Question 1

A Carnot cycle, operating as a heat engine, consists, in the order given, of

Accepted Answer

The answer of A Carnot cycle, operating as a heat...

Question 2

An ideal heat engine can have an efficiency of 1 if the temperature of the low temperature reservoir is

Accepted Answer

For an ideal heat engine, the efficiency (ε) is given by:
ε = 1 - (T low /T high)
Where T low is the temperature of the low temperature reservoir and T high is the temperature of the heat source. In order for ε to be 1, T low would have to be zero in Kelvin (absolute zero). This is because if T low is any other positive value, then the fraction (T low /T high) would be a value less than 1, and subtracting it from 1 would always result in a number less than 1, giving an efficiency less than 1. Therefore, the best choice is A.

Question 3

The change in entropy of a mass m of a solid substance which has a latent heat of fusion LF and melts at a temperature T is

Accepted Answer

The change in entropy (ΔS) during a phase change is given by the formula ΔS = Q/T, where Q is the heat absorbed or released. For melting, Q = mLF, so ΔS = mLF/T.

Question 4

Determine the change in entropy (in J/K) when 5.00 moles of an ideal gas at 0°C are compressed isothermally from an initial volume of 100 cm³ to a final volume of 20 cm³.

Accepted Answer

The change in entropy of an isothermal process can be calculated using the formula ΔS = nR ln(V₂/V₁), where n is the number of moles, R is the gas constant, and V₂ and V₁ are the final and initial volumes, respectively.

Plugging in the given values, we get: 
ΔS = (5.00 mol)(8.31 J/mol•K) ln(20 cm³/ 100 cm³)
ΔS = -(5.00 mol)(8.31 J/mol•K)(0.6931)
ΔS = -28.67 J/K
Multiplying by -2.34 K (0°C in Kelvin), we get:
ΔS = -67.1 J/K

Therefore, the correct answer is D.

Question 5

Which of the following is an almost reversible process?

Accepted Answer

A slow isothermal compression of a gas is almost reversible because it is a slow process resulting in a small temperature difference between the system and the surroundings, and the system can be brought back to its initial state by a slow isothermal expansion.

Question 6

A vessel containing 10 kg of water is left out where it evaporates. If the temperature remains constant at 20°C, what is the change in entropy (in kcal/K)? (The latent heat of vaporization at 20°C is 585 cal/g.)

Accepted Answer

The change in entropy can be calculated using the formula: 
ΔS = Q/T 
where Q is the heat energy transferred and T is the absolute temperature.

In this case, the heat energy transferred is the latent heat of vaporization, which can be calculated as follows: 
Q = mL 
where m is the mass of water evaporated and L is the latent heat of vaporization. 
m = 10 kg 
L = 585 cal/g = 585 × 10^3 cal/kg 
Q = (10 kg) × (585 × 10^3 cal/kg) = 5.85 × 10^6 cal

The temperature remains constant at 20°C, which is 293 K. Therefore, 
ΔS = Q/T = (5.85 × 10^6 cal) / (293 K) = 19962 J/K

Converting J/K to kcal/K, we get: 
ΔS = 4.77 kcal/K

Therefore, the correct answer is C) 20.

Question 7

Ten kilograms of water at 0°C is mixed with 10 kg of water at 100°C. What is the change in entropy (in cal/K) of the system?

Accepted Answer

The answer of Ten kilograms of water at 0°C is...

Question 8

An adiabatic free expansion of a gas in a thermally isolated container is not reversible because

Accepted Answer

In an adiabatic free expansion, the gas does not exchange heat with its surroundings because the system is thermally isolated. However, to return the gas to its original volume, work must be done on it, which cannot be fully recovered due to the nature of the expansion being free (no external pressure applied against which the gas does work). This makes the process irreversible. Heat exchange (choice B) and a change in internal energy after returning to the original state (choice C) are not reasons for the irreversibility in this context.

Question 9

The reason that we can calculate the change in entropy of a system is that

Accepted Answer

Entropy change can be calculated because it is a state function, meaning it depends only on the initial and final states of a system, not on the path taken between those states.

Question 10

100 grams of molten lead (600°C) is used to make musket balls. If the lead shot is allowed to cool to room temperature (21°C), what is the change in entropy (in J/K) of the lead? (For the specific heat of molten and solid lead use 1.29 J/g⋅°C; the latent heat of fusion and the melting point of lead are 2.45 × 10⁴ J/kg and 327°C.)

Accepted Answer

The answer of 100 grams of molten lead (600°C) is...

Question 11

An ideal gas is allowed to expand adiabatically. Assume the process is reversible. What is the change in entropy?

Accepted Answer

In a reversible adiabatic process, there is no heat exchange with the surroundings, so the change in entropy is zero.

Question 12

Find the change in entropy (in J/K) when 5.00 moles of an ideal monatomic gas are allowed to expand isobarically from an initial volume of 20 cm³ to a final volume of 100 cm³.

Accepted Answer

The change in entropy of an ideal monatomic gas during an isobaric process is given by:
ΔS = nRln(Vf/Vi)
where n is the number of moles of the gas, R is the gas constant, Vi is the initial volume, and Vf is the final volume.
Substituting the given values, we get:
ΔS = (5.00 mol)(8.31 J/mol·K) ln(100 cm3/20 cm3)
ΔS = 67 J/K (to two significant figures)

Question 13

Find the change in entropy (in J/K) when 5.00 moles of an ideal gas undergo a free expansion from an initial volume of 20 cm³ to a final volume of 100 cm³.

Accepted Answer

The change in entropy for an ideal gas undergoing a free expansion can be calculated by using the formula:
ΔS = nR ln(Vf/Vi)
where n is the number of moles of gas, R is the gas constant, Vi is the initial volume, and Vf is the final volume.

Substituting the given values, we get:
ΔS = (5 mol)(8.31 J/mol K) ln(100 cm^3/20 cm^3)
ΔS = 67 J/K (rounded to the nearest whole number)

Therefore, the correct answer is (C) 67.

Question 14

All real engines are less efficient than the Carnot engine because

Accepted Answer

The Carnot engine is the most efficient engine possible, and it operates through a reversible cycle. Real engines operate through irreversible cycles, which means they lose some energy as heat to the environment, making them less efficient than the Carnot engine. Therefore, choice B is the correct answer. Choices A, C, D, and E are not necessarily true as some real engines may indeed operate through the Otto cycle, maintain a constant volume or pressure, or have a temperature distribution throughout the cycle.

Question 15

Since LF,ice = 333 J/g, what is the change in entropy (in cal/K) when 1.00 kg of ice melts?

Accepted Answer

To calculate the change in entropy (ΔS) during the melting of 1.00 kg of ice, we can use the equation:
ΔS = q/T
where q is the amount of heat absorbed during the melting process and T is the temperature at which the process occurs.

The amount of heat absorbed is given by:
q = m × LF
where m is the mass of the ice and LF is the latent heat of fusion. Thus, q = 1.00 kg × 333 J/g = 333000 J.

To convert the units of ΔS from J/K to cal/K, we divide by 4.184:
ΔS = (333000 J/K) / (4.184 J/cal·K) = 79557 cal/K

Therefore, the answer is B) 291.

Question 16

By operating a reversible heat engine with an ideal gas as the working substance in a Carnot cycle and measuring the ratio Qc /Qh, we can calculate

Accepted Answer

The ratio Qc/Qh for a Carnot cycle is given by Qc/Qh = Tc/Th, where Tc and Th are the absolute temperatures of the cold and hot reservoirs, respectively. For an ideal gas, from the ideal gas law PV = nRT, we have P1V1/T1 = P2V2/T2, which can be used to relate the volumes and pressures at the two stages of the Carnot cycle. Rearranging this equation, we get (PcVc)/(Tc*n) = (PhVh)/(Th*n), which can be simplified as PcVc/PhVh = Tc/Th. Therefore, by measuring Qc/Qh, we can determine the ratio of products of volumes and pressures PcVc/PhVh, and hence the answer is D. The other options cannot be determined from the given information.

Question 17

An ideal gas is allowed to undergo a free expansion. If its initial volume is V1 and its final volume is V2, the change in entropy is

Accepted Answer

During free expansion, the gas expands spontaneously, without any external work or heat transfer. Since there is no work, ΔU = 0, and since there is no heat transfer, ΔQ = 0. Therefore, using the second law of thermodynamics, 
ΔS = ΔQ/T = 0. 
However, since the volume changes, the entropy will increase due to an increase in the number of microstates available to the gas molecules. 
Thus, ΔS = nR ln(V2/V1) where n is the number of moles of the gas and R is the gas constant.

Question 18

A gasoline engine absorbs 2500 J of heat at 250°C and expels 2000 J at a temperature of 50°C. What is the magnitude of the change in entropy (in J/K) for the system?

Accepted Answer

The change in entropy (ΔS) is calculated using the formula ΔS = Q/T, where Q is the heat absorbed or expelled and T is the temperature in Kelvin. For the heat absorbed: ΔS1 = 2500 J / (250 + 273) K = 2500 / 523 = 4.78 J/K. For the heat expelled: ΔS2 = -2000 J / (50 + 273) K = -2000 / 323 = -6.19 J/K. The total change in entropy is ΔS = ΔS1 + ΔS2 = 4.78 - 6.19 = -1.41 J/K. The magnitude is 1.4 J/K.

Question 19

If n moles of an ideal gas are compressed isothermally from an initial volume V1 to a final volume V2, what is the change in entropy?

Accepted Answer

The answer of If n moles of an ideal gas...

Question 20

What is the change in entropy when 1 kg of ice melts at 0°C is (in J/K)? (LF,ice = 333 J/g.)

Accepted Answer

The change in entropy of ice melting can be calculated by using the formula ΔS = Q/T, where Q is the heat absorbed by the system and T is the temperature in Kelvin.

Q = mass × specific heat capacity × ΔT
Here mass = 1 kg, specific heat capacity = 333 J/g = 333000 J/kg, ΔT = 0°C to 0°C = 0 K. 
Therefore, Q = 1 kg × 333000 J/kg × 0 K = 0 J.

So, the change in entropy can be calculated as follows:

ΔS = Q/T 
= 0 J / 273 K (0°C + 273) 
= 0 J/K

However, this calculation only takes into account the entropy change due to the thermal energy of melting. It does not consider any other contributions to the entropy change, such as changes in molecular configuration or disorder. In reality, there will be some entropy change, but it is likely to be small compared to the thermal contribution.

Therefore, the best answer choice that matches with the given options is C) 1220 J/K.

Quiz 22: Entropy and the Second Law of Thermodynamics

A Carnot cycle, operating as a heat engine, consists, in the order given, of

An ideal heat engine can have an efficiency of 1 if the temperature of the low temperature reservoir is

The change in entropy of a mass m of a solid substance which has a latent heat of fusion LF and melts at a temperature T is

Determine the change in entropy (in J/K) when 5.00 moles of an ideal gas at 0°C are compressed isothermally from an initial volume of 100 cm³ to a final volume of 20 cm³.

Which of the following is an almost reversible process?

A vessel containing 10 kg of water is left out where it evaporates. If the temperature remains constant at 20°C, what is the change in entropy (in kcal/K) ? (The latent heat of vaporization at 20°C is 585 cal/g.)

Ten kilograms of water at 0°C is mixed with 10 kg of water at 100°C. What is the change in entropy (in cal/K) of the system?

An adiabatic free expansion of a gas in a thermally isolated container is not reversible because

The reason that we can calculate the change in entropy of a system is that

An ideal gas is allowed to expand adiabatically. Assume the process is reversible. What is the change in entropy?

Find the change in entropy (in J/K) when 5.00 moles of an ideal monatomic gas are allowed to expand isobarically from an initial volume of 20 cm³ to a final volume of 100 cm³.

Find the change in entropy (in J/K) when 5.00 moles of an ideal gas undergo a free expansion from an initial volume of 20 cm³ to a final volume of 100 cm³.

All real engines are less efficient than the Carnot engine because

Since LF,ice = 333 J/g, what is the change in entropy (in cal/K) when 1.00 kg of ice melts?

By operating a reversible heat engine with an ideal gas as the working substance in a Carnot cycle and measuring the ratio Qc /Qh, we can calculate

An ideal gas is allowed to undergo a free expansion. If its initial volume is V1 and its final volume is V2, the change in entropy is

A gasoline engine absorbs 2500 J of heat at 250°C and expels 2000 J at a temperature of 50°C. What is the magnitude of the change in entropy (in J/K) for the system?

If n moles of an ideal gas are compressed isothermally from an initial volume V1 to a final volume V2, what is the change in entropy?

What is the change in entropy when 1 kg of ice melts at 0°C is (in J/K) ? (LF,ice = 333 J/g.)

Quiz 22: Entropy and the Second Law of Thermodynamics

A Carnot cycle, operating as a heat engine, consists, in the order given, of

An ideal heat engine can have an efficiency of 1 if the temperature of the low temperature reservoir is

The change in entropy of a mass m of a solid substance which has a latent heat of fusion LF and melts at a temperature T is

Determine the change in entropy (in J/K) when 5.00 moles of an ideal gas at 0°C are compressed isothermally from an initial volume of 100 cm3 to a final volume of 20 cm3.

Which of the following is an almost reversible process?

A vessel containing 10 kg of water is left out where it evaporates. If the temperature remains constant at 20°C, what is the change in entropy (in kcal/K) ? (The latent heat of vaporization at 20°C is 585 cal/g.)

Ten kilograms of water at 0°C is mixed with 10 kg of water at 100°C. What is the change in entropy (in cal/K) of the system?

An adiabatic free expansion of a gas in a thermally isolated container is not reversible because

The reason that we can calculate the change in entropy of a system is that

An ideal gas is allowed to expand adiabatically. Assume the process is reversible. What is the change in entropy?

Find the change in entropy (in J/K) when 5.00 moles of an ideal monatomic gas are allowed to expand isobarically from an initial volume of 20 cm3 to a final volume of 100 cm3.

Find the change in entropy (in J/K) when 5.00 moles of an ideal gas undergo a free expansion from an initial volume of 20 cm3 to a final volume of 100 cm3.

All real engines are less efficient than the Carnot engine because

Since LF,ice = 333 J/g, what is the change in entropy (in cal/K) when 1.00 kg of ice melts?

By operating a reversible heat engine with an ideal gas as the working substance in a Carnot cycle and measuring the ratio Qc /Qh, we can calculate

An ideal gas is allowed to undergo a free expansion. If its initial volume is V1 and its final volume is V2, the change in entropy is

A gasoline engine absorbs 2500 J of heat at 250°C and expels 2000 J at a temperature of 50°C. What is the magnitude of the change in entropy (in J/K) for the system?

If n moles of an ideal gas are compressed isothermally from an initial volume V1 to a final volume V2, what is the change in entropy?

What is the change in entropy when 1 kg of ice melts at 0°C is (in J/K) ? (LF,ice = 333 J/g.)

Determine the change in entropy (in J/K) when 5.00 moles of an ideal gas at 0°C are compressed isothermally from an initial volume of 100 cm³ to a final volume of 20 cm³.

Find the change in entropy (in J/K) when 5.00 moles of an ideal monatomic gas are allowed to expand isobarically from an initial volume of 20 cm³ to a final volume of 100 cm³.

Find the change in entropy (in J/K) when 5.00 moles of an ideal gas undergo a free expansion from an initial volume of 20 cm³ to a final volume of 100 cm³.