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.

Short Answer

Expert verified
The equation of the adiabatic surfaces is \(\frac{C_{V,m}}{nR}\text{\textln}T + \text{\textln} V = \frac{\text{\textmu}_{0} m^2}{2 n R C_{\text{C}}} + \text{\textln} A\). The TV diagram shows a curved surface.

Step by step solution

01

Understand the Given Problem

Read through the problem carefully. You need to show the equation of the family of adiabatic surfaces for an ideal paramagnetic gas, and sketch one of these surfaces on a TV diagram. Given: internal energy as function of temperature only, Curie's law, and specific heat capacity.
02

Derive the General Formula for Adiabatic Process

For an adiabatic process, the change in internal energy (U) equals the work done on or by the system: \(dU = T \frac{dS}{dt} = -P dV\)
03

Express Internal Energy

Since internal energy is a function of temperature only (U = U(T)), given: \(dU = C_{V,m} dT\), substitute it into the first law equivalence equation: \(C_{V,m} dT = -P dV\)
04

Utilize Ideal Gas Law

For an ideal gas: \(P V = n R T\). Differentiate: \(V dP + P dV = n R dT\). Substitute \(dT\) from adiabatic equation: \(C_{V,m} n R dT = -nR P dV\)
05

Rearrange the Equation

Rearrange to solve for \(dT\): \(dT = - \frac{P dV}{C_{V,m}}\). Substitute into differential equation derived: \(P dV + \frac{P V}{C_{V,m}} dT = 0\). This leads to: \(dV/V + \frac{dT}{C_{V,m}} = 0\)
06

Integrate Both Sides

On integrating: \(\frac{dV}{V} = -\frac{dT}{C_{V,m}}\), we obtain: \(\frac{C_{V,m}}{nR} \text{ln} T + \text{ln}V = \text{const.}\)
07

Add Curie's Law

Using Curie's law for ideal paramagnetic gas: \(U = \frac{\text{\textmu}_{0} m^2}{2 C_{\text{C}}}\). The constant becomes: \(const = \frac{\text{\textmu}_{0} m^2}{2 n R C_{\text{C}}} + \text{\textln} A\)
08

Form the Equation

Combine all the constants and express the equation of adiabatic surfaces: \( \frac{C_{V,m}}{n R} \text{\textln} T + \text{\textln} V = \frac{\text{\textmu}_{0} m^2}{2 n R C_{\text{C}}} + \text{\textln} A\)
09

Sketch TV Diagram

To sketch the diagram, express surface on (T, V) coordinates. Since \(T\) increases exponentially with \(V\), the \(TV\) diagram shows a curved surface.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation in thermodynamics that describes the relationship between the pressure (P), volume (V), and temperature (T) of an ideal gas. Mathematically, it is defined as: \[ PV = nRT \] Here, \(n\) is the number of moles of the gas, and \(R\) is the ideal gas constant. This law helps us understand how gases behave under different conditions of temperature and pressure. For instance, if the pressure is held constant, increasing the temperature will cause the volume to increase. Similarly, if the volume is kept constant, increasing the temperature will increase the pressure. It is essential for deriving other relations in thermodynamics, such as the ones involving adiabatic processes.
Curie's Law
Curie's law describes the behavior of paramagnetic materials in response to applied magnetic fields and temperature. It states that the magnetization (M) of a paramagnetic material is directly proportional to the applied magnetic field (H) and inversely proportional to the absolute temperature (T):
\[ M = \frac{C}{T} H \]
Here, \(C\) is the Curie constant which depends on the material's nature. This principle helps in understanding how the internal energy of a paramagnetic gas is affected by temperature, which is crucial for deriving the equation of adiabatic surfaces in the provided exercise. Utilizing Curie’s law, we can identify how magnetic properties change, influencing the system's overall behavior during thermodynamic processes.
Adiabatic Surface Equation
In an adiabatic process, there is no heat exchange with the surroundings. That means the system's internal energy changes solely due to the work done. For an ideal gas undergoing an adiabatic process, the change in internal energy (\(U\)) equals the work done on or by the system. The equation derived in the exercise, \ \frac{C_{V, m}}{n R} \text{ln} T + \text{ln} V = \frac{\text{\textmu}_{0} m^2}{2 n R C_{\text{C}}} + \text{\textln} A \ \, represents a family of surfaces where each surface corresponds to a specific adiabatic process.
By plotting one of these surfaces on a \(TV\) diagram, you can visualize how temperature and volume vary under adiabatic conditions. This equation encapsulates the combined effects of temperature, volume, and internal properties defined by Curie's law and the ideal gas constant.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the law of energy conservation, states that the total energy in an isolated system remains constant. It is formulated as:
\[ dU = dQ - dW \] ewline Here, \(dU\) represents the change in internal energy, \(dQ\) is the heat added to the system, and \(dW\) is the work done by the system. In the context of adiabatic processes, \(dQ = 0\), meaning no heat is exchanged. Thus, \(dU = -dW\).
For an ideal paramagnetic gas with internal energy as a function of temperature only, this relationship simplifies further. By integrating over relevant variables, like temperature and volume, and applying specific constraints, we derive key thermodynamic equations useful for practical applications, such as the adiabatic surface equation derived in the exercise.

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Most popular questions from this chapter

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.)

Expand the following equations in the form $$ P_{v}=R T\left(1+B P+C P^{2}+\cdots\right) $$ and determine the second virial coefficient \(B\) in each case: (a) \(\left(P+\frac{a}{v^{2}}\right)(v-b)=R T \quad\) (van der Waals equation of state). (b) \(\left(P e^{a / R T}\right)(\nu-b)=R T \quad\) (Dieterici equation of state). (c) \(\left(P+\frac{a}{v^{2} T}\right)(\nu-b)=R T \quad\) (Berthelot equation of state). (d) \(\left[P+\frac{a}{(v+c)^{2} T}\right](v-b)=R T \quad\) (Clausius equation of state). (e) \(P v=R T\left(1+\frac{B^{\prime}}{v}+\frac{C^{\prime}}{v^{2}}+\cdots\right) \quad\) (another type of virial expansion).

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, \(P V=n R T\)

(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 steel ball of mass \(10 \mathrm{~g}\) is placed in the tube of cross-sectional area \(1 \mathrm{~cm}^{2}\) in Rüchhardt's apparatus. The tube is connected to a jar of air having a capacity of 5 liters, the pressure of the air being \(76 \mathrm{~cm} \mathrm{Hg}\). (a) What is the period of vibration for the ball? (b) If the ball is held initially at a position where the air pressure is exactly atmospheric and then allowed to fall, how far will the ball drop before it starts to come up?

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