Problem 31
Question
According to the ideal gas law, the pressure, temperature, and volume of a gas are related by \(P V=k T\), where \(k\) is a constant. Find the rate of change of pressure (pounds per square inch) with respect to temperature when the temperature is \(300^{\circ} \mathrm{K}\) if the volume is kept fixed at 100 cubic inches.
Step-by-Step Solution
Verified Answer
The rate of change is \(\frac{P}{100}\) psi per \(^\circ\mathrm{K}\).
1Step 1: Formulate the Ideal Gas Law Equation
The ideal gas law is given by the equation \(PV = kT\), where \(P\) is the pressure, \(V\) is the volume, \(k\) is the constant, and \(T\) is the temperature.
2Step 2: Differentiate with Respect to Temperature
Differentiate both sides of the equation \(PV = kT\) with respect to \(T\) while keeping \(V\) constant. This gives us \(\frac{d}{dT}(PV) = \frac{d}{dT}(kT)\). Since \(V\) is constant, the differentiation gives \(V \frac{dP}{dT} = k\).
3Step 3: Solve for the Rate of Change of Pressure
Rearrange the equation from Step 2 to solve for \(\frac{dP}{dT}\): \(\frac{dP}{dT} = \frac{k}{V}\).
4Step 4: Substitute Known Values
Substitute the known values into the equation: \(V = 100\) cubic inches, \(T = 300^{\circ} \mathrm{K}\), and since \(PV = kT\), solve for \(k = \/PV\/300\). However, \(\frac{dP}{dT} = \frac{k}{100}\).
5Step 5: Express \(k\) in Terms of Known Quantities
Since \(k = PV/T\) at \(T = 300\), we substitute: \(k = 300 \times \frac{P}{300} = P\). Simplify the expression: \(\frac{dP}{dT} = \frac{P}{100}\).
6Step 6: Final Answer for Rate of Change
The rate of change of pressure with respect to temperature is \(\frac{dP}{dT} = \frac{P}{100}\) psi per \(^\circ\mathrm{K}\).
Key Concepts
Rate of ChangePressureTemperatureVolume
Rate of Change
In the context of the ideal gas law, understanding the rate of change is essential to determining how one variable, such as pressure, changes in response to another, like temperature. This concept involves calculus, specifically differentiation, as it measures how a function's value changes over incremental shifts in another parameter. In the problem, we aim to find how pressure (\(P\)) varies as temperature (\(T\)) changes when the volume (\(V\)) is held steady.
Differentiating the ideal gas equation, \(PV = kT\), with respect to \(T\) gives \( V \frac{dP}{dT} = k \). This result shows that the change in pressure is directly proportional to the constant \(k\) and inversely proportional to volume. By isolating \( \frac{dP}{dT} \), we find the precise mathematical way to calculate the rate of change of pressure with respect to temperature, crucial for physics and engineering problems involving gases.
Differentiating the ideal gas equation, \(PV = kT\), with respect to \(T\) gives \( V \frac{dP}{dT} = k \). This result shows that the change in pressure is directly proportional to the constant \(k\) and inversely proportional to volume. By isolating \( \frac{dP}{dT} \), we find the precise mathematical way to calculate the rate of change of pressure with respect to temperature, crucial for physics and engineering problems involving gases.
Pressure
Pressure is a measure of the force exerted per unit area and is a fundamental concept in understanding gases. In the ideal gas equation, \(P\) depicts the pressure, which is affected by changes in temperature or volume.
When analyzing how pressure shifts with temperature change, it's important to grasp that, if volume is fixed, pressure will rise if temperature increases. This is based on kinetic molecular theory, where higher temperatures mean gas molecules move faster, colliding with container walls more frequently and forcefully, thus raising pressure.
When analyzing how pressure shifts with temperature change, it's important to grasp that, if volume is fixed, pressure will rise if temperature increases. This is based on kinetic molecular theory, where higher temperatures mean gas molecules move faster, colliding with container walls more frequently and forcefully, thus raising pressure.
- The ideal gas law ties together pressure, temperature, and volume in a simple yet powerful way.
- In our exercise, focusing on pressure's response to temperature shifts is central.
Temperature
In the realm of gases, temperature (\(T\)) is a measure of the average kinetic energy of gas molecules. Within the ideal gas law, temperature plays a critical role; changes in temperature directly impact pressure and volume when expressed through the relationship \(PV = kT\).
When temperature is altered in a system where volume is constant, pressure changes in a predictable manner. This concept, evident in the ratio \( \frac{dP}{dT} = \frac{P}{100} \), shows a linear relationship between temperature and pressure changes. For example, at 300 Kelvin, a rise in temperature will lead to a rise in pressure, provided the gas behaves ideally.
When temperature is altered in a system where volume is constant, pressure changes in a predictable manner. This concept, evident in the ratio \( \frac{dP}{dT} = \frac{P}{100} \), shows a linear relationship between temperature and pressure changes. For example, at 300 Kelvin, a rise in temperature will lead to a rise in pressure, provided the gas behaves ideally.
- Understanding temperature's role offers predictive power in solving real-world issues.
- Temperature adjustments have direct implications for gas-driven processes.
Volume
Volume (\(V\)) is the amount of space a gas occupies and is a vital component of the ideal gas law. The equation \(PV = kT\) indicates that, if temperature is constant, pressure varies inversely with volume. However, in this exercise, volume is constant, simplifying how we view its role in relation to the other variables.
A fixed volume implies that any change in temperature leads solely to changes in pressure. When analyzing changes, \(\frac{dP}{dT} = \frac{k}{V}\), volume plays a key role as a denominator, affecting the sensitivity of pressure changes to temperature shifts.
A fixed volume implies that any change in temperature leads solely to changes in pressure. When analyzing changes, \(\frac{dP}{dT} = \frac{k}{V}\), volume plays a key role as a denominator, affecting the sensitivity of pressure changes to temperature shifts.
- Volume’s constancy simplifies calculations and study of pressure-variable relations.
- Holding volume steady aids clear insight into how gases behave under temperature shifts.
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