Applying the laws of thermodynamics : Exercises

Introduction

In the questions that follow we are going to use the laws of thermodynamics to calculate the amount the entropy changes by in an ideal gas. The focus here is on solving the partial differential equations. We will leave deriving the equation of state for the ideal gas, $PV=Nk_B T$, and deriving an expression for the internal energy of an ideal gas, $E=\frac{3}{2} Nk_BT$ to later in the course once we are more familiar with the statistical concepts that are required to derive and understand these results. For now we will simply note that, in the two expressions that have been given in this paragraph and in the questions that follow, $k_B$ is Boltzmann's constant, $T$ is the absolute temperature, $V$ is the volume and $P$ is the pressure.

Example problems

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Problem 1

A sample of (ideal) gas is compressed from the volume $2V$ to $V$ at constant temperature. If the gas is isolated from the rest of the universe what is the change in the internal energy?

We know from the description above that: $$ E = \frac{3}{2} Nk_B T \qquad \rightarrow \qquad \left( \frac{\partial E}{\partial V} \right)_T = 0 $$ Hence there is no change in internal energy. The value of the internal energy does not depend on the volume (for an ideal gas)

During the compression the gas experiences an external pressure from the surroundings of $P$. Write an expression for the work done, $\textrm{d}w$, during during an infinititisimal change in volume $\textrm{d}V$? Show how you arrived at this expression.

The work done on the gas is given by: $$ \textrm{d}w = -P_{\textrm{ext}}\textrm{d}V $$ There are long discussions on the derivation of this expression in the videos.

Rearrange the equation of state for an ideal gas and obtain an expression that allows one to calculate the pressure, $P$, from the temperature $T$ and the volume $V$. Use the expression that you get for $P$ to calculate the work done on the gas during the compression from $2V$ to $V$.

Rearranging the equation of state gives us: $$ P = \frac{N k_B T}{V} $$ To calculate the work done during the compression we integrate the infinitessimal that was discussed in the previous part between $2V$ and $V$. We use the above expression in place of $P$ as shown below, however. $$ \Delta w = -\int_{2V}^V P\textrm{d}V' = -\int_{2V}^V \frac{N k_B T}{V'} \textrm{d}V' = -\left[ N k_B T \ln \left( V' \right) \right]_{2V}^V = - N k_B T \ln \left( \frac{V}{2V} \right) = N k_B T \ln ( 2) $$

Use what you know about the change in energy that occurs as an ideal gas is compressed from a volume of $2V$ to a volume of $V$ to determine how much heat must be emitted/absorbed during the compression of the gas?

The first law of thermodynamics tells us that: $$ \Delta E = \Delta q + \Delta w $$ We know, however, that when the ideal gas is compressed $\Delta E = 0$. By remembering that $\Delta w = N k_B T \ln ( 2)$ of work has been done on the gas during the compression we thus realise that, in order to ensure that the total energy of the gas does not change, the following amount of heat must be output: $$ \Delta q = - N k_B T \ln ( 2) $$

Calculate how much the entropy of an ideal gas changes by during a compression from $2V$ to $V$.

The second law of thermodynamics tells us that if the system is at equilibrium at all points during the transition: $$ \Delta S = \frac{\Delta q}{T} $$ The change in entropy in this case is thus equal to: $$ \Delta S = \frac{- N k_B T \ln ( 2)}{T} = - N k_B \ln ( 2) $$

Problem 2

Starting from the following statements of the first and second laws of equilibrium thermodynamics $\textrm{d}E = \textrm{d} q + \textrm{d} w$ and $\textrm{d}S = \frac{\textrm{d}q}{T}$ and assuming only $PV$-work is done on the system show that: $$ \textrm{d}S = \frac{1}{T} \textrm{d}E + \frac{P}{T} \textrm{d}V $$

Consider a transition involving an infinitesimal change in volume. The work done during such a transition will be: $$ \textrm{d}w_{rev} = -P\textrm{d}V $$ Notice, however, that for the above equation to hold the system must be in equilibrium with its environment at all points during the transition. That is to say the internal pressure in the system must be equal to the external pressure upon it. We write $\textrm{d}w_{rev}$ rather than simply $\textrm{d}w$ to make it clear that we have taken this particular path in deriving this expression. We can calculate the heat output/input during this transition using the second law of thermodynamics: $$ \textrm{d}q_{rev} = T\textrm{d}S $$ Again though we must assume, when using this result, that the system is in equilibrium with its environment at all points during the transition That is to way the temperature in the system must be equal to the external temperature. This is why we again have used the symbol $\textrm{d}q_{rev}$ in the above expression. Inserting these two results into the first law of thermodynamics gives: $$ \textrm{d}E_{rev} = \textrm{d}q_{rev} + \textrm{d}w_{rev} = T\textrm{d}S - P\textrm{d}V $$ However, the internal energy is an exact differential the change in the internal energy only depends on the initial and final states. It does not depend on the path taken. When using this result we are thus not required to assume that the system is at equilibrium at all points during the transition. We can thus write $\textrm{d}E_{rev}=\textrm{d}E$. We thus have: $$ \textrm{d}E = T\textrm{d}S - P\textrm{d}V $$ which can be rearranged to give the answer required for this question.

The fact that that the heat output and the work done depend on the path the transition takes and that the internal energy does not depend on the path taken is a subtle and important point. You must mention it in your solutions.

$\textrm{d}S$ is an exact differential. Use this information to write an expression for $\textrm{d}S$ in terms of $\left( \frac{\partial S}{\partial E} \right)_V$, $\left( \frac{\partial S}{\partial V} \right)_E$, $\textrm{d}E$ and $\textrm{d}V$ and hence show that for an ideal gas: $$ \left( \frac{\partial S}{\partial E} \right)_V = \frac{3Nk_B}{2E} \qquad \textrm{and} \qquad \left( \frac{\partial S}{\partial V} \right)_E = \frac{Nk_B}{V} $$ You will need the expression for the internal energy of an ideal monoatomic gas $E= \frac{3}{2} N k_B T$ and the mechanical equation of state for an ideal gas $PV = Nk_B T$.

The fact that $S$ is an exact differential ensures that we can write: $$ \textrm{d}S = \left( \frac{\partial S}{\partial E} \right)_V \textrm{d}E + \left( \frac{\partial S}{\partial V} \right)_E \textrm{d}V $$ If we equate the coefficients of $\textrm{d}E$ and $\textrm{d}V$ in this expression and the expression that we derived in the first part we find: $$ \begin{aligned} \left( \frac{\partial S}{\partial E} \right)_V & = \frac{1}{T} \\ \left( \frac{\partial S}{\partial V} \right)_E & = \frac{P}{T} \end{aligned} $$ The equation for the internal energy of an ideal gas that was given in the question can be rearranged to give $\frac{1}{T} = \frac{3 N k_B }{2E}$. Furthermore, the equation of state for the ideal gas can be rearranged to give $\frac{P}{T} = \frac{N k_B }{V}$. When the factors of $\frac{1}{T}$ and $\frac{P}{T}$ in the right hand sides of the above expressions are replaced by the right hand sides of these expressions we get to the desired results.

Integrate the partial differential equations that you derived in the previous question and demonstrate that: $$ S(E,V) = N k_B\ln(E^{\frac{3}{2}} V) + k $$ for an ideal gas.

The first result that was derived in the previous question tells us that: $$ \left( \frac{\partial S}{\partial E} \right)_V = \frac{3 N k_B }{2E} \quad \rightarrow \quad S(E,V) = \frac{3N k_B}{2} \ln(E) + f(V) $$ We arrive at the second result above by integrating along the $E$ axis (so holding V fixed). Furthermore, in writing this expression we introduce $f(V)$ as some as yet unknown function of $V$.

The second result that was derived in the previous question tells us that: $$ \left( \frac{\partial S}{\partial V} \right)_E = \frac{N k_B }{V} \quad \rightarrow \quad S(E,V) = N k_B \ln(V) + g(E) $$ We arrive at the econd result above by integrating along the $V$ axis (so holding E fixed). Much as in the previous part we introduce $g(E)$ as some as yet unknown function of $E$.

We now need to ensure that the two results we have obtained are consistent. To guarantee this we must set $g(E)$ and $f(V)$ as shown below: $$ g(E) = \frac{3N k_B}{2} \ln(E) \qquad \textrm{and} \qquad f(V) = N k_B \ln(V) $$ When these results are inserted into either of the two integrals we have defined we thus find that: $$ S(E,V) = \frac{3N k_B}{2} \ln(E) + N k_B \ln(V) + k = N k_B\ln(E^{\frac{3}{2}} V) + k $$ as required.

Problem 3

A quantity of water, initially at 10$^\circ$C, is brought into content with a heat reservoir at a temperature of 90$^\circ$C. The constant volume heat capacity, $C_v$ of the water is defined as $\textrm{d} q=C_v\textrm{d}T$. If the waters volume is fixed (i.e. if no work is done on or by the system) explain why the change in the internal energy during an infinitesimal change in temperature is given by $\textrm{d}E = C_v \textrm{d}T$. Use this result, and the second law of (equilibrium) thermodynamics, $\textrm{d}S = \frac{\textrm{d}q}{T}$, to demonstrate that the change in entropy during an infinitesimal change in the temperature of the water is given by: $$ \textrm{d}S = \frac{C_v \textrm{d}T }{T} $$

Recall the first law of thermodynamics tells us that: $$ \textrm{d}E = \textrm{d} q + \textrm{d} w $$ In addition the question tells us that no work is done during the transition so $\textrm{d} w = 0$, $\textrm{d} E = \textrm{d} q$ and (using the information in the question) we thus have: $$ \textrm{d}E = \textrm{d} q = C_v \textrm{d}T $$ This result is important as the first law of thermodynamics tells us that the internal energy is an exact differential. In other words, the change in internal energy that occurs during a transition depends only on the initial and final states of the system - not on the path taken. The result above suggests that the heat input/output must always be the same as $\textrm{d} q=\textrm{d} E$. It is critical to realise that the manner in which we heat the water is irelevant. We can thus substitute the equation that we just arrived at, $\textrm{d} q = C_v \textrm{d}T$, into the second law of (equilibrium) thermodynamics, $\textrm{d}S = \frac{\textrm{d}q}{T}$ and get the required differential for the change in the entropy: $$ \textrm{d}S = \frac{\textrm{d} q}{T} = \frac{ C_v \textrm{d}T}{T} $$

Determine the change in the entropy of the water when it is heated from 10$^\circ$C to 90$^\circ$C.

To calculate the change in entropy required we must integrate the result we arrived at in the previous question as shown below: $$ \Delta S_{water} = \int_{10}^{90} \frac{C_v \textrm{d}T}{T} = \left[ C_v \ln (T) \right]_{10}^{90} = 2C_v \ln(3) $$

Now determine how much heat must flow from the reservoir to the water in order to heat the water from 10$^\circ$C to 90$^\circ$C and the change in the entropy of the reservoir that occurs as a consequence of this outflow of heat.

We determine the ammount of heat that must flow by integrating $\textrm{d} q = C_v \textrm{d} T$ as shown below: $$ \Delta q = -\int_{10}^{90} C_v \textrm{d}T = -C_v \int_{10}^{90} \textrm{d}T = - C_v ( 90 - 10 ) = -80C_v $$ There is a minus sign here because we are considering the heat that is lost from the reservoir and given to the water.

To determine the change in entropy we use the fact that we know from the first law of (equilibrium) thermodynatims that $\Delta S = \frac{\Delta q}{T}$. We can thus use the value that we just obtained for $\Delta q$ to calculate the value of $\Delta S$ as shown below: $$ \Delta S_{reservoir} = \frac{\Delta q_{rev}}{T} = -\frac{80C_v}{90} = -\frac{8}{9}C_v $$

What are the changes of entropy for the water and the reservoirs if two reservoirs are used to heat the system rather than one? The first of these reservoirs is at 50$^\circ$C and so it heats the system from 10$^\circ$C to 50$^\circ$C, while the second is at 90$^\circ$C and it heats the system from 50$^\circ$C to 90$^\circ$C.

To tackle this part of the question we have to recognise that the answers to the questions above tell us that if we change the temperature from $T_i$ to $T_f$ the change in the entropy of the water will be: $$ \Delta S_{water} = C_v \ln\left( \frac{T_f}{T_i} \right) $$ while the change in the entropy of the reservoir will be: $$ \Delta S_{reservoir} = \frac{C_v}{T_f}(T_f - T_i) $$ Here $T_{f}$ is the temperature of the reservoir and $T_i$ is the temperature of the system when it is first placed in contact with the reservoir. With this in mind we can calculate the change in entropy for the various parts of thermodynamic transition described in the question: $$ \begin{aligned} \textrm{Change in entropy of water when in contact with 50$^{\circ}$ reservoir} \qquad & \Delta S_{wc} = C_v \ln(5) \\ \textrm{Change in entropy of water when in contact with 90$^{\circ}$ reservoir} \qquad& \Delta S_{wh} = C_v \ln\left( \frac{9}{5} \right) \\ \textrm{Change in entropy of 50$^{\circ}$ reservoir } \qquad & \Delta S_{rc} = -\frac{4}{5} C_v \\ \textrm{Change in entropy of 90$^{\circ}$ reservoir } \qquad & \Delta S_{rh} = -\frac{4}{9} C_v \end{aligned} $$ When these are added together we find that the change in the entropy for the water is given by: $$ \Delta S_{water} = \Delta S_{wc} + \Delta S_{wh} = C_v \ln(5) + C_V \ln\left( \frac{9}{5} \right) = C_v \ln(9) = 2C_v\ln(3) $$ This is the same as it was in the previous part, which should not be surprising - entropy is a state function. The fact that we used two reservoirs is irrelevant the entropy change in the water (a closed system) depends only on the initial and final states.

The total change in the entropy of the reservoirs is: $$ \Delta S_{reservoir} = \Delta S_{rc} + \Delta S_{rh} = -\frac{4}{5} C_v -\frac{4}{9} C_v = - \frac{56}{45} C_v $$ This is larger than it was in the previous part. Noice, however, that we have changed this part of the the system (we now have two reservoirs). The fact that we have a different entropy change is a consequence of this change.

Now suppose that an infinite number of reservoirs are used. What is the entropy change for the entire system (water+all reservoirs) when the system is heated from 10$^\circ$C to 90$^\circ$C?

Lets do the easy part of this question first. We know entropy is a state function so we know that, even with this infinite number of reservoirs, the entropy change for the water (a closed system) is going to be given by: $$ \Delta S_{water} = C_v \ln\left( \frac{T_f}{T_i} \right) $$ what will be different is the entropy change for the set of reservoirs. Using our now familiar equations we can write the total entropy change for all the reservoirs as: $$ \Delta S_{reservoir} = \sum_{i=1}^N \Delta S_{ri} = -C_v \sum_{i=1}^N \frac{ T_i - T_{i-1} }{ T_i } = -C_v \sum_{i=1}^N \frac{ \frac{T_N - T_0}{N} }{ T_0 + \frac{i(T_N-T_0)}{N} } = -C_v \sum_{i=1}^N \frac{\delta T}{ T_0 + i(\delta T)} $$ here $T_0$ is the initial temperature of the water and $T_N$ is the final temperature of the water. Our reservoirs are at equally spaced temperatures between $T_N$ and $T_0$ so if there are $N$ of them in total the difference in temperature between the temperature of the $i$th and $(i+1)$th reservoir is $\delta T = \frac{T_N - T_0}{N}$.

In the limit of infinitely many reservoirs the sum from the above equation becomes an integral and we thus have: $$ \Delta S_{reservoir} = -C_v \int_{T_0}^{T_N} \frac{\textrm{d}T}{T} = -[C_v \ln\left( T \right) ]_{T_0}^{T_N} = -C_v \ln\left( \frac{T_N}{T_0} \right) $$ In other words the entropy change for the reservoirs is equal to minus the entropy change for the water. This makes a lot of sense. When we have an infinite number of reservoirs the water and the reservoirs are in thermodynamic equilibrium at all points during the transition as the internal temperature of the water is the same as the external temperature of the reservoir.

Notice that the maths here works even if $C_v$ is not temperature independent. If $C_v$ did depend on temperature the entropy change for the water would be an integral involving the function $C_v(T)$ and the entropy change for the reservoirs would just be minus that integral.

Applying the laws of thermodynamics : Exercises

Introduction

Example problems

Problem 1

Problem 2

Problem 3

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