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All Solutions of Type: Thermodynamics
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GR8677 #14

Thermodynamics}Exact differentials

The key equation is: PV=nRT, and its players, P,V,n,R,T, are terms one should be able to guess.

(A) True, according to the ideal gas law. (This is also the final step in deriving Mayer's Equation, as shown below.)

(B) This translates into the statement \left.\frac{dq}{dT}\right|_V=\left.\frac{dq}{dT}\right|_P\Rightarrow c_V=c_P.The problem gives away the fact that for an ideal gas C_P \neq C_V. B can't be right.

(C) According to the ideal gas law, the volume might change.

(D) False. An ideal gas's internal energy is dependent only on temperature. More elegantly, u=u(T).

(E) Heat needed for what?

If one is interested in the formal proof of the relation c_p=c_v+nR, read on about Mayer's equation:

For thermo, in general, there's an old slacker's pride line that goes like, ``When in doubt, write a bunch of equations of states and mindlessly begin taking exact differentials. Without exerting much brainpower, one will quickly arrive at a brilliant result." Doing this,

Plugging in the first law of thermodynamics into the U equation of state, one gets dQ=\left.\partial_T U\right|_T+\left(\left.\partial_T U\right|_V+P\right)dV=\left.\partial_T U\right|_T+PdV, where the last simplification is made by remembering the fact that the internal energy of an ideal gas depends only on temperature.

(Taking the derivative with respect to T at constant volume, one gets \left.\frac{dQ}{dT}\right|_V=\left.\partial_T U\right|_T=C_V.)

Plugging in the simplified result for dQ=... into the third equation of state, the ideal gas equation, one gets: dQ-C_vdT+VdP=nRdT. Taking the derivative at constant pressure, one gets:

So, one sees that it is the ideal gas equation that makes the final difference. The work of an ideal gas changes when temperature is varied.

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GR8677 #66

Thermodynamics}First Law

Recall that dQ=TdS=PdV+dU, where work done by the system is positive and heat input into the system is positive. For constant volume, the equation becomes dQ=TdS=dU.

 \Rightarrow T=\left(\frac{\partial U}{\partial S}\right)_V. The reciprocol gives the right answer.

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GR8677 #95

Thermodynamics}Carnot Cycle

The Carnot cycle is the cycle of the most efficient engine, which does NOT have e=1 (unless T_1=0), but rather, dS=0---this means that entropy stays constant. Choice (C) is thus false.

The efficiency of the Carnot cycle is dependent on the temperature of the hot and cold reservoir. The hot reservoir has decreasing entropy because it gets cooler as the cycle proceeds. From writing down the thermodynamic relations for isothermal and adiabatic paths and matching P-V boundary conditions, one can determine that Q_1/Q_2 = T_1/T_2. The efficiency is thus e=\frac{Q_2-Q_1}{Q_2}=1-Q_1/1_2 = 1-T_1/T_2.

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GR8677 #13


Given P=100 W and V=1L=1m^3=1kg for water, one can chunk out the specific heat equation for heat, Q=mc\Delta T=Pt\Rightarrow 4200(1^\circ) = 100t\Rightarrow t \approx 40 s, as in choice (B).

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GR8677 #14


The final temperature is 50^\circ C. The heat exchanged from the hot block to the cool block is Q=mc\Delta T = 5 kcal, as in choice (D).

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GR8677 #15

Thermodynamics}Phase Diagram

Recall that for an ideal gas U=C_v \Delta T and PV=nRT. Don't forget the first law of thermodynamics, Q=W+U.

For A\rightarrow B, U=0, since the temperature is constant. Thus, Q=W=RT_H\ln V_2/V_1.

For B\rightarrow C, W=P_2(V_1-V_2)=R(T_c-T_h). U=C_v (T_c - T_h), and thus Q=W+U=C_v (T_c - T_h)-R(T_h-T_c).

For C\rightarrow A, W=0, U=C_v(T_h-T_c), thus Q=U=C_v(T_h-T_c).

Add up all the Q's from above, cancel the C_v term, to get Q_tot=RT_h\ln(V_2/V_1)-R(T_h-T_c), as in choice (E).s

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GR8677 #16

Thermodynamics}Mean Free Path

Air is obviously less dense than the atomic radius 10^{-10}, thus choices (C), (D), and (E) are out. Air is not dilute enough that the distance between particles is actually within human visible range, as in (A)! Thus, the answer must be (B). (Note how this problem exemplifies the usefulness of common sense.)

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GR8677 #73

Thermodynamics}Adiabatic Work

One should recall the expression for work done by an ideal gas in an adiabatic process. But, if not, one can easily derive it from the condition given in the problem, viz., PV^\gamma =C \Rightarrow P=C/V^\gamma.

Recall that the definition of work is W=\int PdV =\int_{V_1}^{V_2} C dV/V^\gamma =\left.-\frac{1}{\gamma-1} C/V^{\gamma -1} \right|_{V_1}^{V_2}, which when one plugs in the endpoint limits, becomes choice (C).

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GR8677 #74


Recall the definition of entropy to be dS = dQ/T. The heat is defined here as dQ = m c dT, and thus S = \int mcdT/T.

One is given two bodies of the same mass. One mass is at T_1=500 and the other is at T_2=100 before they're placed next to each other. When they're put next to each other, one has the net heat transferred being 0, thus Q_1 = -Q_2 \Rightarrow T_f = (T_1+T_2)/2=300.

The entropy is thus S =  \int^{T_f}_{T_1} mcdT/T +  \int^{T_f}_{T_2} mcdT/T = mc \left ( \ln(3/5) + \ln(3) \right) = 2mc \ln 3 - mc \ln 5 = mc(\ln 9 - \ln 5) = mc \ln(9/5), as in choice (B).

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GR8677 #75

Thermodynamics}Fourier's Law

Recall Fourier's Law q = -k\nabla T, where q is the heat flux vector (rate of heat flowing through a unit area) and T is the temperature and k is the thermal conductivity. (One can also derive it from dimensional analysis, knowing that the energy flux has dimensions of J/(s m^2))

Fourier's Law implies the following simplification: q = -k \frac{\Delta T}{\Delta l}

The problem wants the ratio of heat flows q_A/q_B=\frac{k_A l_B}{k_B l_A}=\frac{0.8 \times 2}{0.025 \times 4}=32/2=16, as in choice (D). (The problem gives l_A = 4, l_B=2, and k_A=0.8, k_B=0.025.)

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GR8677 #91

Thermodynamics}Second Law

The Second Law of thermodynamics has to do with entropy; that entropy can never decrease in the universe. One form of it states that from hot to cold things flow. A cooler body can thus never heat a hotter body. Since the oven is at a much lower temperature than the wanted sample temperature, the oven can only heat the sample to a maximum of 600K without violating the Second Law.
(This solution is due to David Latchman.)

(Also, since the exam is presumably written by theorists, one can narrow down the choices to either (D) or (E), since the typical theorist's stereotype of experimenters usually involves experimenters attempting to violate existing laws of physics---usually due to naivity.)

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GR8677 #5

Thermodynamics}Degree of Freedom

Acording to the Equipartition Theorem, there is a $kt/2$ contribution to the energy from each degree of quadratic freedom in the Hamiltonian. In equation form, the average total energy is , where s is the degrees of freedom.

For a n-dimensional 1-particle system, the Hamiltonian is , where () refer, respectively, to the component of momentum (position).

Thus, for a 3-dimensional 1-particle system, one has 6 quadratic terms in the Hamiltonian, as in .

Plugging in s=6, one finds that the average energy is 3kT.

(This revised solution is due to the GREPhysics.NET user kolndom.)

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GR8677 #6


The work done by an adiabatic process is W=-\frac{1}{\gamma - 1}(P_2V_2-P_1V_1). (One can quickly derive this from noting that PV^\gamma = const in an adiabatic process and W=\int P dV.)

The work done by an isothermal process is W=nRT_iln(V_2/V_1)=P_iV_iln(V_2/V_1) (One can quickly derive this from noting that P_2V_2=P_2V_2=nRT_1=nRT_2 for an isothermal process.)

From the above formulae, one can immediately eliminate choice (A).

One can calculate the isothermal work to be W_i = nRT_1 ln 2 = P_1V_1 ln 2.

One can calculate the adiabatic work to be:

W_a = \frac{1}{1-\gamma}(P_1-2P_2)V_1=\frac{1}{1-\gamma}(P_1-2/2^\gamma P_1)V_1 = \frac{1}{1-\gamma}(1-2^{1-\gamma})P_1V_1.

For a monatomic gas, \gamma=5/3, and one finds that 0 <W_a < W_i. Choice (E).

(Also, in general, an adiabatic process always does less work than an isothermal process in a closed cycle.)

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GR8677 #36


In an adiabatic expansion, dS=0 (entropy), since dQ=0 (heat). No heat flows out, by definition, and thus choice (A) is out, as well as choice (B). Choice (C) is true since by the first law, one has Q=U+W \Rightarrow U=-W, and the given integral is just the definition of work. Choice (D) defines work. Choice (E) remains---so take that.

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GR8677 #37

Thermodynamics}Cycle Analysis

For path C to A, one has W=0 for an isochoric (constant volume) process.

For path A to B, one has just W=P\Delta V = 200(V_B-2) for an isobaric (constant pressure) process.

For path B to C, one has W=P_c V_c ln(V_C/V_B) = 1000 ln(2/V_b).

One can figure out V_B from the isothermal condition P_C V_C = P_B V_B \rightarrow V_B = P_C/P_B V_C = 5/2 \times 2 = 5. Plug that in above to get,



W_{BC}=1000 ln(2/5) > -1000

\sum W \approx -400, which is closest to choice (D).

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GR8677 #47


Entropy is given as dS=\int dQ/T. Since the volume expands by a factor of 2, the work in the isothermal process is W=nRTln(V_2/V_1)=nRTln2. But, for an ideal gas, the internal energy change in an isothermal process is 0, thus from the first law of Thermodynamics, one has dQ=dW+dU \Rightarrow dQ=dW. The temperature cancels out in the entropy integral, and thus the entropy is just nR ln2, as in choice (B).

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GR8677 #48

Thermodynamics}RMS Speed

In case one forgets the RMS speed, one does not need to go through the formalism of deriving it with the Maxwell-Boltzmann distribution. Instead, one can approximate its dependence on mass and temperature by 3/2 kT = 1/2 m v^2. One thus has v \propto \sqrt{kT/m}. For the ratio of velocities, one has \frac{v_N}{v_O}=\frac{m_O}{m_N}. Plug in the given molecular masses for Oxygen and Nitrogen to get choice (C).

Incidentally, the trick for memorizing the diatomic gasses is Have No Fear Of Ice Cold Beer (Hydrogen, Nitrogen, Florine, Oxygen, Iodine, Chlorine, Bromine).

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GR8677 #16

Thermodynamics}Carnot Engine

Recall the common-sense definition of the efficiency e of an engine,

where one can deduce from the requirements of a Carnot process (i.e., two adiabats and two isotherms), that it simplifies to

for Carnot engines, i.e., engines of maximum possible efficiency. (Q_{input} is heat put into the system to get stuff going, Wis work done by the system and T_{low} (T_{high}) is the isotherm of the Carnot cycle at lower (higher) temperature.)

The efficiency of the Carnot engine is thus e=1-\frac{800}{1000}=0.2, where one needs to convert the given temperatures to Kelvin units. (As a general rule, most engines have efficiencies lower than this.) The heat input in the system is Q_{input}=2000J, and thus W_{accomplished}=400 J, as in choice (A).

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GR8677 #46

Thermodynamics}Critical Isotherm

The critical isotherm is the (constant temperature) line that just touches the critical liquid-vapor region, explained in the next question. The condition for the critical isotherm is \left(\frac{dP}{dV}\right)_c=0 and \left(\frac{d^2P}{dV^2}\right)_c=0, where c denotes the critical point.

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GR8677 #47

Thermodynamics}Liquid-Vapor Equilibrium

The liquid-vapor region is where the substance can coexist as both a liquid and vapor. (A gas is just a vapor at normal temperatures.)

In this region, the liquid and vapor are in equilibrium, hence their coexistence. Equilibrium occurs when P_v=P_l and \mu_v=\mu_l, i.e., when the pressure and chemical potential of the liquid and vapor are equal to each other.

Since region B shows a constant pressure behavior, despite the volume-decrease, it is the region of liquid-vapor equilibrium.

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GR8677 #62


The work done by a gas in an isothermal expansion is related to the log of the volumes. If one forgets this, one can quickly derive it from recalling the definition of work W=\int P dV and the ideal gas law equation of state PV=nRT \Rightarrow P=nRT/V.

One has W=\int nRTdV/V = nRT ln(V_1/V_0). For 1 mole, one has n=1, which yields choice (E).

(And the condition for isothermality P_1V_1=P_0V_0=nRT_1=nRT_0 allows one to change the argument in the log.)

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