101. Work in thermodynamics is best described as energy transfer that occurs through
ⓐ. organized force-displacement effects
ⓑ. random molecular naming only
ⓒ. colour change without boundary interaction
ⓓ. the amount of substance measured in \(\text{mol}\)
Correct Answer: organized force-displacement effects
Explanation: Work is a mode of energy transfer between a system and its surroundings. In thermodynamics, it commonly appears through organized mechanical effects such as expansion or compression of a gas. For example, a gas pushing a piston outward transfers energy as work. Work is not a property stored inside the system like internal energy \(U\). It also cannot be identified from colour change alone. The important feature is energy crossing the boundary through a mechanical interaction.
102. A gas pushes a movable piston outward during expansion. For the gas chosen as the system, the work sign in the chemistry convention is
ⓐ. \(w \gt 0\)
ⓑ. \(w \lt 0\)
ⓒ. \(w=0\) always
ⓓ. \(w\) has no sign
Correct Answer: \(w \lt 0\)
Explanation: When a gas expands, it pushes the surroundings by moving the piston outward. The system is doing work on the surroundings. In the chemistry sign convention, work done by the system is negative, so \(w \lt 0\). The sign is assigned from the viewpoint of the system, not from the viewpoint of the surroundings. Work is not automatically zero just because the gas remains inside the cylinder. Expansion against an opposing force is the common situation where negative work for the system appears.
103. A piston compresses a gas slowly into a smaller volume. For the gas as the system, this process has
ⓐ. \(w \lt 0\), because work is done by the system
ⓑ. \(w=0\), because the piston is outside the gas
ⓒ. \(w \gt 0\), because work is done on the system
ⓓ. \(w\) equal to temperature in \(\text{K}\)
Correct Answer: \(w \gt 0\), because work is done on the system
Explanation: During compression, the surroundings push the piston inward and do work on the gas. The gas is the system, so energy enters the system through work. In the chemistry convention, work done on the system is positive, so \(w \gt 0\). This does not require heat to enter at the same time; work and heat are separate modes of energy transfer. The piston may be outside the gas, but it acts through the boundary and changes the gas volume. Compression is therefore associated with positive work for the gas system.
104. The statement that separates work from a thermodynamic state property is
ⓐ. work is stored in a system exactly like \(U\)
ⓑ. work is always equal to pressure only
ⓒ. work has the unit \(\text{K}\)
ⓓ. work is path-dependent energy transfer
Correct Answer: work is path-dependent energy transfer
Explanation: Work \(w\) is a path function because its value depends on how the process occurs. A gas may reach the same final state through different paths and involve different work values. Internal energy \(U\), in contrast, is a state function whose change depends only on the initial and final states. Work is measured in energy units such as \(\text{J}\), not in \(\text{K}\). It is not stored inside the system as a property at a given state. This distinction prevents treating \(w\) as something a system “contains.”
105. A system expands in Case 1 and is compressed in Case 2. The chemistry sign convention for work is best represented by
| Case | Process | Sign of \(w\) for the system |
|---|
| P | Expansion | \(w \lt 0\) |
| Q | Compression | \(w \gt 0\) |
ⓐ. P only
ⓑ. Q only
ⓒ. both P and Q
ⓓ. neither P nor Q
Correct Answer: both P and Q
Explanation: Expansion means the system pushes the surroundings, so work is done by the system. In the chemistry convention, work done by the system is negative, so \(w \lt 0\). Compression means the surroundings push the system, so work is done on the system. Work done on the system is positive, so \(w \gt 0\). Both rows therefore follow the same system-based sign convention. The direction of energy transfer must be read relative to the chosen system.
106. A rigid sealed vessel contains a gas. The gas is heated, but the wall does not move. The work associated with volume change is
ⓐ. positive because heat is supplied
ⓑ. negative because temperature rises
ⓒ. zero because volume does not change
ⓓ. equal to the pressure of the gas
Correct Answer: zero because volume does not change
Explanation: Expansion-compression work requires a change in volume through boundary movement. In a rigid vessel, the volume of the gas does not change because the wall does not move. Therefore, the \(pV\)-type work is zero for this constant-volume situation. Heat may still enter the gas if the wall conducts heat, but heat transfer is not the same as work. A rise in temperature does not by itself prove that work was done. The rigid boundary blocks volume work even when energy transfer as heat is possible.
107. A gas expands by pushing a piston upward while lifting a small load. The most suitable energy-transfer description is that
ⓐ. the gas does work on the surroundings
ⓑ. the gas receives work from the load
ⓒ. no energy transfer occurs because the gas remains inside the cylinder
ⓓ. the work must be positive for the gas
Correct Answer: the gas does work on the surroundings
Explanation: The gas pushes the piston and lifts the load, so the system transfers energy to the surroundings through mechanical motion. This is work done by the gas. In the chemistry convention, this work has negative sign for the gas system. The gas remaining inside the cylinder does not prevent work transfer, because work crosses the boundary through piston movement. The load gaining height is evidence that energy has been transferred mechanically. The sign would be positive only if the surroundings were doing work on the gas.
108. A claim says, “If work is done in a process, the system must also exchange matter.” The best evaluation is that the claim is
ⓐ. valid because work is a kind of matter flow
ⓑ. valid only for a rigid vessel
ⓒ. not valid because work is measured only in \(\text{mol}\)
ⓓ. not valid; work can occur without matter transfer
Correct Answer: not valid; work can occur without matter transfer
Explanation: Work is energy transfer, not matter transfer. A closed system can exchange energy as work while preventing matter from crossing the boundary. A gas in a cylinder with a movable piston is a common example: the gas amount remains fixed, but expansion or compression work can occur. A rigid vessel prevents volume work, but that is due to the fixed boundary, not because work requires matter flow. Work is measured in energy units such as \(\text{J}\). Matter exchange and work exchange are separate boundary interactions.
109. Read the situation and answer the question.
A gas is enclosed by a movable piston. In Step 1, the piston is pushed inward by an external weight. In Step 2, the gas pushes the piston outward after the weight is reduced.
For the gas system, the sign sequence of work in Step 1 and Step 2 is
ⓐ. \(w \lt 0\), then \(w \gt 0\)
ⓑ. \(w \gt 0\), then \(w \lt 0\)
ⓒ. \(w=0\), then \(w=0\)
ⓓ. \(w \gt 0\), then \(w \gt 0\)
Correct Answer: \(w \gt 0\), then \(w \lt 0\)
Explanation: In Step 1, the external weight pushes the piston inward, so work is done on the gas. For the gas system, work done on the system is positive, giving \(w \gt 0\). In Step 2, the gas pushes the piston outward, so the gas does work on the surroundings. Work done by the system is negative, giving \(w \lt 0\). The same piston can therefore correspond to opposite signs depending on the direction of motion. The sign is decided by energy transfer direction relative to the gas system.
110. A gas undergoes two expansions between the same initial and final volumes, but the external conditions during expansion are different. The work values may differ mainly because
ⓐ. work is an intensive property
ⓑ. work is always equal to \(0\)
ⓒ. work depends only on the final volume and never on the process
ⓓ. work is a path function
Correct Answer: work is a path function
Explanation: Work depends on the way the process is carried out. Even when the initial and final volumes are the same, different external pressures or different expansion paths can give different work values. This is why work is classified as a path function. It is not an intensive property describing the state of the system. Work is not always zero; it becomes zero for volume work only when volume change or opposing pressure conditions make it zero. The route of expansion matters for \(w\), unlike changes in state functions.
111. The first law of thermodynamics for a closed system in chemistry is commonly written as
ⓐ. \(\Delta U=q+w\)
ⓑ. \(\Delta U=q-w\) always
ⓒ. \(q=0\) for every process
ⓓ. \(w=0\) for every process
Correct Answer: \(\Delta U=q+w\)
Explanation: The first law is an energy conservation statement. In the chemistry sign convention, it is written as \(\Delta U=q+w\), where \(q\) is heat exchanged by the system and \(w\) is work exchanged by the system. Positive \(q\) means heat absorbed by the system, and positive \(w\) means work done on the system. The relation does not require heat or work to be zero in every process. It says that the change in internal energy is the net result of heat and work interactions. The signs of \(q\) and \(w\) must be chosen consistently before substitution.
112. A system absorbs \(50\,\text{J}\) of heat and has \(20\,\text{J}\) of work done on it. The change in internal energy is
ⓐ. \(+30\,\text{J}\)
ⓑ. \(+70\,\text{J}\)
ⓒ. \(-30\,\text{J}\)
ⓓ. \(-70\,\text{J}\)
Correct Answer: \(+70\,\text{J}\)
Explanation: \( \textbf{Given heat:} \) The system absorbs heat, so \(q=+50\,\text{J}\).
\( \textbf{Given work:} \) Work is done on the system, so \(w=+20\,\text{J}\).
\( \textbf{First-law relation:} \)
\[
\Delta U=q+w
\]
\( \textbf{Substitution:} \)
\[
\Delta U=+50\,\text{J}+20\,\text{J}
\]
\( \textbf{Calculation:} \)
\[
\Delta U=+70\,\text{J}
\]
\( \textbf{Sign meaning:} \) The positive value means internal energy of the system increases.
\( \textbf{Final answer:} \) \(\Delta U=+70\,\text{J}\). Both heat absorption and compression-type work add energy to the system in this case.
113. During a process, a system releases \(80\,\text{J}\) of heat and does \(30\,\text{J}\) of work on the surroundings. The value of \(\Delta U\) is
ⓐ. \(-50\,\text{J}\)
ⓑ. \(+50\,\text{J}\)
ⓒ. \(-110\,\text{J}\)
ⓓ. \(+110\,\text{J}\)
Correct Answer: \(-110\,\text{J}\)
Explanation: \( \textbf{Heat sign:} \) Heat released by the system gives \(q=-80\,\text{J}\).
\( \textbf{Work sign:} \) Work done by the system gives \(w=-30\,\text{J}\).
\( \textbf{Energy balance:} \)
\[
\Delta U=q+w
\]
\( \textbf{Substitute values:} \)
\[
\Delta U=-80\,\text{J}+(-30\,\text{J})
\]
\( \textbf{Simplify:} \)
\[
\Delta U=-110\,\text{J}
\]
\( \textbf{Interpretation:} \) Energy leaves the system both as heat and as work.
\( \textbf{Final answer:} \) \(\Delta U=-110\,\text{J}\). Adding magnitudes without signs would hide the fact that both transfers reduce the system's internal energy.
114. A process has \(q=+40\,\text{kJ}\) and \(w=-15\,\text{kJ}\). The internal energy change is
ⓐ. \(+55\,\text{kJ}\)
ⓑ. \(+25\,\text{kJ}\)
ⓒ. \(-25\,\text{kJ}\)
ⓓ. \(-55\,\text{kJ}\)
Correct Answer: \(+25\,\text{kJ}\)
Explanation: \( \textbf{Given data:} \) \(q=+40\,\text{kJ}\) and \(w=-15\,\text{kJ}\).
\( \textbf{Meaning of signs:} \) Heat enters the system, while work is done by the system.
\( \textbf{First law:} \)
\[
\Delta U=q+w
\]
\( \textbf{Substitute:} \)
\[
\Delta U=+40\,\text{kJ}+(-15\,\text{kJ})
\]
\( \textbf{Calculate:} \)
\[
\Delta U=+25\,\text{kJ}
\]
\( \textbf{Energy interpretation:} \) More energy enters as heat than leaves as work.
\( \textbf{Final answer:} \) \(\Delta U=+25\,\text{kJ}\). The negative work term reduces the energy gain but does not make the net change negative here.
115. A system has \(\Delta U=-25\,\text{kJ}\) and \(q=-10\,\text{kJ}\). The work \(w\) is
ⓐ. \(+15\,\text{kJ}\)
ⓑ. \(-35\,\text{kJ}\)
ⓒ. \(+35\,\text{kJ}\)
ⓓ. \(-15\,\text{kJ}\)
Correct Answer: \(-15\,\text{kJ}\)
Explanation: \( \textbf{Known data:} \) \(\Delta U=-25\,\text{kJ}\), \(q=-10\,\text{kJ}\).
\( \textbf{Required quantity:} \) Work \(w\).
\( \textbf{First-law relation:} \)
\[
\Delta U=q+w
\]
\( \textbf{Rearrange:} \)
\[
w=\Delta U-q
\]
\( \textbf{Substitute values:} \)
\[
w=-25\,\text{kJ}-(-10\,\text{kJ})
\]
\( \textbf{Simplify signs:} \)
\[
w=-25\,\text{kJ}+10\,\text{kJ}
\]
\[
w=-15\,\text{kJ}
\]
\( \textbf{Final answer:} \) \(w=-15\,\text{kJ}\). The negative sign means the system does work on the surroundings.
116. A system is compressed adiabatically, and \(60\,\text{J}\) of work is done on it. Its \(\Delta U\) is
ⓐ. \(-60\,\text{J}\)
ⓑ. \(0\,\text{J}\)
ⓒ. \(+120\,\text{J}\)
ⓓ. \(+60\,\text{J}\)
Correct Answer: \(+60\,\text{J}\)
Explanation: \( \textbf{Adiabatic condition:} \) No heat is exchanged, so \(q=0\).
\( \textbf{Work information:} \) Work is done on the system, so \(w=+60\,\text{J}\).
\( \textbf{Apply first law:} \)
\[
\Delta U=q+w
\]
\( \textbf{Substitute:} \)
\[
\Delta U=0+60\,\text{J}
\]
\[
\Delta U=+60\,\text{J}
\]
\( \textbf{Physical meaning:} \) Compression transfers energy into the system as work.
\( \textbf{Final answer:} \) \(\Delta U=+60\,\text{J}\). Adiabatic means \(q=0\), not \(\Delta U=0\).
117. A gas expands adiabatically and does \(90\,\text{J}\) of work on the surroundings. The internal energy change of the gas is
ⓐ. \(-90\,\text{J}\)
ⓑ. \(+90\,\text{J}\)
ⓒ. \(0\,\text{J}\)
ⓓ. \(+180\,\text{J}\)
Correct Answer: \(-90\,\text{J}\)
Explanation: \( \textbf{Adiabatic condition:} \) \(q=0\).
\( \textbf{Work direction:} \) The gas does work on the surroundings.
\( \textbf{Work sign:} \) Work done by the system gives \(w=-90\,\text{J}\).
\( \textbf{First-law relation:} \)
\[
\Delta U=q+w
\]
\( \textbf{Substitution:} \)
\[
\Delta U=0+(-90\,\text{J})
\]
\[
\Delta U=-90\,\text{J}
\]
\( \textbf{Interpretation:} \) With no heat input, the energy used to do work comes from the system's internal energy.
\( \textbf{Final answer:} \) \(\Delta U=-90\,\text{J}\). The internal energy falls because the system supplies work energy.
118. The table gives heat and work data for three processes of the same system.
| Process | \(q\) | \(w\) |
|---|
| P | \(+20\,\text{J}\) | \(+30\,\text{J}\) |
| Q | \(+20\,\text{J}\) | \(-30\,\text{J}\) |
| R | \(-20\,\text{J}\) | \(+30\,\text{J}\) |
The process with the greatest increase in internal energy is
ⓐ. Q
ⓑ. P
ⓒ. R
ⓓ. Q and R equally
Correct Answer: P
Explanation: \( \textbf{Relation used:} \) \(\Delta U=q+w\).
\( \textbf{For process P:} \)
\[
\Delta U_P=+20\,\text{J}+30\,\text{J}=+50\,\text{J}
\]
\( \textbf{For process Q:} \)
\[
\Delta U_Q=+20\,\text{J}+(-30\,\text{J})=-10\,\text{J}
\]
\( \textbf{For process R:} \)
\[
\Delta U_R=-20\,\text{J}+30\,\text{J}=+10\,\text{J}
\]
\( \textbf{Comparison:} \) \(+50\,\text{J}\) is the largest increase.
\( \textbf{Final answer:} \) Process P gives the greatest increase in internal energy. Positive heat and positive work both add energy to the system in P.
119. Assertion: If \(q\) and \(w\) are both positive for a process, \(\Delta U\) must be positive.
Reason: According to \(\Delta U=q+w\), both positive heat absorbed and positive work done on the system increase the system's internal energy.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: If \(q \gt 0\), heat enters the system. If \(w \gt 0\), work is done on the system. The first law in chemistry form is \(\Delta U=q+w\). Adding two positive quantities gives a positive \(\Delta U\). The Reason directly connects the sign convention with the energy balance. This conclusion depends on using the same system-based sign convention for both heat and work.
120. A process is reported as \(\Delta U=0\), but \(q=+35\,\text{J}\). The work term must be
ⓐ. \(+35\,\text{J}\)
ⓑ. \(0\,\text{J}\)
ⓒ. \(-35\,\text{J}\)
ⓓ. \(+70\,\text{J}\)
Correct Answer: \(-35\,\text{J}\)
Explanation: \( \textbf{Given data:} \) \(\Delta U=0\), \(q=+35\,\text{J}\).
\( \textbf{Relation:} \)
\[
\Delta U=q+w
\]
\( \textbf{Substitute the values:} \)
\[
0=+35\,\text{J}+w
\]
\( \textbf{Solve for \(w\):} \)
\[
w=-35\,\text{J}
\]
\( \textbf{Meaning of result:} \) The system absorbs \(35\,\text{J}\) as heat but gives out the same amount as work.
\( \textbf{Final answer:} \) \(w=-35\,\text{J}\). A zero \(\Delta U\) can occur even when heat and work are both nonzero.
