101. Heat \(Q\) in thermodynamics is best described as
ⓐ. energy stored permanently inside a body because it is hot
ⓑ. the total microscopic energy of all molecules in a system
ⓒ. the same physical quantity as temperature \(T\)
ⓓ. energy transfer due to temperature difference
Correct Answer: energy transfer due to temperature difference
Explanation: Heat \(Q\) is energy in transit, not energy stored as a property of the system. It crosses the boundary when there is a temperature difference between the system and surroundings. A hot body may have internal energy, but it does not “contain heat” in the thermodynamic sense. Temperature \(T\) describes the thermal state and is measured in \(\text{K}\), while heat is measured in \(\text{J}\). The word heat should be used for transfer during a process, not for the current energy content of a body.
102. A metal block at \(350\,\text{K}\) is placed in contact with another block at \(300\,\text{K}\). Before thermal equilibrium is reached, heat transfer is mainly from
ⓐ. the \(350\,\text{K}\) block to the \(300\,\text{K}\) block
ⓑ. the lighter block to the heavier block, regardless of temperature
ⓒ. the \(300\,\text{K}\) block to the \(350\,\text{K}\) block
ⓓ. the block with greater volume to the block with smaller volume
Correct Answer: the \(350\,\text{K}\) block to the \(300\,\text{K}\) block
Explanation: Heat transfer due to thermal contact occurs from higher temperature to lower temperature. The block at \(350\,\text{K}\) is hotter than the block at \(300\,\text{K}\), so the net transfer is from the first block to the second. Mass and volume may affect the final equilibrium temperature, but they do not decide the initial direction of heat flow. The transfer continues until the temperature difference is removed. The sign or amount of heat depends on the chosen system, but the natural direction here is fixed by temperature difference.
103. A sealed gas is taken from state P to state Q by two different processes. In one process, \(600\,\text{J}\) of heat enters the gas; in another, \(400\,\text{J}\) of heat enters it. What does this show about heat \(Q\)?
ⓐ. \(Q\) is measured in kelvin during heating.
ⓑ. \(Q=0\) for every sealed gas process.
ⓒ. \(Q\) is fixed by the final state alone.
ⓓ. \(Q\) is path-dependent energy transfer.
Correct Answer: \(Q\) is path-dependent energy transfer.
Explanation: Heat \(Q\) is not a property fixed by the initial and final states alone. It depends on the process followed between those states. Two paths can connect the same states while involving different heat transfers. This is why \(Q\) is called a path-dependent transfer quantity rather than a state variable. A sealed system can still exchange heat through its boundary if the wall allows thermal energy transfer.
104. A cup of hot water is allowed to cool in a room. If the water is chosen as the system, the sign of heat \(Q\) under the usual convention is
ⓐ. positive only if the room is colder than the water
ⓑ. zero, because the water remains a system
ⓒ. negative, because heat leaves the system
ⓓ. positive, because the water is hot
Correct Answer: negative, because heat leaves the system
Explanation: Under the usual school convention, heat supplied to a system is taken as positive. If energy leaves the system as heat, \(Q\) is negative for that system. In this case, the water is hotter than the room, so it transfers energy to the surroundings. The water being hot does not automatically mean \(Q>0\); the sign depends on the direction of energy crossing the boundary. Choosing the room as the system would reverse the sign assigned to the same heat transfer.
105. Read the situation below.
A gas is kept in a container with conducting walls. The gas and surrounding water bath are initially at different temperatures. After some time, the gas and water bath reach the same temperature.
During the approach to equilibrium, the heat transfer depends directly on
ⓐ. mass difference between gas and water bath
ⓑ. temperature difference between gas and bath
ⓒ. pressure difference between gas and water bath
ⓓ. volume difference between gas and water bath
Correct Answer: temperature difference between gas and bath
Explanation: Heat transfer takes place because there is a temperature difference between the gas and the water bath. Conducting walls allow energy to cross the boundary by thermal interaction. Equal volumes or equal masses are not required for heat transfer to occur. When the gas and bath reach the same temperature, net heat transfer stops. The example shows that heat is linked with temperature difference and boundary crossing, not with the size labels of the bodies.
106. The unit of heat \(Q\) is \(\text{J}\), the same as the unit of work \(W\), because heat is
ⓐ. proportional to volume in every process
ⓑ. a measure of temperature
ⓒ. a form of energy transfer
ⓓ. proportional to pressure in every process
Correct Answer: a form of energy transfer
Explanation: Heat \(Q\) is measured in \(\text{J}\) because it represents energy transferred across a boundary. Work \(W\) is also measured in \(\text{J}\) because it is another mode of energy transfer. Temperature \(T\) is measured in \(\text{K}\), so heat and temperature cannot be treated as the same quantity. Pressure is measured in \(\text{Pa}\), and volume is measured in \(\text{m}^3\). Sharing the unit \(\text{J}\) with work shows an energy connection, not identical physical meaning.
107. A statement says, “The heat of a gas at a state is \(500\,\text{J}\).” The main problem with this statement is that
ⓐ. heat is not a state property of a gas
ⓑ. heat is stored as absolute temperature
ⓒ. gases cannot exchange heat
ⓓ. heat cannot be measured in \(\text{J}\)
Correct Answer: heat is not a state property of a gas
Explanation: Heat \(Q\) is defined for energy transfer during a process, not as a property possessed by a system at a state. A gas has internal energy \(U\), pressure \(P\), volume \(V\), and temperature \(T\) as state-related quantities. It does not have a stored property called “heat of the gas” in the same way. Heat can be measured in \(\text{J}\), but only as energy transferred. A better statement would describe how much heat enters or leaves the gas during a specified process.
108. Consider the following statements about heat.
I. Heat transfer can occur because of temperature difference.
II. Heat is a state variable like pressure \(P\).
III. Heat supplied to a system is usually taken as positive.
ⓐ. II and III only
ⓑ. I and III only
ⓒ. I and II only
ⓓ. I, II, and III
Correct Answer: I and III only
Explanation: Statement I is true because temperature difference is the cause of heat transfer. Statement III is true under the usual thermodynamic sign convention, where heat supplied to the system is positive. Statement II is false because heat is not a state variable; it depends on the process path. Pressure \(P\) can describe a state, but heat \(Q\) describes energy crossing the boundary during a process. The distinction becomes especially important when two different paths connect the same initial and final states.
109. A system absorbs \(250\,\text{J}\) of heat and later loses \(90\,\text{J}\) of heat in a separate cooling step. Taking heat supplied to the system as positive, what are the signs and values of \(Q\) in the two steps?
ⓐ. \(-250\,\text{J}\) and \(+90\,\text{J}\)
ⓑ. \(-250\,\text{J}\) and \(-90\,\text{J}\)
ⓒ. \(+250\,\text{J}\) and \(+90\,\text{J}\)
ⓓ. \(+250\,\text{J}\) and \(-90\,\text{J}\)
Correct Answer: \(+250\,\text{J}\) and \(-90\,\text{J}\)
Explanation: \( \textbf{Sign convention:} \) Heat supplied to the system is taken as positive.
\( \textbf{First step:} \) The system absorbs \(250\,\text{J}\) of heat.
\[
Q_1=+250\,\text{J}
\]
\( \textbf{Second step:} \) The system loses \(90\,\text{J}\) of heat to the surroundings.
\[
Q_2=-90\,\text{J}
\]
\( \textbf{Reason for sign change:} \) The direction of energy crossing the system boundary is opposite in the two steps.
\( \textbf{Unit check:} \) Both values are in \(\text{J}\), since heat is energy transfer.
\( \textbf{Interpretation:} \) A hot system can still have negative \(Q\) if heat leaves it.
\( \textbf{Final answer:} \) The heat values are \(+250\,\text{J}\) and \(-90\,\text{J}\).
110. Assertion: Heat \(Q\) is not a state variable.
Reason: The heat exchanged between two given states may depend on the path followed.
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, and Reason explains Assertion
ⓒ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The Assertion is true because heat is not a property that belongs to a thermodynamic state. It is energy transferred during a process due to temperature difference. The Reason is also true because different paths between the same states can involve different heat transfers. A state variable would have a value fixed by the state alone, but \(Q\) does not behave that way. The Reason gives the path-dependence explanation for why heat is not a state variable.
111. Study the table and identify the row with the most accurate classification.
| Row | Quantity | Best description |
|---|
| P | \(Q\) | Energy transfer due to temperature difference |
| Q | \(Q\) | State variable stored inside the system |
| R | \(T\) | Energy transfer measured in \(\text{J}\) |
| S | \(U\) | Heat crossing the boundary only |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: Row P correctly describes heat \(Q\) as energy transfer due to temperature difference. Row Q is wrong because heat is not a state variable stored inside the system. Row R is wrong because temperature \(T\) is a state variable measured in \(\text{K}\), not an energy transfer measured in \(\text{J}\). Row S is wrong because internal energy \(U\) is a state property, not heat crossing a boundary. Keeping \(Q\), \(T\), and \(U\) separate prevents many thermodynamic confusions.
112. Work \(W\) in thermodynamics is most directly associated with energy transfer caused by
ⓐ. equality of temperatures between two bodies only
ⓑ. chemical composition remaining unchanged
ⓒ. microscopic random motion with no boundary effect
ⓓ. macroscopic force and boundary displacement
Correct Answer: macroscopic force and boundary displacement
Explanation: Thermodynamic work is energy transfer through macroscopic force and displacement. In gas processes, this often appears as the movement of a piston or boundary. Heat transfer is different because it is driven by temperature difference, not by boundary displacement. Work does not require the gas molecules to leave the system; the boundary interaction is enough. This definition prepares for expansion and compression work in \(P\)-\(V\) processes.
113. A gas expands slowly against a movable piston. Under the usual school convention, the work done by the gas is
ⓐ. equal to temperature in \(\text{K}\)
ⓑ. positive during expansion of the gas
ⓒ. zero because the gas remains inside the cylinder
ⓓ. negative because the gas volume increases
Correct Answer: positive during expansion of the gas
Explanation: Under the usual convention used with \(Q=\Delta U+W\), \(W\) is positive when work is done by the system. During expansion, the gas pushes the piston outward and transfers energy to the surroundings as work. The gas need not leave the cylinder for work to be done. The sign is decided by the direction of boundary displacement relative to the gas force. Expansion work by the gas is therefore positive in this convention.
114. A gas is compressed by pushing the piston inward. Under the convention that \(W\) means work done by the gas, \(W\) is
ⓐ. negative because work is done on the gas
ⓑ. positive because the external agent pushes
ⓒ. zero because pressure increases
ⓓ. equal to heat supplied in every compression
Correct Answer: negative because work is done on the gas
Explanation: If \(W\) is defined as work done by the gas, compression gives a negative value of \(W\). In compression, the piston moves inward and the surroundings do work on the gas. The gas force and boundary displacement are opposite in the work-by-gas sign convention. Pressure may increase during compression, but pressure increase alone does not decide the sign. The sign comes from whether the gas is doing work on the surroundings or receiving work from them.
115. The same gas is compressed from volume \(V_1\) to \(V_2\) by two different paths on a \(P\)-\(V\) diagram. The final state is the same in both cases, but the areas under the curves differ. What can be concluded about work?
ⓐ. Work is the same because the final volume is the same.
ⓑ. Work is path-dependent for compression.
ⓒ. Work is fixed by the final volume alone.
ⓓ. Work is a state variable like \(P\) and \(T\).
Correct Answer: Work is path-dependent for compression.
Explanation: Work in a volume-change process is linked with the area under the path on a \(P\)-\(V\) diagram. If two paths between the same volumes have different areas, the work values can differ. This shows that work is not fixed by the initial and final states alone. Compression does not automatically mean zero work; it usually gives negative work done by the gas under the standard convention. Work is a process quantity, unlike state variables such as \(P\), \(V\), and \(T\).
116. In a piston-cylinder system, the gas pressure is \(2.0\times10^5\,\text{Pa}\), and the piston moves outward so that the volume increases by \(3.0\times10^{-4}\,\text{m}^3\) at approximately constant pressure. The work done by the gas is
ⓐ. \(-60\,\text{J}\)
ⓑ. \(0.60\,\text{J}\)
ⓒ. \(60\,\text{J}\)
ⓓ. \(6.0\times10^8\,\text{J}\)
Correct Answer: \(60\,\text{J}\)
Explanation: \( \textbf{Given data:} \) \(P=2.0\times10^5\,\text{Pa}\) and \(\Delta V=3.0\times10^{-4}\,\text{m}^3\).
\( \textbf{Process clue:} \) The pressure is approximately constant.
\( \textbf{Work relation:} \)
\[
W=P\Delta V
\]
\( \textbf{Substitution:} \)
\[
W=(2.0\times10^5)(3.0\times10^{-4})\,\text{J}
\]
\( \textbf{Power calculation:} \)
\[
10^5\times10^{-4}=10
\]
\( \textbf{Number calculation:} \)
\[
(2.0)(3.0)(10)=60
\]
\( \textbf{Sign check:} \) The piston moves outward, so \(\Delta V>0\) and work done by the gas is positive.
\( \textbf{Final answer:} \) The work done by the gas is \(60\,\text{J}\).
117. A process takes a gas from state I to state II without any change in volume. What can be said about boundary work due to expansion or compression?
ⓐ. It is zero because \(\Delta V=0\).
ⓑ. It is equal to the temperature change in \(\text{K}\).
ⓒ. It is negative because the process occurs in a container.
ⓓ. It is positive because the gas has pressure.
Correct Answer: It is zero because \(\Delta V=0\).
Explanation: Boundary work in a volume-change process requires displacement of the system boundary. If the volume does not change, then \(\Delta V=0\). For pressure-volume work, no volume change means no expansion or compression work. The gas may still have pressure, but pressure alone does not create boundary work without displacement. Energy can still enter as heat in such a process, but the expansion work part is zero.
118. Use the graph description below.
A \(P\)-\(V\) graph shows two expansion paths from the same initial volume to the same final volume. Path P lies above Path Q at every intermediate volume.
For the same volume limits, the work done by the gas is
ⓐ. zero along both paths because only volume changes
ⓑ. greater along Path Q because lower pressure always means more work
ⓒ. equal along both paths because endpoints are the same
ⓓ. greater along Path P because the area under the curve is larger
Correct Answer: greater along Path P because the area under the curve is larger
Explanation: Work in a quasistatic volume change is represented by area under the \(P\)-\(V\) curve. If Path P lies above Path Q for the same volume interval, the pressure is greater along Path P at each intermediate volume. A larger pressure over the same change in volume gives a larger area. Since the paths are expansions, the work done by the gas is positive for both, but larger for Path P. The graph comparison shows path dependence of work even when the initial and final volumes match.
119. Consider the statements about work in thermodynamics.
I. Work can be energy transfer through a moving boundary.
II. Work done by a gas during expansion is usually taken as positive.
III. Work is a state variable because it is measured in \(\text{J}\).
ⓐ. I, II, and III
ⓑ. II and III only
ⓒ. I and II only
ⓓ. I and III only
Correct Answer: I and II only
Explanation: Statement I is true because thermodynamic work can occur through macroscopic boundary displacement. Statement II is true under the usual school convention where work done by the gas is positive during expansion. Statement III is false because being measured in \(\text{J}\) does not make a quantity a state variable. Heat and work are both measured in \(\text{J}\), yet both are path-dependent transfer quantities. The unit tells the energy nature of work, while path dependence tells its thermodynamic classification.
120. A gas expands from \(1.0\,\text{L}\) to \(4.0\,\text{L}\) against constant external pressure \(1.0\times10^5\,\text{Pa}\). What is the work done by the gas?
ⓐ. \(300\,\text{J}\)
ⓑ. \(30\,\text{J}\)
ⓒ. \(-300\,\text{J}\)
ⓓ. \(3.0\times10^5\,\text{J}\)
Correct Answer: \(300\,\text{J}\)
Explanation: \( \textbf{Given pressure:} \) \(P=1.0\times10^5\,\text{Pa}\).
\( \textbf{Initial volume:} \) \(V_i=1.0\,\text{L}\).
\( \textbf{Final volume:} \) \(V_f=4.0\,\text{L}\).
\( \textbf{Volume conversion:} \) \(1\,\text{L}=10^{-3}\,\text{m}^3\).
\( \textbf{Volume change:} \)
\[
\Delta V=(4.0-1.0)\times10^{-3}\,\text{m}^3
\]
\[
\Delta V=3.0\times10^{-3}\,\text{m}^3
\]
\( \textbf{Constant-pressure work:} \)
\[
W=P\Delta V
\]
\( \textbf{Substitution:} \)
\[
W=(1.0\times10^5)(3.0\times10^{-3})\,\text{J}
\]
\( \textbf{Calculation:} \)
\[
W=3.0\times10^2\,\text{J}=300\,\text{J}
\]
\( \textbf{Sign check:} \) The gas expands, so work done by the gas is positive.
\( \textbf{Final answer:} \) The work done by the gas is \(300\,\text{J}\).
