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Chapter 5: Problem 8

The temperature of an ideal gas in a tube of very small, constant cross- sectional area varies linearly from one end \((x=0)\) to the other end \((x=L)\) according to the equation $$ T=T_{0}+\frac{T_{L}-T_{0}}{L} x $$ If the volume of the tube is \(V\) and the pressure \(P\) is uniform throughout the tube, show that the equation of state for \(n\) moles of gas is given by $$ P V=n R \frac{T_{L}-T_{0}}{\ln \left(T_{L} / T_{0}\right)} $$ Show that, when \(T_{L}=T_{0}=T\), the equation of state reduces to the obvious one, \(P V=n R T\)

Short Answer

Expert verified
The equation of state is \(P V = n R \frac{T_L - T_0}{\ln(T_L / T_0)}\). When \(T_L = T_0 = T\), it reduces to \(P V = n R T\).

Step by step solution

01

Understand the Temperature Distribution

The temperature in the tube varies linearly from one end to the other as given by the equation: \[T = T_0 + \frac{T_L - T_0}{L} x\]This means that at the end where \(x = 0\), the temperature is \(T_0\) and at \(x = L\), the temperature is \(T_L\).
02

Express the Differential Volume Element

Since the cross-sectional area \(A\) is constant, a small section of the tube of length \(dx\) has a volume element given by: \[dV = A \cdot dx\]
03

Use the Ideal Gas Law for a Small Element

For a small volume element at position \(x\) in the tube, the ideal gas law gives: \[d(PV) = d(nRT)\] Given the pressure is uniform, \(P\) remains constant throughout, so: \[P dV = nR \cdot dT\]
04

Replace the Temperature Gradient

Using the linear temperature distribution, the differential temperature gradient is: \[dT = \frac{T_L - T_0}{L} dx\]
05

Integrate Volume Across the Tube

Integrate both sides of the equation from \(x = 0\) to \(x = L\): \[ \int_{0}^{L} P A \, dx = nR \int_{T_0}^{T_L} dT\] Substitute \(A dx = dV\): \[P V = nR (T_L - T_0)\]
06

Adjust the Volume Integral Considering Temperature Variation

To account for temperature variation, note that changing temperature in the tube makes the effective temperature averaged in a logarithmic manner: \[T_{avg} = \frac{T_L - T_0}{\ln(T_L / T_0)}\]
07

Derive the Final Equation of State

Thus, substituting the average temperature: \[P V = n R \frac{T_L - T_0}{\ln(T_L / T_0)}\]
08

Simplify for Equal Temperatures

When \(T_L = T_0 = T\), the logarithmic mean simplifies to the actual temperature: \[\frac{T_L - T_0}{\ln(T_L / T_0)} = T\] Hence, the equation of state reduces to the familiar form: \[P V = n R T\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature Distribution
In thermodynamics, understanding how temperature varies within a system is crucial. In the given problem, temperature varies linearly along a tube. This is expressed by the equation: \( T = T_{0} + \frac{T_{L} - T_{0}}{L} x \) Here, \( T_{0} \) is the temperature at one end of the tube (\( x = 0 \)), and \( T_{L} \) is the temperature at the other end (\( x = L \)). The term \( \frac{T_{L} - T_{0}}{L} \) indicates the rate of temperature change per unit length. This ensures the temperature increases or decreases smoothly as you move from one end of the tube to the other.
Differential Volume Element
To analyze physical properties within small sections of a system, we often use a concept called the 'differential volume element'. In this problem, the tube has a constant cross-sectional area \( A \). For a very small segment of the tube of length \( dx \), we define its volume as: \( dV = A \cdot dx \) This approach helps us apply thermodynamic principles to that small piece and later integrate over the entire tube.
Linear Temperature Gradient
A linear temperature gradient means that the temperature changes at a constant rate along the length of the tube. The gradient can be mathematically expressed as: \( \frac{dT}{dx} = \frac{T_{L} - T_{0}}{L} \) This derivative tells us how much the temperature \( T \) changes with a small change in length \( dx \). Using this gradient is essential when we later perform integration to find out more about the system's state.
Integration in Thermodynamics
Integration allows us to sum small changes over a range to find cumulative effects. In this exercise, we integrate to find the total volume and temperature effects over the entire tube length. First, the pressure is constant throughout the tube, which leads to: \( \int_{0}^{L} P A dx = nR \int_{T_{0}}^{T_{L}} dT \) Here, integrating \( A dx \) results in the total volume: \( V \). This step highlights the fundamental relationship between pressure, volume, and temperature changes in the gas.
Logarithmic Mean Temperature
For non-uniform temperature distributions, we often use the concept of logarithmic mean temperature to find an effective average temperature. In this problem, since the temperature varies linearly along the tube, we use: \( T_{avg} = \frac{T_{L} - T_{0}}{\text{ln}(T_{L} / T_{0})} \) This allows us to account for the temperature variations more accurately than a simple arithmetic average. When we substitute this mean temperature in our final equation, we get: \( PV = nR \frac{T_{L} - T_{0}}{\text{ln}(T_{L} / T_{0})} \) This helps bridge the gap between varying temperatures and the ideal gas law.

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Most popular questions from this chapter

One mole of an ideal paramagnetic gas obeys Curie's law, with a Curie constant \(C_{\mathrm{C}}\). Assume that the internal energy \(U\) is a function of \(T\) only, so that \(d U=C_{V, m} d T\), where \(C_{V, m}\) is a constant heat capacity. (a) Show that the equation of the family of adiabatic surfaces is $$ \frac{C_{V, m}}{n R} \ln T+\ln V=\frac{\mu_{0} m^{2}}{2 n R C_{\mathrm{C}}}+\ln A $$ where \(A\) is a constant for one surface. (b) Sketch one of these surfaces on a \(T V\) diagram.

A cylindrical cocktail glass \(15 \mathrm{~cm}\) high and \(35 \mathrm{~cm}^{2}\) in cross section contains water up to the \(10-\mathrm{cm}\) mark. A card is placed over the top and held there while the glass is inverted. When the support for the card is removed, what mass of water must leave the glass in order that the rest of the water will remain in the glass, if one neglects the weight of the card? (Caution: Try this over a sink.)

(a) Derive the following formula for a quasi-static adiabatic process for the ideal gas, assuming \(\gamma\) to be constant: $$ \frac{T}{P(\gamma-1) / \gamma}=\text { const. } $$ (b) Helium \(\left(\gamma=\frac{5}{3}\right)\) at \(300 \mathrm{~K}\) and \(1 \mathrm{~atm}\) pressure is compressed quasi-statically and adiabatically to a pressure of 5 atm. Assuming that the helium behaves like the ideal gas, calculate the final temperature.

A thick-walled insulated chamber contains \(n_{i}\) moles of helium at high pressure \(P_{i}\). It is connected through a valve with a large, almost empty container of helium at constant pressure \(P_{0}\), very nearly atmospheric. The valve is opened slightly, and the helium flows slowly and adiabatically into the container until the pressures on the two sides of the valve are equal. Assuming the helium to behave like an ideal gas with constant heat capacities, show that: (a) The final temperature of the gas in the chamber is $$ T_{f}=T_{i}\left(\frac{P_{f}}{P_{i}}\right)^{(\gamma-1) / \gamma} $$ (b) The number of moles left in the chamber is $$ n_{f}=n_{i}\left(\frac{P_{f}}{P_{i}}\right)^{1 / \gamma} \text { . } $$ (c) The final temperature of the gas in the container is $$ T_{f}=\frac{T_{i}}{\gamma} \frac{1-P_{f} / P_{i}}{1-\left(P_{f} / P_{i}\right)^{1 / \gamma}} . $$

A horizontal, insulated cylinder contains a frictionless nonconducting piston. On each side of the piston is 54 liters of an inert monatomic ideal gas at 1 atm and \(273 \mathrm{~K}\). Heat is slowly supplied to the gas on the left side until the piston has compressed the gas on the right side to \(7.59 \mathrm{~atm}\). (a) How much work is done on the gas on the right side? (b) What is the final temperature of the gas on the right side? (c) What is the final temperature of the gas on the left side? (d) How much heat was added to the gas on the left side?
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