Problem 53
Question
The volume \(V\) (in liters) of a certain mass of gas is related to its pressure \(P\) (in millimeters of mercury) and its temperature \(T\) (in degrees Kelvin) by the law $$V=\frac{30.9 T}{P}$$ Compute \(\partial V / \partial T\) and \(\partial V / \partial P\) when \(T=300\) and \(P=800\). Interpret your results.
Step-by-Step Solution
Verified Answer
The partial derivatives of V with respect to T and P at the given point (T=300, P=800) are \(\frac{\partial V}{\partial T} = 0.038625\) and \(\frac{\partial V}{\partial P} = -0.00014428125\). This means that at this point, a small increase in temperature while keeping pressure constant would increase the volume by 0.038625 liters, and a small increase in pressure while keeping temperature constant would decrease the volume by 0.00014428125 liters.
1Step 1: Calculate the partial derivatives of V with respect to T and P
\[
\frac{\partial V}{\partial T} = \frac{\partial}{\partial T} \Big(\frac{30.9T}{P}\Big)
\]
and
\[
\frac{\partial V}{\partial P} = \frac{\partial}{\partial P} \Big(\frac{30.9T}{P}\Big)
\]
2Step 2: Evaluate the partial derivatives
To evaluate the partial derivatives, we need to apply the rules of differentiation. For the first one:
\[
\frac{\partial V}{\partial T} = \frac{30.9}{P}
\]
For the second one, we can rewrite the equation as \(V = 30.9T P^{-1}\), and obtain:
\[
\frac{\partial V}{\partial P} = 30.9T \cdot (-1)P^{-2} = -\frac{30.9T}{P^2}
\]
3Step 3: Calculate the partial derivatives at the given point (T=300, P=800)
Now, let's evaluate the partial derivatives at the given point (T=300, P=800):
\[
\frac{\partial V}{\partial T}\Big|_{T=300, P=800} = \frac{30.9}{800} = 0.038625
\]
\[
\frac{\partial V}{\partial P}\Big|_{T=300, P=800} = -\frac{30.9(300)}{800^2} = -0.00014428125
\]
4Step 4: Interpret the results
The partial derivative \(\frac{\partial V}{\partial T}\) at point (T=300, P=800) is equal to 0.038625. This means that for a small increase in the temperature T, while keeping the pressure P constant, the volume V would increase by approximately 0.038625 liters.
The partial derivative \(\frac{\partial V}{\partial P}\) at point (T=300, P=800) is equal to -0.00014428125. This means that for a small increase in the pressure P, while keeping the temperature T constant, the volume V would decrease by approximately 0.00014428125 liters. This negative value indicates that the relationship between pressure and volume is inversely proportional, which is consistent with the gas laws.
Key Concepts
Volume of GasPressure-Volume RelationshipDifferentiation RulesInterpretation of Derivatives
Volume of Gas
In the study of gases, volume is an important characteristic that can change depending on other variables such as pressure and temperature. The volume of gas, in a given cylindrical container, can be found using the formula: \[ V = \frac{30.9 T}{P} \] In this equation, \(V\) stands for volume, \(T\) is the temperature in Kelvin, and \(P\) is the pressure in millimeters of mercury.
This formula derives from ideal gas laws, which describe how gases behave under different conditions. As either pressure or temperature changes, the volume of the gas adjusts accordingly.
This formula derives from ideal gas laws, which describe how gases behave under different conditions. As either pressure or temperature changes, the volume of the gas adjusts accordingly.
- A higher temperature results in greater volume, assuming pressure is constant.
- Conversely, a higher pressure compresses the gas volume, given a constant temperature.
Pressure-Volume Relationship
The relationship between pressure and volume of a gas is often described by Boyle's Law. It states that pressure and volume are inversely proportional when temperature is kept constant.
In other words, as pressure increases, volume decreases, and vice versa. For our formula \(V = \frac{30.9 T}{P}\), you can see that the pressure \(P\) appears in the denominator.
This placement suggests that as pressure \(P\) rises, the overall volume \(V\) should decrease.
It is essential to understand that this inverse relationship doesn't mean the volume will shrink infinitely or become negative. As the formula implies, for a fixed temperature \(T\), if the pressure doubles, the volume is halved.
Such practical applications of these principles help us predict the behavior of gases in real-world scenarios, like understanding airbags in vehicles or industrial gas storage.
In other words, as pressure increases, volume decreases, and vice versa. For our formula \(V = \frac{30.9 T}{P}\), you can see that the pressure \(P\) appears in the denominator.
This placement suggests that as pressure \(P\) rises, the overall volume \(V\) should decrease.
It is essential to understand that this inverse relationship doesn't mean the volume will shrink infinitely or become negative. As the formula implies, for a fixed temperature \(T\), if the pressure doubles, the volume is halved.
Such practical applications of these principles help us predict the behavior of gases in real-world scenarios, like understanding airbags in vehicles or industrial gas storage.
Differentiation Rules
Differentiating functions is a fundamental calculus skill that helps us understand how one variable changes concerning another one. In this problem, we are asked to find the partial derivatives of \(V\) with respect to \(T\) and \(P\).
Partial differentiation allows us to examine how the volume changes when we alter either the temperature or the pressure while keeping the other constant.
To compute \(\frac{\partial V}{\partial T}\), we treat \(P\) as a constant and differentiate \(\frac{30.9T}{P}\) with respect to \(T\), resulting in \(\frac{30.9}{P}\).
This derivative indicates changes in volume with a change in temperature.
For \(\frac{\partial V}{\partial P}\), \(T\) is constant, and differentiation involves treating pressure \(P\) as a variable. Rewriting the volume function as \( 30.9T P^{-1} \), differentiating with respect to \(P\) gives \(-\frac{30.9T}{P^2}\).
This is significant as it shows the change in volume concerning pressure, confirming the inverse relation as proposed by gas laws.
Partial differentiation allows us to examine how the volume changes when we alter either the temperature or the pressure while keeping the other constant.
To compute \(\frac{\partial V}{\partial T}\), we treat \(P\) as a constant and differentiate \(\frac{30.9T}{P}\) with respect to \(T\), resulting in \(\frac{30.9}{P}\).
This derivative indicates changes in volume with a change in temperature.
For \(\frac{\partial V}{\partial P}\), \(T\) is constant, and differentiation involves treating pressure \(P\) as a variable. Rewriting the volume function as \( 30.9T P^{-1} \), differentiating with respect to \(P\) gives \(-\frac{30.9T}{P^2}\).
This is significant as it shows the change in volume concerning pressure, confirming the inverse relation as proposed by gas laws.
Interpretation of Derivatives
Interpreting derivatives helps us comprehend the rate at which one variable impacts another.
In our specific example, two partial derivatives offer insights into how temperature and pressure changes affect the volume of gas.
Let's look at \(\frac{\partial V}{\partial T}\). The evaluated value of 0.038625 informs us that if the temperature increases by 1 Kelvin while maintaining constant pressure, the volume increases by approximately 0.038625 liters.
This positive value shows a direct proportion between temperature and volume.
Meanwhile, for \(\frac{\partial V}{\partial P}\), the result is -0.00014428125. This negative rate indicates a decrease in volume when pressure increases by 1 mmHg, maintaining constant temperature.
The negative sign here emphasizes the inverse relationship expected between pressure and volume.
This application of derivative interpretation thereby solidifies our theoretical understanding with practical numerics, showcasing fundamental physical principles at work.
In our specific example, two partial derivatives offer insights into how temperature and pressure changes affect the volume of gas.
Let's look at \(\frac{\partial V}{\partial T}\). The evaluated value of 0.038625 informs us that if the temperature increases by 1 Kelvin while maintaining constant pressure, the volume increases by approximately 0.038625 liters.
This positive value shows a direct proportion between temperature and volume.
Meanwhile, for \(\frac{\partial V}{\partial P}\), the result is -0.00014428125. This negative rate indicates a decrease in volume when pressure increases by 1 mmHg, maintaining constant temperature.
The negative sign here emphasizes the inverse relationship expected between pressure and volume.
This application of derivative interpretation thereby solidifies our theoretical understanding with practical numerics, showcasing fundamental physical principles at work.
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