]$^2$[/tex]?). Use base units to show that the equation is homogeneous.
Answer (2)
Determine the base units of pressure p : kg ⋅ m − 1 ⋅ s − 2 .
Determine the base units of density ρ : kg ⋅ m − 3 .
Determine the base units of mean-square-speed ⟨ c 2 ⟩ : m 2 ⋅ s − 2 .
Substitute the base units of ρ and ⟨ c 2 ⟩ into the expression 3 1 ρ ⟨ c 2 ⟩ and verify that they are the same as the base units of pressure p . The equation is homogeneous because both sides have units of kg ⋅ m − 1 ⋅ s − 2 .
Explanation
Understanding the Problem We want to show that the equation p = 3 1 ρ ⟨ c 2 ⟩ is homogeneous using base units. This means we need to show that the units on both sides of the equation are the same.
Base Units of Pressure First, let's determine the base units of pressure p . Pressure is defined as force per unit area, p = A F . Force, according to Newton's second law, is mass times acceleration, F = ma . Therefore, the units of force are kg ⋅ m/s 2 . Area has units of m 2 . Thus, the units of pressure are: m 2 kg ⋅ m/s 2 = m ⋅ s 2 kg = kg ⋅ m − 1 ⋅ s − 2
Base Units of Density Next, let's determine the base units of density ρ . Density is defined as mass per unit volume, ρ = V m . The units of mass are kg, and the units of volume are m 3 . Thus, the units of density are: m 3 kg = kg ⋅ m − 3
Base Units of Mean-Square-Speed Now, let's determine the base units of the mean-square-speed ⟨ c 2 ⟩ . Since c is a speed, its units are m/s. Therefore, the units of c 2 are (m/s) 2 = m 2 /s 2 . The mean-square-speed ⟨ c 2 ⟩ is just the average of the square of the speeds, so it has the same units as c 2 : ⟨ c 2 ⟩ = s 2 m 2 = m 2 ⋅ s − 2
Combining the Units Now, let's substitute the base units of ρ and ⟨ c 2 ⟩ into the expression 3 1 ρ ⟨ c 2 ⟩ : 3 1 ρ ⟨ c 2 ⟩ = 3 1 ( kg ⋅ m − 3 ) ( m 2 ⋅ s − 2 ) = 3 1 kg ⋅ m − 1 ⋅ s − 2 Since 3 1 is a dimensionless constant, it doesn't affect the units. Therefore, the units of 3 1 ρ ⟨ c 2 ⟩ are kg ⋅ m − 1 \cdot s ^{-2}$.
Conclusion Finally, let's compare the base units of p with the base units of 3 1 ρ ⟨ c 2 ⟩ . We found that the units of pressure p are kg ⋅ m − 1 \cdot s ^{-2} , an d t h e u ni t so f \frac{1}{3} \rho \langle c^2 \rangle a re a l so k g \cdot m ^{-1} ⋅ s − 2 . Since the units on both sides of the equation are the same, the equation is homogeneous.
Examples
Understanding the homogeneity of equations is crucial in physics and engineering. For instance, when designing a bridge, engineers must ensure that all terms in their structural equations have consistent units. If the equation isn't homogeneous, it indicates a fundamental error in the model, potentially leading to catastrophic failures. Verifying homogeneity helps ensure the reliability and accuracy of physical models used in real-world applications.
Both sides of the equation p = 3 1 ρ ⟨ c 2 ⟩ have the same units of kg ⋅ m − 1 ⋅ s − 2 , confirming its homogeneity. This indicates that the equation is valid in terms of dimensional analysis. Therefore, the dimensions are consistent on both sides of the equation.
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