Problem 252 - Calculus 3

Question

The pressure \(P\) of a gas is related to the volume and temperature by the formula \(P V=k T,\) where temperature is expressed in kelvins. Express the pressure of the gas as a function of both \(V\) and \(T .\) Find \(\frac{d P}{d t}\) when \(k=1\) \(\frac{d V}{d t}=2 \mathrm{cm}^{3} / \mathrm{min}, \frac{d T}{d t}=\frac{1}{2} \quad \mathrm{K} / \mathrm{min}, \quad V=20 \mathrm{cm}^{3},\) and \(T=20^{\circ} \mathrm{F}\)

Step-by-Step Solution

Answer

\(\frac{dP}{dt} = -1.3074\, \mathrm{K/cm}^3 \text{ per minute}\).

Step 1: Convert Temperature to Kelvin

First, convert the temperature from Fahrenheit to Kelvin. Use the conversion formula:\[T(K) = \frac{5}{9}(T(\degree F) - 32) + 273.15\]Given \(T = 20\degree F\), substitute into the formula:\[T(K) = \frac{5}{9}(20 - 32) + 273.15 = \frac{5}{9}(-12) + 273.15 = -\frac{60}{9} + 273.15 = -6.67 + 273.15 \approx 266.48 K\]

Step 2: Express Pressure as a Function of Volume and Temperature

The relationship between pressure \(P\), volume \(V\), and temperature \(T\) is given by \(PV = kT\). To express the pressure \(P\) as a function of \(V\) and \(T\), solve for \(P\) in terms of \(V\) and \(T\):\[P = \frac{kT}{V}\]where \(k\) is a constant.

Step 3: Differentiate the Pressure with Respect to Time

To find \(\frac{dP}{dt}\), we first differentiate the expression \(P = \frac{kT}{V}\) implicitly with respect to time \(t\). Using the chain rule and product rule, we get:\[\frac{dP}{dt} = \frac{d}{dt}\left(\frac{kT}{V}\right) = \frac{1}{V} \cdot k \cdot \frac{dT}{dt} - \frac{kT}{V^2} \cdot \frac{dV}{dt}\]This gives:\[\frac{dP}{dt} = \frac{k}{V} \cdot \frac{dT}{dt} - \frac{kT}{V^2} \cdot \frac{dV}{dt}\]

Step 4: Substitute Given Values and Solve

Substitute \(k = 1\), \(\frac{dV}{dt} = 2\, \mathrm{cm}^3/\mathrm{min}\), \(\frac{dT}{dt} = \frac{1}{2}\, \mathrm{K/min}\), \(V = 20\, \mathrm{cm}^3\), and \(T = 266.48\, \mathrm{K}\) into the formula for \(\frac{dP}{dt}\):\[\frac{dP}{dt} = \frac{1}{20} \times \frac{1}{2} - \frac{266.48}{20^2} \times 2\]Calculate:\[= \frac{1}{40} - \frac{266.48}{400} \cdot 2\]\[= 0.025 - \frac{266.48}{200}\]\[= 0.025 - 1.3324 = -1.3074 \mathrm{K/cm}^3 \text{ per minute}\]

Step 5: Result Interpretation

The result of \(\frac{dP}{dt} = -1.3074\, \mathrm{K/cm}^3 \text{ per minute}\) indicates that the pressure of the gas is decreasing at this rate with respect to time given the conditions specified.

Key Concepts

Pressure-Volume-Temperature Relation

The pressure-volume-temperature relationship for gases is often summarized by the formula \( PV = kT \), where:\

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• \( P \) is the pressure of the gas.
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• \( V \) represents the volume.
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• \( T \) is the temperature in kelvins.
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• \( k \) is a constant, specific to the particular process or gas being considered.
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This equation comes from the ideal gas law, which describes how gases should perfectly behave under "ideal" conditions. In many real-world situations, scientists use this equation to approximate gas behavior when deviations are tolerable.
By rearranging the equation to \( P = \frac{kT}{V} \), you can see that pressure depends directly on both temperature and inversely on volume. As temperature increases, pressure tends to increase, provided volume stays constant; conversely, if volume is decreased while temperature is constant, pressure will increase. This relationship is crucial in thermodynamics and various engineering applications.

Chain Rule

In calculus, the chain rule is a fundamental technique for differentiating composite functions. It lets us find the derivative of a function that consists of other functions nested within each other.
When working with variables dependent on time, such as temperature or volume changing over time, we use the chain rule to differentiate them with respect to time as well.
The formula is:\

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• If you have a function \( y = f(u) \), and \( u = g(x) \), then \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
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In the context of gas laws, when differentiating the pressure \( P \) of a gas expressed with respect to both \( V \) and \( T \) as \( P = \frac{kT}{V} \), the chain rule helps us find \( \frac{dP}{dt} \), which shows how pressure changes over time.

Implicit Differentiation

Implicit differentiation is applied when it is difficult to solve for a variable explicitly. Instead of expressing one variable purely in terms of the other, both are interlinked in an equation. When differentiating, we treat each variable as implicitly dependent on a third variable, like time.
For the given gas law \( PV = kT \), implicit differentiation allows us to find how each component, such as pressure or volume, changes over time without solving for one variable solely.
Here is why it is useful in our context:

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• We need to determine \( \frac{dP}{dt} \), which involves both \( \frac{dV}{dt} \) and \( \frac{dT}{dt} \).
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• Implicit differentiation provides a way to handle these rates of change without fully separating or solving the original equation for \( P \) or \( V \).
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Temperature Conversion

Temperature conversion is often necessary because various equations or systems might require temperature in specific units. In our case, to use the gas law efficiently, temperature should be converted into kelvins, the SI unit for temperature.
For converting from Fahrenheit to Kelvin, we use the formula:

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• \( T(K) = \frac{5}{9}(T(\degree F) - 32) + 273.15 \)
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This formula first converts Fahrenheit to Celsius by using \( \frac{5}{9}(T(\degree F) - 32) \), then adds 273.15 to convert Celsius to Kelvin.
In our example, given \( T = 20\degree F \), the conversion showed \( T \approx 266.48 \space \text{K} \). Converting temperature is crucial because different units might not directly correspond to the specific concepts or calculations involving gases and other physical processes.