Problem 55 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

According to the ideal gas law, the volume \(V\) (in liters) of an ideal gas is related to its pressure \(P\) (in pascals) and temperature \(T\) (in degrees Kelvin) by the formula $$V=\frac{k T}{P}$$ where \(k\) is a constant. Show that $$\frac{\partial V}{\partial T} \cdot \frac{\partial T}{\partial P} \cdot \frac{\partial P}{\partial V}=-1$$

Step-by-Step Solution

Answer

We have the ideal gas law formula \(V=\frac{kT}{P}\). After finding the partial derivatives \(\frac{\partial V}{\partial T}\), \(\frac{\partial T}{\partial P}\), and \(\frac{\partial P}{\partial V}\), we obtain the following expressions: \(\frac{\partial V}{\partial T} = \frac{k}{P}\), \(\frac{\partial T}{\partial P} = \frac{V}{k}\), and \(\frac{\partial P}{\partial V} = \frac{-kT}{V^2}\). Next, we multiply these partial derivatives: \(\frac{\partial V}{\partial T} \cdot \frac{\partial T}{\partial P} \cdot \frac{\partial P}{\partial V} = \frac{k}{P} \cdot \frac{V}{k} \cdot \frac{-kT}{V^2}\). After simplifying, we obtain: \(\frac{\partial V}{\partial T} \cdot \frac{\partial T}{\partial P} \cdot \frac{\partial P}{\partial V} = -1\), as required.

Step 1: Find the partial derivative of V with respect to T.

To find the partial derivative of V with respect to T, we treat P as a constant and differentiate V with respect to T: \(\frac{\partial V}{\partial T} = \frac{\partial}{\partial T} (\frac{kT}{P})\) Recall that k is a constant. Since P is treated as a constant, \(\frac{k}{P}\) is also a constant (let's call it \(c_1\)). So, \(\frac{\partial V}{\partial T} = \frac{\partial}{\partial T} (c_1T)\) \(\frac{\partial V}{\partial T} = c_1\)

Step 2: Find the partial derivative of T with respect to P.

To find the partial derivative of T with respect to P, we rewrite the ideal gas law formula to isolate T: \(T = \frac{PV}{k}\) Now, differentiate T with respect to P, treating V as a constant: \(\frac{\partial T}{\partial P} = \frac{\partial}{\partial P} (\frac{PV}{k})\) Since V and k are treated as constants, \(\frac{V}{k}\) is also a constant (let's call it \(c_2\)). So, \(\frac{\partial T}{\partial P} = \frac{\partial}{\partial P} (c_2P)\) \(\frac{\partial T}{\partial P} = c_2\)

Step 3: Find the partial derivative of P with respect to V.

To find the partial derivative of P with respect to V, we rewrite the ideal gas law formula to isolate P: \(P = \frac{kT}{V}\) Now, differentiate P with respect to V, treating T as a constant: \(\frac{\partial P}{\partial V} = \frac{\partial}{\partial V} (\frac{kT}{V})\) To differentiate this expression, we will use the Quotient Rule: \(\frac{\partial P}{\partial V} = \frac{-kT}{V^2}\) Since T and k are treated as constants, -kT is also a constant (let's call it \(c_3\)). So, \(\frac{\partial P}{\partial V} = \frac{c_3}{V^2}\)

Step 4: Multiply the partial derivatives and simplify.

Now, multiply the three partial derivatives: \(\frac{\partial V}{\partial T} \cdot \frac{\partial T}{\partial P} \cdot \frac{\partial P}{\partial V} = c_1 \cdot c_2 \cdot \frac{c_3}{V^2}\) Recall that \(c_1\) is \(\frac{k}{P}\), \(c_2\) is \(\frac{V}{k}\), and \(c_3\) is \(-kT\). Substitute these values and simplify the expression: \(\frac{k}{P} \cdot \frac{V}{k} \cdot \frac{-kT}{V^2} = -\frac{kTV}{PV^2}\) Cancel out terms: \(-\frac{kTV}{PV^2} = -1\) We have shown that \(\frac{\partial V}{\partial T} \cdot \frac{\partial T}{\partial P} \cdot \frac{\partial P}{\partial V} = -1\), as required.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They help us understand how a multivariable function changes when we vary one of its input variables while holding others constant. In situations involving physics and engineering, partial derivatives are used to examine changes in systems with several interdependent variables. These derivatives provide a snapshot of how, for example, temperature or pressure affects volume in the Ideal Gas Law.
When we compute a partial derivative, we denote it with a unique symbol like \( \frac{\partial V}{\partial T} \). Here, it indicates the change in volume \( V \) with respect to temperature \( T \) while keeping pressure \( P \) constant. This method allows us to isolate and study the effect of one variable at a time, making it simpler to predict system behaviors.

Ideal Gas Law

The Ideal Gas Law is a crucial principle in thermodynamics. It establishes a relationship between pressure \( P \), volume \( V \), and temperature \( T \) of an ideal gas through the formula \( V = \frac{kT}{P} \), where \( k \) is a constant. This equation is particularly helpful in predicting the behavior of gases under different conditions.

• Pressure \( (P) \): The force exerted by gas particles against the walls of its container.
• Volume \( (V) \): The space occupied by the gas.
• Temperature \( (T) \): A measure of the kinetic energy of gas particles.

By manipulating these variables, we can analyze various thermodynamic processes. The Ideal Gas Law is essential for understanding real-world scenarios like engine operations and atmospheric pressure changes.

Quotient Rule

In calculus, the Quotient Rule is used to differentiate functions that are ratios, helping us find the derivative of a function formed by dividing two other functions. It's particularly useful in cases like the Ideal Gas Law's rearrangement of formulas.
The Quotient Rule states that if you have a function \( f(x) = \frac{u(x)}{v(x)} \), the derivative \( f'(x) \) is given by:
\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \]
In the context of the Ideal Gas Law, when we isolated \( P \) as \( \frac{kT}{V} \), we needed to apply the Quotient Rule to find the partial derivative \( \frac{\partial P}{\partial V} \). Understanding this rule aids in differentiating complex expressions accurately.

Differentiation

Differentiation is a core mathematical technique used to determine the rate at which a function is changing at any given point. In single-variable calculus, differentiation provides the slope of a curve or the rate of change of a function.
In multivariable calculus, differentiation extends to partial derivatives, where each variable can be examined individually. This allows for precise modeling of more complex systems like those dealt with using the Ideal Gas Law. Differentiation is essential for not just determining changes but also for optimizing and controlling processes efficiently in applied mathematics.
Mastering differentiation is key in fields ranging from engineering to economics, where change needs to be quantified and understood.