Question 1

A cyclical process that transfers energy as heat from a high temperature reservoir to a low temperature reservoir with no other change would violate:

Accepted Answer

The answer of A cyclical process that transfers energy as...

Question 2

Let k be the Boltzmann constant.If the configuration of molecules in a gas changes from one with a multiplicity of M₁ to one with a multiplicity of M₂, then entropy changes by:

Accepted Answer

The entropy change can be calculated using the formula ΔS = k ln(W2/W1), where W1 and W2 are the initial and final multiplicities of the system, respectively. Therefore, the correct choice is E: "ΔS = k ln(M₂/M₁)".

Question 3

The thermodynamic state of gas changes configuration from one with 3.8 *10¹⁸ microstates to one with 7.9 * 10¹⁹ microstates.The Boltzmann constant is 1.38 * 10^-23 J/K.The change in entropy is:

Accepted Answer

The change in entropy ($\Delta S$) can be calculated using the formula $\Delta S = k \ln(\frac{W_f}{W_i})$, where $k$ is the Boltzmann constant, $W_f$ is the final number of microstates, and $W_i$ is the initial number of microstates. Substituting $k = 1.38 \times 10^{-23} \, J/K$, $W_f = 7.9 \times 10^{19}$, and $W_i = 3.8 \times 10^{18}$ into the formula gives a positive change in entropy, indicating an increase in disorder. The calculation results in $\Delta S = 4.19 \times 10^{-23} \, J/K$.

Question 4

According to the second law of thermodynamics:

Accepted Answer

The second law of thermodynamics states that heat energy cannot be completely converted into work in a cyclic process, which makes option A correct. Options B, C, and D are incorrect interpretations or applications of the second law.

Question 5

A refrigerator absorbs energy of magnitude  as heat from a low temperature reservoir and rejects energy of magnitude  as heat to a high temperature reservoir.Work W is done on the working substance.The coefficient of performance is given by:

Accepted Answer

The coefficient of performance (COP) for a refrigerator is defined as the ratio of the heat removed from the low temperature reservoir (Q_L) to the work done on the system (W), which is correctly represented in option A.

Question 6

A Carnot heat engine and an irreversible heat engine both operate between the same high temperature and low temperature reservoirs.They absorb the same heat from the high temperature reservoir as heat.The irreversible engine:

Accepted Answer

According to the Second Law of Thermodynamics, the efficiency of an irreversible heat engine must be less than that of a Carnot heat engine operating between the same two temperatures. Therefore, the irreversible engine will reject more energy to the low temperature reservoir as heat. The amount of work done by the irreversible engine depends on its overall efficiency, which is always less than that of the Carnot engine. Hence, we cannot conclude that the irreversible engine will do more work. Therefore, B is the correct choice.

Question 7

A Carnot refrigerator runs between a cold reservoir at temperature T_L and a hot reservoir at temperature T_H.You want to increase its coefficient of performance.Of the following, which change results in the greatest increase in the coefficient? The value of $\Delta$T is the same for all changes.

Accepted Answer

The answer of A Carnot refrigerator runs between a cold...

Question 8

On a warm day a pool of water transfers energy to the air as heat and freezes.This is a direct violation of:

Accepted Answer

This violates the second law of thermodynamics, which states that heat always flows from a hot object to a cold object, never the reverse. Heat is transferred from the warm water to the colder air, but in this scenario, the water freezes, which requires the removal of more heat from the water. This can only happen if the air is even colder than the freezing point of water, which is unlikely on a warm day.

Question 9

An inventor claims to have a heat engine that has efficiency of 40% when it operates between a high temperature reservoir of 150°C and a low temperature reservoir of 30°C.This engine:

Accepted Answer

The second law of thermodynamics states that no engine can be more efficient than a Carnot engine operating between the same two reservoirs. The maximum efficiency of a Carnot engine operating between 150F1F1F1S1F1F1F10C and 30F1F1F1S1F1F1F10C is given by:
Efficiency = 1 - (T_low / T_high)
where T_low and T_high are the absolute temperatures of the low and high temperature reservoirs, respectively. Plugging in the values, we get:
Efficiency = 1 - (303 K / 394 K) = 0.233
Since the claimed efficiency of the engine is higher than the maximum efficiency of a Carnot engine operating between the same two reservoirs, it must violate the second law of thermodynamics.

Question 10

For a system of molecules,

Accepted Answer

The answer of For a system of molecules,
...

Question 11

Twenty-five identical molecules are in a box.Microstates are designated by identifying the molecules in the left and right halves of the box.The multiplicity of the configuration with 15 molecules in the right half and 10 molecules in the left half is:

Accepted Answer

The multiplicity of a configuration with 15 molecules in the right half and 10 in the left half of the box can be calculated using the binomial coefficient, which is given by the formula $ \binom{N}{n} = \frac{N!}{n!(N-n)!} $, where $N$ is the total number of molecules, and $n$ is the number of molecules in one half of the box. For 25 molecules with 15 in one half, the calculation is $ \binom{25}{15} $, which results in a value that corresponds to choice B when calculated correctly.

Question 12

A heat engine operates between 200 K and 100 K.In each cycle it takes 100 J from the hot reservoir, loses 25 J to the cold reservoir, and does 75 J of work.This heat engine violates:

Accepted Answer

According to the second law of thermodynamics, no heat engine can have a greater efficiency than a Carnot engine operating between the same two temperatures. The efficiency of a Carnot engine operating between 200 K and 100 K is 1 - (100/200) = 0.5 or 50%. The efficiency of this engine is (100-25)/100 = 75%, which is higher than the Carnot efficiency. Therefore, this engine violates the second law of thermodynamics. The first law of thermodynamics is not violated as the amount of energy input and output are accounted for.

Question 13

A perfectly reversible heat pump with a coefficient of performance of 14 supplies energy to a building as heat to maintain its temperature at 27°C.If the pump motor does work at the rate of 1 kW, at what rate does the pump supply energy to the building?

Accepted Answer

The coefficient of performance (COP) for a heat pump is defined as the ratio of the heat supplied to the work input. Given a COP of 14 and a work input of 1 kW, the heat supplied to the building is COP × work input = 14 × 1 kW = 14 kW. However, since the heat pump is perfectly reversible and supplies energy, the actual rate at which energy is supplied to the building is the sum of the work input and the heat supplied, which equals 1 kW (work input) + 14 kW (heat supplied) = 15 kW.

Question 14

Is it possible to transfer energy from a low-temperature reservoir to a high-temperature reservoir?

Accepted Answer

Transferring energy from a low-temperature reservoir to a high-temperature reservoir is the principle behind refrigeration and heat pumps, which requires work to be done on the system, in accordance with the second law of thermodynamics.

Question 15

For a system of molecules,

Accepted Answer

Configurations with more microstates have more ways to arrange the molecules, which means there are more ways for the system to exist in that way. Therefore, these configurations are more probable than configurations with fewer microstates.

Question 16

A heat engine in each cycle absorbs energy of magnitude  as heat from a high temperature reservoir, does work of magnitude , and then absorbs energy of magnitude  as heat from a low temperature reservoir.If  this engine violates:

Accepted Answer

The given equation violates the second law of thermodynamics, which states that the total entropy of a closed system always increases over time or remains constant in processes that are reversible. The given equation indicates that the engine is producing work without any decrease in entropy, which is not possible as it would violate the second law. Therefore, the correct answer is C.

Question 17

A reversible refrigerator operates between a low temperature reservoir at T_L and a high temperature reservoir at T_H.Its coefficient of performance is given by:

Accepted Answer

The coefficient of performance (COP) of a reversible refrigerator is given by the ratio of the temperature of the low-temperature reservoir (T_L) to the difference in temperatures between the high-temperature (T_H) and low-temperature reservoirs, which is T_L / (T_H - T_L).

Question 18

Consider the following processes:  Which are never found to occur?

Accepted Answer

The answer of Consider the following processes:  Which are...

Question 19

Let k be the Boltzmann constant.If the configuration of the molecules in a gas changes so that the multiplicity is reduced to one-third its previous value, the entropy of the gas changes by:

Accepted Answer

The entropy change, ΔS, when the multiplicity of a system changes can be calculated using the formula ΔS = k ln(Ω_final/Ω_initial). Here, if the multiplicity is reduced to one-third, Ω_final/Ω_initial = 1/3, leading to ΔS = k ln(1/3).

Question 20

Twenty-five identical molecules are in a box.Microstates are designated by identifying the molecules in the left and right halves of the box.The Boltzmann constant is 1.38 *10^-23 J/K.The entropy associated with the configuration for which 15 molecules are in the left half and 10 molecules are in the right half is:

Accepted Answer

The entropy associated with the configuration for which 15 molecules are in the left half and 10 molecules are in the right half is given by:S = k * ln(W)where k is the Boltzmann constant and W is the number of microstates.In this case, there are 25! ways to distribute the molecules in the box, and 15! ways to distribute the molecules in the left half of the box. Therefore, the number of microstates is:W = 25! / (15! * 10!) = 3003Substituting this into the equation for entropy, we get:S = 1.38 * 10^-23 J/K * ln(3003) = 2.07 * 10^-22 J/KTherefore, the correct answer is A.

Quiz 20: Entropy and the Second Law of Thermodynamics

A cyclical process that transfers energy as heat from a high temperature reservoir to a low temperature reservoir with no other change would violate:

Let k be the Boltzmann constant.If the configuration of molecules in a gas changes from one with a multiplicity of M₁ to one with a multiplicity of M₂, then entropy changes by:

The thermodynamic state of gas changes configuration from one with 3.8 10¹⁸ microstates to one with 7.9 10¹⁹ microstates.The Boltzmann constant is 1.38 * 10^-23 J/K.The change in entropy is:

According to the second law of thermodynamics:

A refrigerator absorbs energy of magnitude as heat from a low temperature reservoir and rejects energy of magnitude as heat to a high temperature reservoir.Work W is done on the working substance.The coefficient of performance is given by:

A Carnot heat engine and an irreversible heat engine both operate between the same high temperature and low temperature reservoirs.They absorb the same heat from the high temperature reservoir as heat.The irreversible engine:

A Carnot refrigerator runs between a cold reservoir at temperature T_L and a hot reservoir at temperature T_H.You want to increase its coefficient of performance.Of the following, which change results in the greatest increase in the coefficient? The value of $\Delta$ T is the same for all changes.

On a warm day a pool of water transfers energy to the air as heat and freezes.This is a direct violation of:

An inventor claims to have a heat engine that has efficiency of 40% when it operates between a high temperature reservoir of 150°C and a low temperature reservoir of 30°C.This engine:

For a system of molecules,

Twenty-five identical molecules are in a box.Microstates are designated by identifying the molecules in the left and right halves of the box.The multiplicity of the configuration with 15 molecules in the right half and 10 molecules in the left half is:

A heat engine operates between 200 K and 100 K.In each cycle it takes 100 J from the hot reservoir, loses 25 J to the cold reservoir, and does 75 J of work.This heat engine violates:

A perfectly reversible heat pump with a coefficient of performance of 14 supplies energy to a building as heat to maintain its temperature at 27°C.If the pump motor does work at the rate of 1 kW, at what rate does the pump supply energy to the building?

Is it possible to transfer energy from a low-temperature reservoir to a high-temperature reservoir?

For a system of molecules,

A heat engine in each cycle absorbs energy of magnitude as heat from a high temperature reservoir, does work of magnitude , and then absorbs energy of magnitude as heat from a low temperature reservoir.If this engine violates:

A reversible refrigerator operates between a low temperature reservoir at T_L and a high temperature reservoir at T_H.Its coefficient of performance is given by:

Consider the following processes: Which are never found to occur?

Let k be the Boltzmann constant.If the configuration of the molecules in a gas changes so that the multiplicity is reduced to one-third its previous value, the entropy of the gas changes by:

Quiz 20: Entropy and the Second Law of Thermodynamics

A cyclical process that transfers energy as heat from a high temperature reservoir to a low temperature reservoir with no other change would violate:

Let k be the Boltzmann constant.If the configuration of molecules in a gas changes from one with a multiplicity of M1 to one with a multiplicity of M2, then entropy changes by:

The thermodynamic state of gas changes configuration from one with 3.8 *1018 microstates to one with 7.9 * 1019 microstates.The Boltzmann constant is 1.38 * 10-23 J/K.The change in entropy is:

According to the second law of thermodynamics:

A refrigerator absorbs energy of magnitude as heat from a low temperature reservoir and rejects energy of magnitude as heat to a high temperature reservoir.Work W is done on the working substance.The coefficient of performance is given by:

A Carnot heat engine and an irreversible heat engine both operate between the same high temperature and low temperature reservoirs.They absorb the same heat from the high temperature reservoir as heat.The irreversible engine:

A Carnot refrigerator runs between a cold reservoir at temperature TL and a hot reservoir at temperature TH.You want to increase its coefficient of performance.Of the following, which change results in the greatest increase in the coefficient? The value of Δ\DeltaΔ T is the same for all changes.

On a warm day a pool of water transfers energy to the air as heat and freezes.This is a direct violation of:

An inventor claims to have a heat engine that has efficiency of 40% when it operates between a high temperature reservoir of 150°C and a low temperature reservoir of 30°C.This engine:

For a system of molecules,

Twenty-five identical molecules are in a box.Microstates are designated by identifying the molecules in the left and right halves of the box.The multiplicity of the configuration with 15 molecules in the right half and 10 molecules in the left half is:

A heat engine operates between 200 K and 100 K.In each cycle it takes 100 J from the hot reservoir, loses 25 J to the cold reservoir, and does 75 J of work.This heat engine violates:

A perfectly reversible heat pump with a coefficient of performance of 14 supplies energy to a building as heat to maintain its temperature at 27°C.If the pump motor does work at the rate of 1 kW, at what rate does the pump supply energy to the building?

Is it possible to transfer energy from a low-temperature reservoir to a high-temperature reservoir?

For a system of molecules,

A heat engine in each cycle absorbs energy of magnitude as heat from a high temperature reservoir, does work of magnitude , and then absorbs energy of magnitude as heat from a low temperature reservoir.If this engine violates:

A reversible refrigerator operates between a low temperature reservoir at TL and a high temperature reservoir at TH.Its coefficient of performance is given by:

Consider the following processes: Which are never found to occur?

Let k be the Boltzmann constant.If the configuration of the molecules in a gas changes so that the multiplicity is reduced to one-third its previous value, the entropy of the gas changes by:

Let k be the Boltzmann constant.If the configuration of molecules in a gas changes from one with a multiplicity of M₁ to one with a multiplicity of M₂, then entropy changes by:

The thermodynamic state of gas changes configuration from one with 3.8 10¹⁸ microstates to one with 7.9 10¹⁹ microstates.The Boltzmann constant is 1.38 * 10^-23 J/K.The change in entropy is:

A Carnot refrigerator runs between a cold reservoir at temperature T_L and a hot reservoir at temperature T_H.You want to increase its coefficient of performance.Of the following, which change results in the greatest increase in the coefficient? The value of $\Delta$ T is the same for all changes.

A reversible refrigerator operates between a low temperature reservoir at T_L and a high temperature reservoir at T_H.Its coefficient of performance is given by: