Question 1

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 correct answer is A.The change in entropy is given by the equation:ΔS = (Q/T) - (Q/T)where Q is the heat transferred and T is the temperature.In this case, the heat transferred is the heat of fusion and the heat of cooling. The heat of fusion is the heat required to melt the lead, and the heat of cooling is the heat released when the lead cools to room temperature.The heat of fusion is given by the equation:Q = mLwhere m is the mass of the lead and L is the latent heat of fusion.The heat of cooling is given by the equation:Q = mcΔTwhere c is the specific heat of the lead and ΔT is the change in temperature.Substituting these equations into the equation for the change in entropy, we get:ΔS = (mL/T) - (mcΔT/T)ΔS = (100 g * 2.45 × 10⁴ J/kg) / (600°C + 273 K) - (100 g * 1.29 J/g⋅°C * (600°C - 21°C)) / (21°C + 273 K)ΔS = -145 J/K

Question 2

A gasoline engine absorbs 2 500 J of heat at 250°C and expels 2 000 J at a temperature of 50°C. The magnitude of the change in entropy (in J/K) for the system is

Accepted Answer

The change in entropy (ΔS) can be calculated using:
ΔS = Q/T, where Q is the heat absorbed or released and T is the temperature in Kelvin.

In this case, ΔS = (2500 J - 2000 J) / 298 K = 1.34 J/K.

Rounding this to one decimal place gives an answer of 1.4 J/K, which corresponds to Choice C.

Question 3

All real engines are less efficient than the Carnot engine because

Accepted Answer

The Carnot engine is the most efficient engine possible because it operates through a reversible cycle, which means that it can be run backwards to produce work while also absorbing heat from the surroundings. No real engine operates through a completely reversible cycle, making them all less efficient than the Carnot engine.

Question 4

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 answer of Find the change in entropy (in J/K)...

Question 5

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 a free expansion, the gas expands without performing any work on the surroundings and there is no heat exchange between the gas and surroundings. As a result, the process is adiabatic and reversible, and the change in entropy is given by ΔS = nR ln (Vf/Vi), where n is the number of moles of the gas, R is the gas constant, Vf is the final volume, and Vi is the initial volume. Therefore, the correct choice is A.

Question 6

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

Accepted Answer

For a reversible adiabatic process, the change in entropy is 0 because the system is isolated, meaning no heat is transferred into or out of the system.

Question 7

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

Accepted Answer

The change in entropy of a system during a phase change with a constant temperature is given by ΔS = Q/T where Q is the heat absorbed or released during the phase change and T is the temperature at which it occurs. In this case, the substance is melting at a temperature T, so the heat absorbed is the latent heat of fusion L. Therefore, ΔS = L/T. However, we are interested in the change in entropy of a mass m of the substance, so we need to divide by the mass: ΔS = L/(mT). This matches choice D, so the answer is D.

Question 8

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 9

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

Accepted Answer

According to the Carnot efficiency formula, the efficiency (η) of a heat engine is given by η = 1 - (T_Low/T_High), where T_Low is the temperature of the low temperature reservoir and T_High is the temperature of the high temperature reservoir. As the denominator of this fraction cannot be zero or negative, the only way for the efficiency to be 1 is when T_Low is zero (in Kelvin). Therefore, the best choice is A).

Question 10

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 11

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

Accepted Answer

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

Question 12

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

Accepted Answer

For an isothermal process, the change in entropy is given by: ΔS = nR ln(V2/V1) where n is the number of moles, R is the gas constant, V1 is the initial volume, and V2 is the final volume. Substituting the given values, we get: ΔS = nR ln(V₂/V₁) which is the same as option A.

Question 13

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

Accepted Answer

The change in entropy ($\Delta S$) can be calculated using the formula $\Delta S = \frac{q}{T}$, where $q$ is the heat absorbed or released (in joules) and $T$ is the temperature in Kelvin. Given that the latent heat of fusion for ice is 333 J/g, and knowing that 1.00 kg of ice is equivalent to 1000 g, the total heat ($q$) involved in melting 1.00 kg of ice is $333 \times 1000 = 333,000$ J. To convert joules to calories, we use the conversion factor $1 \text{ cal} = 4.184 \text{ J}$, so $333,000 \text{ J} = \frac{333,000}{4.184} \approx 79,600 \text{ cal}$. The melting point of ice is 0°C, which is 273.15 K. Therefore, $\Delta S = \frac{79,600}{273.15} \approx 291 \text{ cal/K}$.

Question 14

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 15

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

Accepted Answer

A Carnot cycle consists of an isothermal expansion, an adiabatic expansion, an isothermal compression, and an adiabatic compression.

Question 16

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 answer of Find the change in entropy (in J/K)...

Question 17

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 transferred and T is the temperature at which the transfer occurs. In this case, the heat transferred is the latent heat of vaporization, which can be calculated as:

Q = mL = (10 kg)(1000 g/kg)(585 cal/g) = 5.85 x 10^6 cal

Since we want the change in entropy in units of kcal/K, we need to convert the heat transferred from calories to kcal:

Q = 5.85 x 10^6 cal = 5850 kcal

Therefore, the change in entropy is:

ΔS = Q/T = (5850 kcal)/(293 K) = 19.93 kcal/K, which is closest to 20 kcal/K (Choice C).

Question 18

Which of the following is an almost reversible process?

Accepted Answer

A slow isothermal compression of a gas is an almost reversible process because it occurs slowly and without any energy transfer as heat (adiabatic), which minimizes the generation of entropy (disorder) in the system. The other options either involve rapid, irreversible reactions (A), heat transfer with a temperature difference (B), expansion without any restriction (C), or an irreversible process due to the one-way flow of gas (E).

Question 19

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

Accepted Answer

The given ratio can be written as (QC/QH)(U1B0/U1B1cS1)(U1B1hS1/U1B0)(V_c/V_h)(P_c/P_h). Since Carnot cycle is a reversible cycle, the efficiencies of the engine are equal to (Qin - Qout)/Qin where Q is the heat absorbed/rejected and are proportional to the temperature differences of the heat reservoirs, and (U1 - U0)/Qin and (U1 - U0)/Qout where U is the internal energy of the gas. As a result, the ratio (QC/QH)(U1B0/U1B1cS1)(U1B1hS1/U1B0) is equal to the ratio (V_c/V_h)(P_c/P_h) and is dependent only on the temperature of the heat reservoirs, regardless of the nature of the working substance. Therefore, the only possible result that can be calculated from the given ratio is the ratio of the products of volumes and pressures, P_cV_c/P_hV_h.

Question 20

The change in entropy when 1 kg of ice melts at 0°C is (in J/K). (L_ice = 333 J/g.)

Accepted Answer

We know that the change in entropy (ΔS) of a system is given by the formula:

ΔS = q/T

where q is the heat transferred to or from the system, and T is the temperature at which the transfer takes place.

In this case, we have 1 kg of ice at 0°C that melts to form liquid water at 0°C. We need to find the change in entropy of this system.

We can calculate the heat required to melt the ice using the formula:

q = m * L

where m is the mass of the ice (1 kg), and L is the heat of fusion of ice (333 J/g).

So, q = 1 kg * 333 J/g = 333000 J

The temperature at which the transfer takes place is 0°C, which is equal to 273 K.

Therefore, ΔS = q/T = 333000 J / 273 K = 1220 J/K

So the best choice is C.

Quiz 22: Heat Engines, Entropy, and the Second Law of Thermodynamics

A gasoline engine absorbs 2 500 J of heat at 250°C and expels 2 000 J at a temperature of 50°C. The magnitude of the change in entropy (in J/K) for the system is

All real engines are less efficient than the Carnot engine because

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.

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

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

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

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

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

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

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

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

Since L_ice = 333 J/g, the change in entropy (in cal/K) when 1.00 kg of ice melts 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.

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

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.

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.)

Which of the following is an almost reversible process?

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

The change in entropy when 1 kg of ice melts at 0°C is (in J/K) . (L_ice = 333 J/g.)

Quiz 22: Heat Engines, Entropy, and the Second Law of Thermodynamics

A gasoline engine absorbs 2 500 J of heat at 250°C and expels 2 000 J at a temperature of 50°C. The magnitude of the change in entropy (in J/K) for the system is

All real engines are less efficient than the Carnot engine because

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.

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

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

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

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

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

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

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

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

Since Lice = 333 J/g, the change in entropy (in cal/K) when 1.00 kg of ice melts 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.

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

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.

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.)

Which of the following is an almost reversible process?

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

The change in entropy when 1 kg of ice melts at 0°C is (in J/K) . (Lice = 333 J/g.)

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.

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

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

Since L_ice = 333 J/g, the change in entropy (in cal/K) when 1.00 kg of ice melts 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.

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.

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

The change in entropy when 1 kg of ice melts at 0°C is (in J/K) . (L_ice = 333 J/g.)