Applications of Thermodynamics Practice Questions
AP (Advanced Placement) · AP Chemistry · 149 free MCQs with instant results and detailed explanations.
149
Total
50
Easy
77
Medium
22
Hard
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Sample Questions from Applications of Thermodynamics
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Q1
Easy
What is the effect of increasing temperature on the entropy of a substance?
A.
It increases the entropy.
B.
It decreases the entropy.
C.
It has no effect on the entropy.
D.
It first increases, then decreases the entropy.
Show Answer & Explanation
Correct Answer: A
Increasing temperature generally increases the entropy of a substance because higher temperatures provide more energy to the particles, leading to greater disorder and randomness in the system.
Q2
Easy
If the enthalpy change (ฮH) of a reaction is -150 kJ and the reaction is carried out at constant pressure, what can be said about the reaction?
A.
It is exothermic.
B.
It is endothermic.
C.
It is at equilibrium.
D.
It requires heat input.
Show Answer & Explanation
Correct Answer: A
A negative enthalpy change (ฮH < 0) indicates that the reaction releases heat to the surroundings, which characterizes it as exothermic.
Q3
Easy
Which of the following processes is an example of an isothermal process?
A.
Melting of ice at 0ยฐC.
B.
Heating water from 20ยฐC to 100ยฐC.
C.
Combustion of gasoline in an engine.
D.
Expansion of a gas in a piston without heat exchange.
Show Answer & Explanation
Correct Answer: A
An isothermal process occurs at a constant temperature. The melting of ice at 0ยฐC maintains a constant temperature while the phase change occurs, hence it is isothermal.
Q4
Medium
The standard enthalpy change for the reaction A + B โ C is -150 kJ/mol. If the reaction is reversed to C โ A + B, what is the standard enthalpy change for the reversed reaction?
A.
150 kJ/mol
B.
-150 kJ/mol
C.
0 kJ/mol
D.
300 kJ/mol
Show Answer & Explanation
Correct Answer: A
When a reaction is reversed, the sign of the standard enthalpy change also reverses. Therefore, the enthalpy change for the reaction C โ A + B is +150 kJ/mol.
Q5
Medium
A system absorbs 200 J of heat and does 50 J of work on the surroundings. What is the change in internal energy (ฮU) of the system?
A.
150 J
B.
250 J
C.
200 J
D.
50 J
Show Answer & Explanation
Correct Answer: A
According to the first law of thermodynamics, ฮU = Q - W. Here, Q = 200 J and W = 50 J. Therefore, ฮU = 200 J - 50 J = 150 J.
Q6
Medium
In an isothermal process for an ideal gas, which of the following is true about the change in internal energy (ฮU)?
A.
ฮU is equal to zero
B.
ฮU is greater than zero
C.
ฮU is less than zero
D.
ฮU depends on the volume change
Show Answer & Explanation
Correct Answer: A
In an isothermal process, the temperature remains constant. For an ideal gas, the internal energy is a function of temperature, hence ฮU = 0.
Q7
Medium
During a phase change at constant temperature and pressure, which of the following is true regarding the enthalpy change?
A.
The enthalpy change is zero.
B.
The enthalpy change is positive if heat is absorbed.
C.
The enthalpy change is negative if heat is absorbed.
D.
The enthalpy change equals the work done on the system.
Show Answer & Explanation
Correct Answer: B
In a phase change at constant temperature and pressure, if heat is absorbed (like during melting), the enthalpy change is positive, indicating a gain in energy.
Q8
Hard
A 1.00 mol sample of an ideal gas expands isothermally at 300 K from a volume of 5.00 L to 15.00 L. Calculate the change in entropy (ฮS) for the gas during this process. (R = 8.314 J/(molยทK))
A.
24.9 J/K
B.
16.9 J/K
C.
8.31 J/K
D.
30.2 J/K
Show Answer & Explanation
Correct Answer: A
The change in entropy for an isothermal process can be calculated using ฮS = nR ln(Vf/Vi). Here, ฮS = 1.00 mol * 8.314 J/(molยทK) * ln(15.00 L / 5.00 L) = 24.9 J/K.
Q9
Hard
The enthalpy change (ฮH) for the reaction 2H2(g) + O2(g) โ 2H2O(g) is -483.6 kJ. If the reaction occurs at constant pressure, what is the maximum work (w) done by the system if it produces 1 mole of H2O at 298 K? Assume that all heat released is converted into work.
A.
-241.8 kJ
B.
-483.6 kJ
C.
-120.9 kJ
D.
-362.7 kJ
Show Answer & Explanation
Correct Answer: A
The maximum work done by the system at constant pressure can be calculated as the change in enthalpy divided by the number of moles produced. Therefore, w = ฮH / 2 = -483.6 kJ / 2 = -241.8 kJ.
Q10
Hard
A system undergoes a process where 300 J of heat is absorbed while doing 150 J of work on the surroundings. What is the change in internal energy of the system?
A.
450 J
B.
150 J
C.
300 J
D.
0 J
Show Answer & Explanation
Correct Answer: B
According to the first law of thermodynamics, ฮU = Q - W. Here, Q = 300 J (heat absorbed) and W = 150 J (work done on the surroundings). Therefore, ฮU = 300 J - 150 J = 150 J.
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Applications of Thermodynamics โ AP (Advanced Placement) AP Chemistry Practice Questions Online
This page contains 149 practice MCQs for the chapter Applications of Thermodynamics in AP (Advanced Placement) AP Chemistry. The questions are organized by difficulty โ 50 easy, 77 medium, 22 hard โ so you can choose the right level for your preparation.
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