Problem 156
Question
The function \(P(T, V)=\frac{n R T}{V}\) gives the pressure at a point in a gas as a function of temperature \(T\) and volume V. The letters \(n\) and \(R\) are constants. Find \(\frac{\partial P}{\partial V}\) and \(\frac{\partial P}{\partial T}\) and explain what these quantities represent.
Step-by-Step Solution
Verified Answer
\(\frac{\partial P}{\partial V} = -\frac{nRT}{V^2}\), pressure decreases with volume. \(\frac{\partial P}{\partial T} = \frac{nR}{V}\), pressure increases with temperature.
1Step 1: Identify the Function
The given function is \( P(T, V) = \frac{nRT}{V} \). This equation represents the pressure of a gas in terms of the temperature \( T \) and volume \( V \).
2Step 2: Partial Derivative with Respect to Volume
To find \( \frac{\partial P}{\partial V} \), we differentiate \( P(T, V) = \frac{nRT}{V} \) with respect to \( V \), treating \( T \) as a constant. Using the power rule, \( \frac{d}{dV} \left( V^{-1} \right) = -V^{-2} \), we have:\[\frac{\partial P}{\partial V} = \frac{nRT}{V^2} (-1) = -\frac{nRT}{V^2}.\]
3Step 3: Interpretation: \(\frac{\partial P}{\partial V}\)
The term \( \frac{\partial P}{\partial V} = -\frac{nRT}{V^2} \) represents the rate of change of pressure with respect to changes in volume, holding temperature constant. Since the expression is negative, it indicates that pressure decreases as volume increases (assuming other factors remain constant).
4Step 4: Partial Derivative with Respect to Temperature
To find \( \frac{\partial P}{\partial T} \), we differentiate \( P(T, V) = \frac{nRT}{V} \) with respect to \( T \), treating \( V \) as a constant. Since \( nR \) and \( V \) are constants, the derivative simplifies to:\[\frac{\partial P}{\partial T} = \frac{nR}{V}.\]
5Step 5: Interpretation: \(\frac{\partial P}{\partial T}\)
The term \( \frac{\partial P}{\partial T} = \frac{nR}{V} \) represents the rate of change of pressure with respect to changes in temperature, holding volume constant. It shows that pressure increases as temperature increases (assuming volume remains constant).
Key Concepts
Pressure EquationRate of ChangeTemperature and Volume
Pressure Equation
The pressure equation for a gas is a fundamental concept in thermodynamics, especially when dealing with ideal gases. The equation we're focusing on is given by the function \( P(T, V) = \frac{nRT}{V} \). This equation allows us to see how the pressure \( P \) changes when the temperature \( T \) and volume \( V \) change, assuming they both contribute to how gases behave under various conditions.
Here, \( n \) denotes the number of moles of the gas and \( R \) is the universal gas constant. Both \( n \) and \( R \) are constants, meaning their values do not change as \( T \) and \( V \) change.
The equation shows that:
Here, \( n \) denotes the number of moles of the gas and \( R \) is the universal gas constant. Both \( n \) and \( R \) are constants, meaning their values do not change as \( T \) and \( V \) change.
The equation shows that:
- If the temperature \( T \) increases while volume \( V \) stays the same, the pressure \( P \) will increase.
- If the volume \( V \) increases while temperature \( T \) remains constant, the pressure \( P \) will decrease.
Rate of Change
The rate of change in mathematics tells us how one variable changes in relation to another. For our gas pressure function, we look at how pressure changes when volume or temperature changes. This involves calculating partial derivatives.
When we take the partial derivative of pressure with respect to volume \( \frac{\partial P}{\partial V} \), it shows the rate at which the pressure changes if only the volume changes, assuming temperature remains constant. For the given function:
When we take the partial derivative of pressure with respect to volume \( \frac{\partial P}{\partial V} \), it shows the rate at which the pressure changes if only the volume changes, assuming temperature remains constant. For the given function:
- The derivative is \( -\frac{nRT}{V^2} \), which signifies that pressure decreases if volume increases, keeping temperature constant.
- \( \frac{nR}{V} \), showing that pressure increases as temperature goes up, with a constant volume.
Temperature and Volume
Temperature and volume are two crucial factors that affect the behavior of gases, covered under the broader Umbrella of the ideal gas law. Here's how they play their roles:Temperature:
- Temperature is a measure of the average kinetic energy of the gas molecules. Higher temperatures mean these molecules move more vigorously, leading to increased pressure.
- According to our pressure function, an increase in temperature \( T \) results in a proportional increase in pressure, provided the volume \( V \) remains fixed.
- The volume of a gas dictates how much space the gas particles have to move around. More volume gives them more space to spread, which decreases the collisions against the container walls, thus, reducing pressure.
- In our function, as volume \( V \) increases, the pressure \( P \) decreases if the temperature \( T \) is kept constant.
Other exercises in this chapter
Problem 154
Find \(\quad \lim _{\Delta x \rightarrow 0} \frac{\Delta f}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x, y)-f(x, y)}{\Delta x}\) for \(f(x, y)=x
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Find \(\lim _{\Delta x \rightarrow 0} \frac{\Delta f}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x, y)-f(x, y)}{\Delta x}\) for \(f(x, y)=\sin (x
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The equation for heat flow in the \(x y\) -plane is \(\frac{\partial f}{\partial t}=\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}\
View solution Problem 158
The basic wave equation is \(f_{t t}=f_{x x} .\) Verify that \(f(x, t)=\sin (x+t) \quad\) and \(\quad f(x, t)=\sin (x-t) \quad\) are solutions.
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