Thermodynamics Exam For 12th Grade! Quiz
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Questions and Answers
- 1.
A well stoppered Thermo flask containing some ice cubes is an example of
- A.
Closed system
- B.
Open system
- C.
Isolated system
- D.
Non thermodynamic system
Correct Answer
C. Isolated systemExplanation
A well stoppered Thermo flask containing some ice cubes is an example of an isolated system because it is completely sealed off from its surroundings, preventing the exchange of matter and energy with the external environment. The ice cubes inside the flask will not melt or transfer heat to the surroundings, and no external factors can affect the system. Therefore, it can be considered an isolated system where no interactions occur with the surroundings.Rate this question:
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- 2.
Identify the intensive quantity from the following.
- A.
Enthalpy and temperature
- B.
Volume and temperature
- C.
Enthalpy and volume
- D.
Temperature and refractive index
Correct Answer
D. Temperature and refractive indexExplanation
The intensive quantity in this case refers to a property that does not depend on the size or amount of the substance. Enthalpy and volume are extensive quantities because they depend on the amount of substance present. Temperature and refractive index, on the other hand, are intensive quantities because they remain the same regardless of the amount of substance.Rate this question:
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- 3.
For an adiabatic process, which of the following is correct?
- A.
Q = 0
- B.
P∆V = 0
- C.
∆E = 0
- D.
Q = + W
Correct Answer
A. Q = 0Explanation
In an adiabatic process, there is no heat exchange with the surroundings, meaning that q (heat) is equal to 0. This is because adiabatic processes are characterized by the absence of heat transfer. Therefore, the correct statement is "q = 0".Rate this question:
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- 4.
In which of the following, work behaves as a state function?
- A.
Isothermal
- B.
Adiabatic
- C.
Isobaric
- D.
Isochoric
Correct Answer
B. AdiabaticExplanation
Work behaves as a state function in an adiabatic process. In an adiabatic process, there is no heat exchange with the surroundings, meaning that the change in internal energy is solely determined by the work done on or by the system. This allows the work to be independent of the path taken and only depend on the initial and final states of the system. Therefore, work is a state function in an adiabatic process.Rate this question:
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- 5.
When a gas is compressed adiabatically and reversibly the final temperature is:
- A.
Higher than the initial temperature
- B.
Lower than the initial temperature
- C.
The same as initial temperature
- D.
Dependent upon the rate of compression
Correct Answer
A. Higher than the initial temperatureExplanation
When a gas is compressed adiabatically and reversibly, the compression process is done without any heat exchange with the surroundings and is also done in a way that the system can be returned to its original state. During adiabatic compression, the gas molecules are forced closer together, increasing their kinetic energy and hence the temperature of the gas. Therefore, the final temperature of the gas will be higher than the initial temperature.Rate this question:
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- 6.
Q = -w is not true for
- A.
Isothermal
- B.
Adiabatic
- C.
Cyclic process
- D.
Both 1 & 3
Correct Answer
B. AdiabaticExplanation
The equation q = -w represents the relationship between heat transfer (q) and work done (w) in a thermodynamic process. In an adiabatic process, there is no heat transfer between the system and its surroundings, meaning q = 0. Therefore, the equation q = -w is not true for an adiabatic process.Rate this question:
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- 7.
The temperature of an ideal gas increases in an:
- A.
Adiabatic expansion
- B.
Isothermal expansion
- C.
Adiabatic compression
- D.
Isothermal compression
Correct Answer
C. Adiabatic compressionExplanation
In an adiabatic compression, the temperature of an ideal gas increases because no heat is exchanged with the surroundings. As the gas is compressed, the work done on the gas increases its internal energy, causing the temperature to rise. This is in contrast to an isothermal compression, where the temperature remains constant as heat is allowed to flow out of the gas to maintain a constant temperature.Rate this question:
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- 8.
Which one is a state function:-
- A.
Heat supplied at constant pressure
- B.
Heat supplied at constant volume
- C.
Enthalpy
- D.
All of the above
Correct Answer
D. All of the aboveExplanation
All of the given options are examples of state functions. A state function is a property that depends only on the current state of the system and not on the path taken to reach that state. Heat supplied at constant pressure and heat supplied at constant volume are both examples of energy transfer, which is a state function. Enthalpy is also a state function that represents the total heat content of a system at constant pressure. Therefore, all of the options listed are state functions.Rate this question:
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- 9.
The work is done by a weightless piston in causing an expansion ΔV (at constant temperature), when the opposing pressure P is variable. What will be the final expression of the work?
- A.
W = - ∫ P∆V
- B.
W= - PΔV
- C.
W = 0
- D.
None of these
Correct Answer
A. W = - ∫ P∆VExplanation
The final expression of the work done by a weightless piston in causing an expansion with a variable opposing pressure is given by W = - ∫ P∆V. This expression represents the negative integral of the product of the opposing pressure P and the change in volume ∆V. The negative sign indicates that the work is done on the system, as the piston is causing the expansion against the opposing pressure.Rate this question:
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- 10.
The work done by 100 calorie of heat in isothermal expansion of ideal gas is:
- A.
418.4 J
- B.
4.184 J
- C.
41.84 J
- D.
None
Correct Answer
A. 418.4 JExplanation
In an isothermal expansion of an ideal gas, the temperature remains constant. The work done by the gas can be calculated using the formula W = nRT ln(V2/V1), where n is the number of moles of gas, R is the gas constant, T is the temperature in Kelvin, V2 is the final volume, and V1 is the initial volume. Since the temperature is constant, the change in volume is the only factor affecting the work done. The work done is directly proportional to the change in volume. Therefore, if the volume increases, the work done by the gas will also increase. Hence, the correct answer is 418.4 J.Rate this question:
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- 11.
Temperature and heat are not:
- A.
Extensive properties
- B.
Intensive properties
- C.
Intensive and extensive properties respectively
- D.
Extensive and intensive properties respectively
Correct Answer
D. Extensive and intensive properties respectivelyExplanation
Temperature and heat are not extensive properties and intensive properties respectively. Extensive properties are dependent on the amount or size of the system, while intensive properties are independent of the system's size or amount. Temperature and heat are intensive properties because they do not depend on the size or amount of the system. Temperature is a measure of the average kinetic energy of the particles in a system, while heat is the transfer of energy between two systems due to a temperature difference.Rate this question:
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- 12.
One mole of gas absorbs 200 J of heat at constant volume. Its temperature rises from 298 K to 308 K. The change in internal energy is:
- A.
-200 J
- B.
400 J
- C.
200 J
- D.
- 400 j
Correct Answer
C. 200 JExplanation
When heat is absorbed by a gas at constant volume, the change in internal energy is equal to the amount of heat absorbed. In this case, the gas absorbs 200 J of heat, so the change in internal energy is also 200 J.Rate this question:
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- 13.
Which statement is true for the reversible process:-
- A.
It takes place in single step
- B.
Driving force is much greater than opposing force
- C.
Work obtain is minimum
- D.
None
Correct Answer
D. NoneExplanation
None of the statements provided are true for a reversible process. A reversible process is a hypothetical ideal process that can be reversed without leaving any trace on the surroundings. It is a continuous and gradual process, not a single step. The driving force and opposing force are equal in a reversible process, ensuring equilibrium. The work obtained in a reversible process is maximum, not minimum. Therefore, the correct answer is "None."Rate this question:
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- 14.
Both q & w are __________ function & q + w is a ___________ function.
- A.
State, State
- B.
State, path
- C.
Path, state
- D.
Path, path
Correct Answer
C. Path, stateExplanation
The given question is asking about the nature of the functions q and w, as well as their sum q + w. From the answer choices, we can see that q and w are described as either "State" or "Path" functions. The correct answer states that q and w are "Path" functions, while their sum q + w is a "State" function. This implies that q and w individually represent paths, while their sum represents a state.Rate this question:
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- 15.
The work done by a system is 8J when 40 J heat is supplied to it. Calculate the increases in internal energy of the system.
- A.
32 J
- B.
42 J
- C.
48 J
- D.
- 32 J
Correct Answer
A. 32 JExplanation
When heat is supplied to a system, it can either be used to do work or increase the internal energy of the system. In this case, the work done by the system is given as 8J. Since work is a form of energy transfer, it means that 8J of the supplied heat is used to do work. Therefore, the remaining heat is used to increase the internal energy of the system, which is calculated by subtracting the work done (8J) from the total heat supplied (40J). Thus, the increase in internal energy of the system is 32J.Rate this question:
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- 16.
Under which of the following conditions is the relation ΔH=ΔE+PΔV valid for a closed system?
- A.
Constant pressure
- B.
Constant temperature
- C.
Constant temperature and pressure
- D.
Constant temperature pressure and composition
Correct Answer
A. Constant pressureExplanation
The relation ΔH=ΔE+PΔV is valid for a closed system under constant pressure. This is because in a closed system, the only work done is due to pressure-volume work, which is given by PΔV. Therefore, the change in enthalpy (ΔH) is equal to the change in internal energy (ΔE) plus the work done on or by the system due to pressure-volume work (PΔV).Rate this question:
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- 17.
ΔvapH for water at 100°C is 40.66 kJ mol-1 The internal energy of vaporization of water at 100°C (in KJ)
- A.
37.53
- B.
39.08
- C.
42.19
- D.
43.73
Correct Answer
A. 37.53Explanation
The internal energy of vaporization of water at 100°C is 37.53 kJ.Rate this question:
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- 18.
The difference between heats of reaction at constant pressure and constant volume for the reaction, 2C6H6 (l)+15 O2 (g) → 12CO2(g)+ 6H2O(l) at 25°C in kJ is:
- A.
7.43
- B.
-7.43
- C.
3.72
- D.
-3.72
Correct Answer
B. -7.43Explanation
The negative sign indicates that the heat of reaction at constant volume is lower than the heat of reaction at constant pressure. This means that more heat is released when the reaction is carried out at constant volume compared to constant pressure. The value of -7.43 kJ represents the difference in heat released between the two conditions.Rate this question:
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- 19.
For a gaseous reaction, A(g) + 3B(g) → 3C(g) + 3D(g) ∆E is 17 kCal at 27°C assuming R = 2 Cal /Kmol, the value of ∆H for the above reaction is:
- A.
15.8 Kcal
- B.
18.2 Kcal
- C.
20.0 Kcal
- D.
16.4 Kcal
Correct Answer
B. 18.2 KcalExplanation
The value of ∆H for a reaction can be calculated using the equation ∆H = ∆E + RT, where ∆E is the change in internal energy, R is the gas constant, and T is the temperature in Kelvin. In this case, ∆E is given as 17 kCal, R is given as 2 Cal/Kmol, and the temperature is 27°C which is 300 K. Plugging these values into the equation, we get ∆H = 17 + (2 * 300) = 17 + 600 = 617 kCal. Converting this to Kcal gives us 18.2 Kcal, which is the correct answer.Rate this question:
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- 20.
For CaCO3(s)→CaO(s)+CO2(g ) at 977°C, ΔH=174 kJ/mol; then ΔE is:
- A.
160 KJ
- B.
163.6 KJ
- C.
186.4 KJ
- D.
180 KJ
Correct Answer
B. 163.6 KJExplanation
The value of ΔE can be calculated using the equation ΔE = ΔH - PΔV, where ΔH is the enthalpy change, P is the pressure, and ΔV is the change in volume. In this case, the reaction is taking place at a constant temperature and pressure, so ΔV is assumed to be zero. Therefore, ΔE = ΔH - PΔV = ΔH. Given that ΔH = 174 kJ/mol, the value of ΔE is also 174 kJ/mol. The answer of 163.6 KJ is incorrect.Rate this question:
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- 21.
The difference in ΔH and ΔU for combustion of methane for the combustion of methane at 250C would be:
- A.
Zero
- B.
2 x 298 x -2 cals
- C.
2 x 298 x -3 cals
- D.
2 x 25 x -3 cals
Correct Answer
B. 2 x 298 x -2 calsExplanation
The difference in ΔH and ΔU for the combustion of methane at 250C would be 2 x 298 x -2 cals. This is because ΔH represents the change in enthalpy, which is the heat released or absorbed during a reaction at constant pressure. ΔU represents the change in internal energy, which is the heat released or absorbed during a reaction at constant volume. In this case, since the combustion of methane is occurring at constant pressure, the difference between ΔH and ΔU would be equal to the change in enthalpy, which is 2 x 298 x -2 cals.Rate this question:
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- 22.
For the system S(s) + O2(g) → SO2(g)
- A.
ΔH = ΔE
- B.
ΔH > ΔE
- C.
ΔE > ΔH
- D.
ΔH = 0
Correct Answer
A. ΔH = ΔEExplanation
The given correct answer is ΔH = ΔE. This means that the enthalpy change (ΔH) is equal to the change in internal energy (ΔE) for the given reaction. Enthalpy (ΔH) represents the heat energy transferred during a chemical reaction, while internal energy (ΔE) represents the overall change in energy of the system. In this case, since the reaction involves the formation of SO2 gas from S and O2, the enthalpy change (ΔH) would be equal to the change in internal energy (ΔE) of the system.Rate this question:
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- 23.
ΔS for the reaction : MgCO3(s) → MgO(s) + CO2(g) will be:
- A.
0
- B.
-ve
- C.
+ve
- D.
Infinity
Correct Answer
C. +veExplanation
The positive value of ΔS for the reaction indicates an increase in the entropy or disorder of the system. In this reaction, a solid (MgCO3) is converted into a gas (CO2) and a solid (MgO). The formation of gas molecules increases the randomness and disorder of the system, leading to a positive ΔS value.Rate this question:
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- 24.
In which of the following cases, entropy will decrease:
- A.
Solid changing to liquid
- B.
Expansion of gas
- C.
Crystal dissolve
- D.
Polymerization
Correct Answer
D. PolymerizationExplanation
Polymerization is a chemical reaction in which small molecules, called monomers, combine to form a larger molecule, called a polymer. This process leads to a decrease in entropy because the number of possible arrangements of molecules decreases as they combine into a larger, more ordered structure. In contrast, the other options (solid changing to liquid, expansion of gas, and crystal dissolve) involve an increase in entropy as the molecules become more disordered or spread out. Therefore, polymerization is the only case in which entropy will decrease.Rate this question:
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- 25.
For the reaction Ag2O(s) → 2Ag(s) + 1/2 O2(g), the value of ΔH=30.56 KJ/mol and ΔS=66 J/Kmol. The temperature at which the free energy change for the reaction will be zero is:
- A.
373 K
- B.
413 K
- C.
463 K
- D.
493 K
Correct Answer
C. 463 KExplanation
At a given temperature, the free energy change for a reaction can be calculated using the equation ΔG = ΔH - TΔS, where ΔG is the free energy change, ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature in Kelvin. In order for the free energy change to be zero, ΔG = 0. Rearranging the equation, we get T = ΔH/ΔS. Plugging in the values given in the question, we find T = 30.56 KJ/mol / (66 J/Kmol) = 463 K. Therefore, the temperature at which the free energy change for the reaction will be zero is 463 K.Rate this question:
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- 26.
Given enthalpy of formation of CO2(g) and CaO(s) are -94.0 KJ and -152 KJ respectively and the enthalpy of the reaction: CaCO3(s) → CaO(s) + CO2(g) is 42 KJ. The enthalpy of formation of CaCO3(s) is
- A.
-42 kJ
- B.
-202 kJ
- C.
+202 kJ
- D.
-288 kJ
Correct Answer
D. -288 kJExplanation
The enthalpy of formation of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states. In this case, we are given the enthalpy of formation for CO2(g) and CaO(s), and the enthalpy change for the reaction CaCO3(s) → CaO(s) + CO2(g). To find the enthalpy of formation of CaCO3(s), we can use the equation:
ΔH = ΣΔHf(products) - ΣΔHf(reactants)
where ΔH is the enthalpy change for the reaction and ΔHf is the enthalpy of formation. Plugging in the given values:
42 kJ = (-152 kJ) + (-94 kJ) - ΔHf
Rearranging the equation, we find that the enthalpy of formation of CaCO3(s) is -288 kJ.Rate this question:
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- 27.
A gas is allowed to expand under reversible adiabatic conditions. What will be zero in this case?
- A.
ΔG=0
- B.
ΔT=0
- C.
ΔS=0
- D.
All of the above
Correct Answer
C. ΔS=0Explanation
When a gas expands under reversible adiabatic conditions, there is no heat exchange with the surroundings, so the change in temperature, ΔT, will be zero. Additionally, since the process is adiabatic, there is no change in entropy, ΔS, because entropy is related to heat transfer. Therefore, the correct answer is ΔS=0.Rate this question:
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- 28.
Two moles of an ideal gas expand spontaneously into vacuum. The work done is
- A.
0
- B.
2J
- C.
8J
- D.
4J
Correct Answer
A. 0Explanation
When an ideal gas expands into a vacuum, it means that there is no external pressure acting on the gas. In this case, no work is done because work is defined as the product of force and displacement, and there is no force acting on the gas. Therefore, the correct answer is 0.Rate this question:
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- 29.
Heat of reaction for CO(g) + 1/2 O2(g) → CO2(g) at constant Volume is -67.71 Kcal at 170C. The heat of reaction at constant P at 170C is:
- A.
-68 K cal
- B.
+68 K cal
- C.
-67.42 K cal
- D.
None
Correct Answer
A. -68 K calExplanation
The heat of reaction at constant volume and constant pressure can be different because they involve different conditions. In this case, the heat of reaction at constant volume is given as -67.71 Kcal. Since the question asks for the heat of reaction at constant pressure, it is reasonable to assume that it would be different from the given value. Therefore, the correct answer is -68 Kcal, which is the closest option to the given value.Rate this question:
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- 30.
The heat of reaction for: A + 1/2O2 → AO is -50 K cal and AO + 1/2 O2 → AO2 is 100 K cal. The heat of reaction for A + O2 → AO2 is:
- A.
-50 Kcal
- B.
+50 Kcal
- C.
100 Kcal
- D.
150 Kcal
Correct Answer
B. +50 KcalExplanation
The heat of reaction for A + O2 → AO2 is +50 Kcal. This can be determined by adding the heats of the two given reactions: -50 Kcal (A + 1/2O2 → AO) and 100 Kcal (AO + 1/2 O2 → AO2). When these two reactions are combined, the heat of reaction for A + O2 → AO2 is the sum of the heats of the individual reactions, which is +50 Kcal.Rate this question:
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