Statistics-lab™可以为您提供cornell.edu ENGRD2210 Thermodynamics热力学课程的代写代考和辅导服务！

ENGRD2210 Thermodynamics课程简介

Thermodynamics is the branch of physics that deals with the study of the relationship between heat, work, and energy. The laws of thermodynamics govern the behavior of all physical systems, including those that involve the transfer of energy as heat or work.

The First Law of Thermodynamics: The law of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another. The first law of thermodynamics is a statement of this principle as applied to thermodynamic systems. It states that the total energy of a closed system remains constant, although it may be converted from one form to another.

The Second Law of Thermodynamics: The second law of thermodynamics states that the entropy of an isolated system never decreases over time. Entropy is a measure of the disorder or randomness of a system, and the second law of thermodynamics tells us that the natural tendency of any system is to become more disordered over time.

Thermodynamic Property Relationships: In thermodynamics, a property is a characteristic of a system that can be measured or calculated. Examples of thermodynamic properties include temperature, pressure, volume, and entropy. The relationships between these properties are described by equations of state, which relate them to each other and to the state of the system.

PREREQUISITES 

Presents the definitions, concepts, and laws of thermodynamics. Topics include the first and second laws, thermodynamic property relationships, and applications to vapor and gas power systems, refrigeration, and heat pump systems. Examples and problems are related to contemporary aspects of energy and power generation and to broader environmental issues.

Outcome 1: Students will be able to choose an appropriate system and identify interactions between system and surroundings.
Outcome 2: Obtain values of thermodynamic properties for a pure substance in a given state, using table, relations for incompressible substances, and relations for gases.

Outcome 3: Apply energy and entropy balances in the control mass (closed system) and control volume formulations to the analysis of devices and cycles.

Cornell students enroll only in ENGRD 2210. MAE 2210 for Non-CU students.

ENGRD2210 Thermodynamics HELP（EXAM HELP， ONLINE TUTOR）

When the piston is fixed, the gas is in equilibrium under its weight, so its pressure is $P_0 = mg/A$, where $m$ is the mass of the gas, $g$ is the acceleration due to gravity, and $A$ is the cross-sectional area of the cylinder. Since the gas is ideal and monatomic, we have $PV = Nk_BT$, where $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $N$ is the number of molecules, and $k_B$ is Boltzmann’s constant.

When the piston is released, the gas expands, pushing the piston upward. The work done by the gas is $W = F\Delta h$, where $F$ is the force on the piston and $\Delta h$ is the change in height of the piston. Since the piston is moving slowly, we can assume that the pressure inside the cylinder is always in equilibrium with the external pressure $P_a$. Therefore, we have $F = P_a A$, and $\Delta h$ is the change in the position of the piston. The work done by the gas is therefore

$$W = P_a A\Delta h.$$

Since the system is thermally isolated, the total energy of the system is conserved. The initial energy of the system is the sum of the gravitational potential energy of the piston, $Mgh$, and the internal energy of the gas, $U_0 = \frac{3}{2}Nk_BT_0$. The final energy of the system is the sum of the gravitational potential energy of the piston at its new position, $Mgh’$, and the internal energy of the gas, $U_f = \frac{3}{2}Nk_BT_f$, where $h’$ is the new height of the piston.

Since the system is in equilibrium at the end, the final pressure of the gas is equal to the external pressure $P_a$. Therefore, we have

$$PV = Nk_BT = P_aV_f,$$

where $V_f$ is the final volume of the gas. Solving for $T_f$ in terms of $V_f$ gives

$$T_f = \frac{P_aV_f}{Nk_B}.$$

Conservation of energy gives

$$Mgh + U_0 = Mgh’ + U_f + W.$$

Substituting in the expressions for $U_0$, $U_f$, and $W$, and solving for $V_f$ gives

$$V_f = V_0\exp\left(\frac{Mgh}{Nk_BT_0}-\frac{Mgh’}{Nk_BT_f}-\frac{P_aA(h’-h)}{Nk_BT_f}\right).$$

Substituting in the expression for $T_f$ gives

$$V_f = V_0\exp\left(\frac{Mgh}{Nk_BT_0}-\frac{Mgh’}{Nk_BP_aV_f}-\frac{Ah’}{V_fk_B}\right).$$

This equation cannot be solved analytically, but it can be solved numerically using iterative methods. Start with an initial guess for $V_f$, and use the above equation to compute a new value of $V_f$. Repeat this process until the value of $V_f$ converges to a fixed point. Once $V_f$ is known, $T_f$ can be computed using the expression for $T_f$ above.

a) The heat capacity at constant volume $C_V$ is given by

$$C_V = \left(\frac{\partial U}{\partial T}\right)_V,$$

where $U$ is the internal energy of the gas. For a van der Waals gas, the internal energy is a function of temperature $T$, volume $V$, and the number of molecules $N$. Taking the partial derivative of $C_V$ with respect to volume $V$ at constant temperature $T$ gives

$$\left(\frac{\partial C_V}{\partial V}\right)_T = \frac{\partial^2 U}{\partial V\partial T}.$$

Using the van der Waals equation of state,

$$\left(P+\frac{aN^2}{V^2}\right)(V-Nb) = Nk_BT,$$

we can write the internal energy as

$$U = \frac{3}{2}Nk_BT – \frac{aN^2}{V},$$

where we have added a constant of integration to ensure that $U$ goes to zero as $V$ goes to infinity. Taking the partial derivative of $U$ with respect to volume $V$ and temperature $T$, we get

$$\frac{\partial U}{\partial V} = \frac{aN^2}{V^2}, \quad \frac{\partial U}{\partial T} = \frac{3}{2}Nk_B.$$

Taking the partial derivative of $\frac{\partial U}{\partial V}$ with respect to temperature $T$ gives

$$\frac{\partial^2 U}{\partial V\partial T} = 0,$$

since $\frac{\partial}{\partial T}(\frac{aN^2}{V^2})=0$. Therefore, we have

$$\left(\frac{\partial C_V}{\partial V}\right)_T = 0.$$

b) Using the result from part (a), we can write the entropy $S$ of a van der Waals gas as

$$S(T,V) = S_0(T) + \int_{V_0}^V \frac{C_V(T)}{T}\mathrm{d}V = S_0(T) + \frac{3}{2}Nk_B\ln\left(\frac{V}{V_0}\right),$$

where $S_0(T)$ is an arbitrary function of temperature that accounts for the entropy of the gas when its volume is $V_0$, and $V_0$ is a reference volume. Note that the constant of integration is chosen so that the entropy goes to zero as $V$ goes to infinity.

Similarly, using the result from part (a) and the van der Waals equation of state, we can write the energy $E$ of the gas as

$$E(T,V) = E_0(T) + \frac{3}{2}Nk_BT – \frac{aN^2}{V},$$

where $E_0(T)$ is an arbitrary function of temperature that accounts for the energy of the gas when its volume is $V_0$. Note that the constant of integration is chosen so that the energy goes to zero as $V$ goes to infinity.

c) When the gas is adiabatically compressed from $\left(V_1, T_1\right)$ to $V_2$, we have

$$P_1V_1^{\gamma} = P_2V_2^{\gamma},$$

where $\gamma = C_P/C_V$ is the ratio of specific heats. For a van der Waals gas, we have $\gamma

Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

Statistics-lab™可以为您提供cornell.edu ENGRD2210 Thermodynamics热力学课程的代写代考和辅导服务！ 请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。

Statistics-lab™可以为您提供cornell.edu ENGRD2210 Thermodynamics热力学课程的代写代考和辅导服务！

ENGRD2210 Thermodynamics课程简介

Thermodynamics is the branch of physics that deals with the study of the relationship between heat, work, and energy. The laws of thermodynamics govern the behavior of all physical systems, including those that involve the transfer of energy as heat or work.

The First Law of Thermodynamics: The law of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another. The first law of thermodynamics is a statement of this principle as applied to thermodynamic systems. It states that the total energy of a closed system remains constant, although it may be converted from one form to another.

The Second Law of Thermodynamics: The second law of thermodynamics states that the entropy of an isolated system never decreases over time. Entropy is a measure of the disorder or randomness of a system, and the second law of thermodynamics tells us that the natural tendency of any system is to become more disordered over time.

Thermodynamic Property Relationships: In thermodynamics, a property is a characteristic of a system that can be measured or calculated. Examples of thermodynamic properties include temperature, pressure, volume, and entropy. The relationships between these properties are described by equations of state, which relate them to each other and to the state of the system.

PREREQUISITES 

Presents the definitions, concepts, and laws of thermodynamics. Topics include the first and second laws, thermodynamic property relationships, and applications to vapor and gas power systems, refrigeration, and heat pump systems. Examples and problems are related to contemporary aspects of energy and power generation and to broader environmental issues.

Outcome 1: Students will be able to choose an appropriate system and identify interactions between system and surroundings.
Outcome 2: Obtain values of thermodynamic properties for a pure substance in a given state, using table, relations for incompressible substances, and relations for gases.

Outcome 3: Apply energy and entropy balances in the control mass (closed system) and control volume formulations to the analysis of devices and cycles.

Cornell students enroll only in ENGRD 2210. MAE 2210 for Non-CU students.

ENGRD2210 Thermodynamics HELP（EXAM HELP， ONLINE TUTOR）

To derive the expression for the Joule-Thompson coefficient, we start with the definition of the enthalpy $H=E+PV$, where $E$ is the internal energy, $P$ is the pressure, $V$ is the volume, and $T$ is the temperature. Taking the differential of $H$ with respect to $V$ at constant $T$, we get:

dH = \left(\frac{\partial H}{\partial T}\right)_V dT + \left(\frac{\partial H}{\partial V}\right)_T dVdH=(∂T∂H​)V​dT+(∂V∂H​)T​dV

But at constant $T$, $dH = dE + PdV$, so we can substitute $dH$ with $dE + PdV$ to get:

dE + PdV = \left(\frac{\partial H}{\partial T}\right)_V dT + \left(\frac{\partial H}{\partial V}\right)_T dVdE+PdV=(∂T∂H​)V​dT+(∂V∂H​)T​dV

Solving for $\left(\frac{\partial E}{\partial V}\right)_T$, we get:

\left(\frac{\partial E}{\partial V}\right)_T = \left(\frac{\partial H}{\partial V}\right)_T – P = T\left(\frac{\partial P}{\partial T}\right)_V – P(∂V∂E​)T​=(∂V∂H​)T​−P=T(∂T∂P​)V​−P

where we have used the fact that $\left(\frac{\partial H}{\partial V}\right)_T = \left(\frac{\partial E}{\partial V}\right)_T + P$. This is the desired expression for the Joule-Thompson coefficient.

For a van der Waals gas, the pressure and volume are related by the equation of state:

\left(P + \frac{a}{V^2}\right)(V – b) = RT(P+V2a​)(V−b)=RT

where $a$ and $b$ are constants related to the intermolecular forces in the gas. Differentiating this equation with respect to $T$ at constant $V$, we get:

\left(\frac{\partial P}{\partial T}\right)_V = \frac{R}{V – b} – \frac{2a}{k_BTV^3}(∂T∂P​)V​=V−bR​−kB​TV32a​

where $k_B$ is the Boltzmann constant. Substituting this expression into the Joule-Thompson coefficient equation, we get:

\left(\frac{\partial E}{\partial V}\right)_T = T\left(\frac{R}{V – b} – \frac{2a}{k_BTV^3}\right) – P(∂V∂E​)T​=T(V−bR​−kB​TV32a​)−P

This expression can be further simplified using the van der Waals equation of state to eliminate $P$ in favor of $V$ and $T$. After some algebraic manipulation, we get:

\left(\frac{\partial E}{\partial V}\right)_T = \frac{3a}{2V^2} – \frac{R}{V – b} + \frac{a}{(V-b)^2} \left(\frac{3V^2}{k_BTV^3}-1\right)(∂V∂E​)T​=2V23a​−V−bR​+(V−b)2a​(kB​TV33V2​−1)

This is the Joule-Thompson coefficient for a van der Waals gas. Note that it depends on the volume, temperature, and the constants $a$ and $b$, which are related to the intermolecular forces in the gas.

a) The heat capacities $C_P$ and $C_V$ are defined as:

C_P = \left(\frac{\partial H}{\partial T}\right)_P = \left(\frac{\partial (E+PV)}{\partial T}\right)_P = C_V + P\left(\frac{\partial V}{\partial T}\right)_PCP​=(∂T∂H​)P​=(∂T∂(E+PV)​)P​=CV​+P(∂T∂V​)P​

where $H$ is the enthalpy and $E$ is the internal energy. Using the chain rule, we can write:

\left(\frac{\partial V}{\partial T}\right)_P = \left(\frac{\partial V}{\partial T}\right)_E + \left(\frac{\partial V}{\partial E}\right)_T \left(\frac{\partial E}{\partial T}\right)_P(∂T∂V​)P​=(∂T∂V​)E​+(∂E∂V​)T​(∂T∂E​)P​

Now, using the definition of the compressibility factor $Z=\frac{PV}{RT}$, we have:

\left(\frac{\partial V}{\partial T}\right)_P = \frac{V}{ZRT}\left[\left(\frac{\partial P}{\partial T}\right)_V – \frac{P}{V}\right] = -\frac{V}{ZRT}\left[\left(\frac{\partial f}{\partial V}\right)_T + \frac{f}{V}\right](∂T∂V​)P​=ZRTV​[(∂T∂P​)V​−VP​]=−ZRTV​[(∂V∂f​)T​+Vf​]

where we have used the fact that $P=f(T,V)$ and the chain rule. Substituting this expression into the equation for $C_P-C_V$, we get:

C_P-C_V = P\left(\frac{\partial V}{\partial T}\right)_P = -\frac{PV}{ZRT}\left[\left(\frac{\partial f}{\partial V}\right)_T + \frac{f}{V}\right]CP​−CV​=P(∂T∂V​)P​=−ZRTPV​[(∂V∂f​)T​+Vf​]

This is the desired expression for $C_P-C_V$ in terms of $f$.

b) For a van der Waals gas, the equation of state is given by:

\left(P + \frac{a}{V^2}\right)\left(V-b\right) = RT(P+V2a​)(V−b)=RT

where $a$ and $b$ are constants related to the intermolecular forces in the gas. Differentiating this equation with respect to $V$ at constant $T$, we get:

\left(\frac{\partial P}{\partial V}\right)_T = -\frac{RT}{(V-b)^2} + \frac{2a}{V^3}(∂V∂P​)T​=−(V−b)2RT​+V32a​

Substituting this expression into the equation for $C_P-C_V$ from part a), we get:

\begin{align} C_P-C_V &= -\frac{PV}{ZRT}\left[\left(\frac{\partial P}{\partial V}\right)_T + \frac{P}{V}\right] \ &= -\frac{PV}{ZRT}\left[-\frac{RT}{(V-b)^2} + \frac{2a}{V^3} + \frac{P}{V}\right] \ &= \frac{a}{R}\left[\frac{3V-b}{(V-b)^3}\right] \end{align}

where we have used the expressions for $Z$ and $P$ in terms of $V$ and $T$ and simplified the algebra. This is the value of $C_P-C_V$ for a van der Waals gas. Note that it is independent of temperature and depends only on the volume and the constants $a$ and $b$.

Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

Statistics-lab™可以为您提供cornell.edu ENGRD2210 Thermodynamics热力学课程的代写代考和辅导服务！ 请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。