101. A gas undergoes two consecutive quasi-static stages. First it expands from \(2.0\,\text{L}\) to \(6.0\,\text{L}\) at \(3.0\times10^5\,\text{Pa}\). It is then compressed from \(6.0\,\text{L}\) to \(3.0\,\text{L}\) at \(5.0\times10^5\,\text{Pa}\). What is the net work done by the gas?
ⓐ. \(-300\,\text{J}\)
ⓑ. \(+300\,\text{J}\)
ⓒ. \(+2700\,\text{J}\)
ⓓ. \(-2700\,\text{J}\)
Correct Answer: \(-300\,\text{J}\)
Explanation: \(\textbf{Expansion stage:}\)
\[
\Delta V_1=(6.0-2.0)\times10^{-3}
=4.0\times10^{-3}\,\text{m}^3
\]
Therefore,
\[
W_1=(3.0\times10^5)(4.0\times10^{-3})
=+1200\,\text{J}
\]
\(\textbf{Compression stage:}\)
\[
\Delta V_2=(3.0-6.0)\times10^{-3}
=-3.0\times10^{-3}\,\text{m}^3
\]
Hence
\[
W_2=(5.0\times10^5)(-3.0\times10^{-3})
=-1500\,\text{J}
\]
\(\textbf{Net work:}\)
\[
W_{\text{net}}=W_1+W_2
\]
\[
W_{\text{net}}=1200-1500=-300\,\text{J}
\]
Although the final volume is greater than the initial volume, the gas does negative net work because the compression occurs at the larger pressure. Thus the sign of total work cannot be inferred from net volume change alone when pressure differs between stages.
\(\textbf{Final answer:}\)
\[
\boxed{W_{\text{net}}=-300\,\text{J}}
\]
102. A gas expands quasi-statically at constant pressure from \(2.5\,\text{L}\) to \(5.5\,\text{L}\) and does \(900\,\text{J}\) of work. What was the constant pressure?
ⓐ. \(1.5\times10^5\,\text{Pa}\)
ⓑ. \(6.0\times10^5\,\text{Pa}\)
ⓒ. \(4.5\times10^5\,\text{Pa}\)
ⓓ. \(3.0\times10^5\,\text{Pa}\)
Correct Answer: \(3.0\times10^5\,\text{Pa}\)
Explanation: \(\textbf{Extract the signed volume change:}\)
The process is an expansion, so
\[
\Delta V=(5.5-2.5)\,\text{L}=3.0\,\text{L}
\]
Converting to SI units,
\[
\Delta V=3.0\times10^{-3}\,\text{m}^3
\]
For constant-pressure work done by the gas,
\[
W=P\Delta V
\]
Hence the pressure is
\[
P=\frac{W}{\Delta V}
\]
Substituting,
\[
P=\frac{900}{3.0\times10^{-3}}
\]
\[
P=3.0\times10^5\,\text{Pa}
\]
The positive work is consistent with the positive volume change. Also,
\[
\text{Pa}\,\text{m}^3=\text{J},
\]
so the units are consistent.
\(\textbf{Final answer:}\)
\[
\boxed{P=3.0\times10^5\,\text{Pa}}
\]
103. A gas executes a rectangular cycle on a \(P-V\) diagram between pressures \(P_1=1.0\times10^5\,\text{Pa}\) and \(P_2=4.0\times10^5\,\text{Pa}\), and volumes \(V_1=2.0\,\text{L}\) and \(V_2=5.0\,\text{L}\). The cycle is traversed counterclockwise. What is the net work done by the gas in one complete cycle?
ⓐ. \(-900\,\text{J}\)
ⓑ. \(+900\,\text{J}\)
ⓒ. \(+1200\,\text{J}\)
ⓓ. \(-1200\,\text{J}\)
Correct Answer: \(-900\,\text{J}\)
Explanation: \(\textbf{Magnitude of the enclosed area:}\)
For a rectangular cycle,
\[
|W_{\text{net}}|=(P_2-P_1)(V_2-V_1)
\]
The pressure difference is
\[
P_2-P_1=(4.0-1.0)\times10^5
=3.0\times10^5\,\text{Pa}
\]
The volume difference is
\[
V_2-V_1=(5.0-2.0)\times10^{-3}
=3.0\times10^{-3}\,\text{m}^3
\]
Therefore,
\[
|W_{\text{net}}|
=(3.0\times10^5)(3.0\times10^{-3})
=900\,\text{J}
\]
\(\textbf{Determine the sign from the direction:}\)
A counterclockwise cycle has compression occurring at the higher-pressure portion relative to expansion. Thus the work done by the gas over the cycle is negative.
\[
W_{\text{net}}=-900\,\text{J}
\]
\(\textbf{Final answer:}\)
\[
\boxed{W_{\text{net}}=-900\,\text{J}}
\]
104. A gas follows a quasi-static path from state \(A\) to state \(B\), doing \(+450\,\text{J}\) of pressure-volume work. The same geometrical \(P-V\) path is then traversed exactly in reverse from \(B\) to \(A\). What is the pressure-volume work done by the gas on the reverse path?
ⓐ. \(+450\,\text{J}\)
ⓑ. \(-450\,\text{J}\)
ⓒ. \(0\)
ⓓ. The value cannot be inferred because work is path dependent.
Correct Answer: \(-450\,\text{J}\)
Explanation: For the forward process,
\[
W_{A\to B}=\int_A^B P\,dV=+450\,\text{J}
\]
The reverse process follows exactly the same geometrical curve but with the integration limits reversed:
\[
W_{B\to A}=\int_B^A P\,dV
\]
Reversing the limits changes the sign:
\[
\int_B^A P\,dV=-\int_A^B P\,dV
\]
Therefore,
\[
W_{B\to A}=-450\,\text{J}
\]
This does not contradict path dependence. Path dependence means different routes between the same states may have different work values. Here the route itself is identical geometrically and simply traversed in the opposite direction. The magnitude is therefore unchanged while the sign reverses.
\(\textbf{Final answer:}\)
\[
\boxed{W_{B\to A}=-450\,\text{J}}
\]
105. Consider the following statements about quasi-static pressure-volume work.
Statement I: The work between two states is fixed once their initial and final volumes are known.
Statement II: A constant-volume segment contributes zero \(P-V\) work even if pressure changes greatly.
Statement III: For the same expansion volume interval, a path lying at higher pressure produces greater work.
Statement IV: A positive geometric area associated with a compression path corresponds to negative work done by the gas under the adopted convention.
Which combination is correct?
ⓐ. I, II and III only
ⓑ. II, III and IV only
ⓒ. I and IV only
ⓓ. I, II, III and IV
Correct Answer: II, III and IV only
Explanation: Statement I is incorrect because pressure-volume work depends on the full process path:
\[
W=\int P\,dV.
\]
Knowing only the endpoint volumes does not specify the pressures encountered during the process. Statement II is correct because at constant volume,
\[
dV=0,
\]
so the pressure-volume work is zero regardless of how much the pressure changes. Statement III is correct because, for the same positive \(dV\), a larger pressure produces a larger area under the \(P-V\) curve and therefore greater positive work. Statement IV is also correct because the ordinary geometric area gives a magnitude, while compression has \(dV\lt 0\). Consequently, work done by the gas during compression is negative. The set therefore tests endpoint reasoning, zero-work conditions, graph-area comparison and work-sign convention together. Hence Statements II, III and IV only are correct.
106. A gas pushes a frictionless piston of area \(A\) outward through distance \(x\) while exerting a constant pressure \(P\). Which pair of expressions correctly gives the same work done by the gas?
ⓐ. \(W=PAx=P\Delta V\), where \(\Delta V=Ax\).
ⓑ. \(W=Px/A\) and \(W=P/\Delta V\)
ⓒ. \(W=PA/x\) and \(W=P\Delta V^2\)
ⓓ. \(W=P/Ax\) and \(W=P/\Delta V\)
Correct Answer: \(W=PAx=P\Delta V\), where \(\Delta V=Ax\).
Explanation: Pressure is force per unit area:
\[
P=\frac{F}{A}.
\]
Hence the force exerted by the gas on the piston is
\[
F=PA.
\]
For a constant force parallel to the piston displacement,
\[
W=Fx.
\]
Therefore,
\[
W=(PA)x=PAx.
\]
The corresponding volume increase of a piston-cylinder system is
\[
\Delta V=Ax.
\]
Substituting this into the thermodynamic expression gives
\[
W=P\Delta V=P(Ax)=PAx.
\]
Thus the mechanical force-displacement description and the pressure-volume description are physically equivalent for this constant-pressure piston motion. The result also shows dimensionally why \(\text{Pa}\,\text{m}^3\) is a joule.
\(\textbf{Final answer:}\)
\[
W=PAx=P\Delta V.
\]
107. A gas is confined by a frictionless piston of area \(100\,\text{cm}^2\). During a slow expansion, the gas pressure increases linearly from \(1.0\times10^5\,\text{Pa}\) to \(3.0\times10^5\,\text{Pa}\) while the piston moves outward by \(10\,\text{cm}\). What is the pressure-volume work done by the gas?
ⓐ. \(100\,\text{J}\)
ⓑ. \(150\,\text{J}\)
ⓒ. \(200\,\text{J}\)
ⓓ. \(300\,\text{J}\)
Correct Answer: \(200\,\text{J}\)
Explanation: \(\textbf{Volume change from piston motion:}\)
Convert the piston area to SI units:
\[
A=100\times10^{-4}=1.0\times10^{-2}\,\text{m}^2
\]
and the displacement is
\[
x=10\,\text{cm}=0.10\,\text{m}
\]
so
\[
\Delta V=Ax=(1.0\times10^{-2})(0.10)=1.0\times10^{-3}\,\text{m}^3
\]
\(\textbf{Average pressure on the linear path:}\)
Because pressure varies linearly with volume during the slow expansion,
\[
P_{\text{avg}}=\frac{1.0\times10^5+3.0\times10^5}{2}
=2.0\times10^5\,\text{Pa}
\]
\(\textbf{Work done by the gas:}\)
The signed pressure-volume work is
\[
W=P_{\text{avg}}\Delta V
\]
Thus,
\[
W=(2.0\times10^5)(1.0\times10^{-3})
=200\,\text{J}
\]
The sign is positive because the gas expands. The unit check is \(\text{Pa}\,\text{m}^3=\text{J}\), as required.
\(\textbf{Final answer:}\)
\[
\boxed{W=200\,\text{J}}
\]
108. The same gas is taken quasi-statically between states \(A\) and \(B\) twice. In the first trial the process is completed slowly in \(10\) minutes. In the second trial the same sequence of equilibrium \(P-V\) states is followed in \(20\) minutes. Neglect all dissipative differences. Which statement about the pressure-volume works \(W_1\) and \(W_2\) is correct?
ⓐ. \(W_2=2W_1\) because the process lasts twice as long.
ⓑ. \(W_1=2W_2\) because the first process is faster.
ⓒ. \(W_1=W_2\) because the same \(P-V\) path is traversed.
ⓓ. The works cannot be compared without knowing the temperatures.
Correct Answer: \(W_1=W_2\) because the same \(P-V\) path is traversed.
Explanation: For quasi-static pressure-volume work,
\[
W=\int P\,dV.
\]
The integral depends on the sequence of pressure and volume states along the path. The problem states that both trials follow the same \(P-V\) path. Therefore each infinitesimal volume change occurs at the same corresponding pressure in both trials. Changing the time taken does not alter the geometrical area under that path. Hence the total pressure-volume work is unchanged:
\[
W_1=W_2.
\]
This does not mean that rate is never physically important in thermodynamics. Rather, under the stated assumptions, the path itself has been held fixed and dissipative differences are excluded. Work is path dependent, not directly time dependent. Therefore completing the same quasi-static path in a different duration does not by itself alter \(P-V\) work.
109. A gas goes from state \(A\) to state \(B\) by an open path on a \(P-V\) diagram. A student claims that the work done equals “the area enclosed by the path.” Which correction is most appropriate?
ⓐ. For an open path, work is the signed area under the \(P-V\) curve over its volume interval.
ⓑ. Work is represented only by the vertical height of the final point.
ⓒ. Work is represented by the area to the left of the path between the pressure axis and the curve.
ⓓ. No graphical interpretation of work is possible unless the path forms a closed cycle.
Correct Answer: For an open path, work is the signed area under the \(P-V\) curve over its volume interval.
Explanation: For a quasi-static process,
\[
W=\int_{V_i}^{V_f}P\,dV.
\]
On a \(P-V\) diagram, this integral corresponds to the signed area under the process curve over the relevant volume interval. An open path need not geometrically enclose a region by itself. The familiar statement that net cyclic work equals enclosed area applies specifically to a closed cycle where expansion and compression paths together bound a region. Applying that cycle rule mechanically to an open path produces the wrong interpretation. The vertical height of the final point gives only the final pressure and cannot determine the accumulated work. Area to the left of a curve corresponds to integrating volume with respect to pressure rather than \(P\,dV\), so it is not the required \(P-V\) work. Thus the correct graphical quantity is the area under the open path between its endpoint volumes, with sign determined by traversal direction.
110. A gas undergoes the following quasi-static sequence on a \(P-V\) diagram.
Stage I: expansion from \(1.0\,\text{L}\) to \(3.0\,\text{L}\) at \(4.0\times10^5\,\text{Pa}\). Stage II: compression from \(3.0\,\text{L}\) to \(2.0\,\text{L}\) at \(2.0\times10^5\,\text{Pa}\). Stage III: expansion from \(2.0\,\text{L}\) to \(5.0\,\text{L}\) while pressure decreases linearly from \(2.0\times10^5\,\text{Pa}\) to \(1.0\times10^5\,\text{Pa}\).
What is the total work done by the gas?
ⓐ. \(650\,\text{J}\)
ⓑ. \(850\,\text{J}\)
ⓒ. \(1050\,\text{J}\)
ⓓ. \(1450\,\text{J}\)
Correct Answer: \(1050\,\text{J}\)
Explanation: \(\textbf{Stage I — constant-pressure expansion:}\)
\[
W_1=P\Delta V
\]
so
\[
W_1=(4.0\times10^5)(3.0-1.0)\times10^{-3}
=800\,\text{J}
\]
\(\textbf{Stage II — constant-pressure compression:}\)
Here the volume change is negative:
\[
W_2=(2.0\times10^5)(2.0-3.0)\times10^{-3}
=-200\,\text{J}
\]
\(\textbf{Stage III — linear-pressure expansion:}\)
The average pressure is
\[
P_{\text{avg}}=\frac{2.0\times10^5+1.0\times10^5}{2}
=1.5\times10^5\,\text{Pa}
\]
Hence,
\[
W_3=(1.5\times10^5)(5.0-2.0)\times10^{-3}
=450\,\text{J}
\]
\(\textbf{Combine the signed works:}\)
\[
W_{\text{total}}=800-200+450
=1050\,\text{J}
\]
The compression contribution must be subtracted because work done by the gas is negative during compression.
\(\textbf{Final answer:}\)
\[
\boxed{W_{\text{total}}=1050\,\text{J}}
\]
111. A gas in a piston-cylinder is in thermal contact with a hotter reservoir while an external agent simultaneously compresses it. During the process, \(250\,\text{J}\) enters the gas because of the temperature difference, while the surroundings mechanically supply \(400\,\text{J}\) through piston motion. Which description is correct under the adopted conventions?
ⓐ. \(Q=+250\,\text{J}\) and work done by the gas is negative.
ⓑ. \(Q=-250\,\text{J}\) and work done by the gas is positive.
ⓒ. The entire \(650\,\text{J}\) is positive heat because all energy enters the gas.
ⓓ. The entire \(650\,\text{J}\) is negative work because compression occurs.
Correct Answer: \(Q=+250\,\text{J}\) and work done by the gas is negative.
Explanation: The \(250\,\text{J}\) transfer is caused by a temperature difference between the hotter reservoir and the gas. It is therefore heat supplied to the gas, so
\[
Q=+250\,\text{J}.
\]
The \(400\,\text{J}\) transfer occurs mechanically through inward piston displacement. It must therefore be classified as work rather than heat. Since the surroundings do work on the gas, work done by the gas has the opposite sign and is negative. Thus the process simultaneously has positive heat supplied and negative work done by the gas. The fact that both transfers deliver energy to the gas does not make them the same thermodynamic transfer mode. Heat and work are classified by the mechanism of transfer, not simply by the direction of energy flow. This distinction becomes essential when thermodynamic energy bookkeeping is later applied. Hence option A is correct.
112. Four quasi-static processes \(P,Q,R,S\) are described over the same positive volume interval \(V_1\rightarrow V_2\).
| Process | Pressure behaviour |
|---|
| \(P\) | Constant at \(4P_0\) |
| \(Q\) | Decreases linearly from \(5P_0\) to \(P_0\) |
| \(R\) | Constant at \(2P_0\) |
| \(S\) | Increases linearly from \(P_0\) to \(5P_0\) |
Which ordering of work done by the gas is correct?
ⓐ. \(W_P\gt W_Q=W_S\gt W_R\)
ⓑ. \(W_Q\gt W_S\gt W_P\gt W_R\)
ⓒ. \(W_P\gt W_Q\gt W_S\gt W_R\)
ⓓ. \(W_Q=W_S\gt W_P\gt W_R\)
Correct Answer: \(W_P\gt W_Q=W_S\gt W_R\)
Explanation: All four processes cover the same positive volume change,
\[
\Delta V=V_2-V_1\gt 0.
\]
Therefore their works can be ranked by the average pressure over that interval. For Process \(P\),
\[
P_{\text{avg},P}=4P_0.
\]
For Process \(Q\), the pressure changes linearly, so
\[
P_{\text{avg},Q}=\frac{5P_0+P_0}{2}=3P_0.
\]
For Process \(R\),
\[
P_{\text{avg},R}=2P_0.
\]
For Process \(S\),
\[
P_{\text{avg},S}=\frac{P_0+5P_0}{2}=3P_0.
\]
Thus
\[
W_P=4P_0\Delta V,
\]
\[
W_Q=W_S=3P_0\Delta V,
\]
and
\[
W_R=2P_0\Delta V.
\]
The direction in which pressure changes along a straight-line path does not affect the total work if the endpoint pressures, and therefore the average pressure, are the same.
\(\textbf{Final answer:}\)
\[
W_P\gt W_Q=W_S\gt W_R.
\]
113. A sealed container of gas is initially at rest and in a definite thermodynamic state. Without deforming or heating the container, an external agent gives the entire container a uniform translational velocity \(v\). Which statement about the gas is correct immediately afterward?
ⓐ. Its internal energy increases by the translational kinetic energy of the container and gas.
ⓑ. Its internal energy decreases because part of its molecular kinetic energy becomes bulk motion.
ⓒ. Its thermodynamic internal energy is unchanged, although the total energy measured in the laboratory frame has increased.
ⓓ. Its internal energy becomes zero in the frame moving with the container.
Correct Answer: Its thermodynamic internal energy is unchanged, although the total energy measured in the laboratory frame has increased.
Explanation: Internal energy \(U\) represents the microscopic energy associated with molecular motion and interactions within the system. The kinetic energy associated with translation of the system as a whole is a macroscopic mechanical energy and is not included in thermodynamic internal energy. Giving the complete container a uniform velocity changes its centre-of-mass kinetic energy. The problem states that the thermodynamic state inside the container is otherwise unchanged, so the random molecular motions and microscopic interaction energies are not altered by this operation. Therefore \(U\) remains unchanged. The total laboratory-frame energy of the moving container-gas system does increase because it now possesses bulk translational kinetic energy. Changing the observer's frame does not make the microscopic internal energy vanish. The essential distinction is between disordered microscopic energy within the system and ordered motion of the complete system. Hence option C is correct.
114. Consider the following statements about thermodynamic internal energy.
Statement I: It includes microscopic kinetic energy associated with random molecular motion.
Statement II: For a system with intermolecular interactions, microscopic potential energy can contribute to internal energy.
Statement III: The centre-of-mass translational kinetic energy of the complete system is part of its thermodynamic internal energy.
Statement IV: Heat and work describe modes of energy transfer, whereas internal energy characterises the thermodynamic state of the system.
Which combination is correct?
ⓐ. I, II and III only
ⓑ. I, II and IV only
ⓒ. II, III and IV only
ⓓ. I, III and IV only
Correct Answer: I, II and IV only
Explanation: Statement I is correct because microscopic molecular motion contributes kinetic energy to the internal energy of the system. Statement II is also correct because internal energy can contain potential energy associated with molecular interactions. Statement III is incorrect because motion of the system as a whole is macroscopic mechanical motion rather than part of its thermodynamic internal energy. Internal energy is considered after separating out such bulk centre-of-mass motion. Statement IV is correct and makes an important thermodynamic distinction. Heat and work refer to ways in which energy crosses the boundary of a system during a process. Internal energy \(U\), in contrast, is a state variable belonging to the system. A system therefore does not “store work” or “store heat” in the same sense in which it possesses internal energy. Hence Statements I, II and IV are correct.
115. A system has internal energy \(U_A=2.4\,\text{kJ}\) in equilibrium state \(A\) and \(U_B=3.1\,\text{kJ}\) in equilibrium state \(B\). It goes from \(A\) to \(B\) by one process and later returns from \(B\) to \(A\) by a completely different process. Which statement is correct?
ⓐ. \(\Delta U_{A\to B}=+0.7\,\text{kJ}\) and \(\Delta U_{B\to A}=+0.7\,\text{kJ}\)
ⓑ. \(\Delta U_{A\to B}=-0.7\,\text{kJ}\) and \(\Delta U_{B\to A}=+0.7\,\text{kJ}\)
ⓒ. The two values of \(\Delta U\) cannot be found unless heat and work are known for both paths.
ⓓ. \(\Delta U_{A\to B}=+0.7\,\text{kJ}\) and \(\Delta U_{B\to A}=-0.7\,\text{kJ}\)
Correct Answer: \(\Delta U_{A\to B}=+0.7\,\text{kJ}\) and \(\Delta U_{B\to A}=-0.7\,\text{kJ}\)
Explanation: \(\textbf{Forward change:}\)
Internal-energy change is defined by the endpoint values:
\[
\Delta U_{A\to B}=U_B-U_A
\]
Therefore,
\[
\Delta U_{A\to B}=3.1-2.4=+0.7\,\text{kJ}
\]
\(\textbf{Reverse change:}\)
For the return process,
\[
\Delta U_{B\to A}=U_A-U_B
\]
Hence,
\[
\Delta U_{B\to A}=2.4-3.1=-0.7\,\text{kJ}
\]
The paths may be completely different, but internal energy is a state function, so its change is fixed by the initial and final states. Heat and work along the two paths may differ without altering these values of \(\Delta U\).
\(\textbf{Final answer:}\)
\[
\boxed{\Delta U_{A\to B}=+0.7\,\text{kJ},\qquad \Delta U_{B\to A}=-0.7\,\text{kJ}}
\]
116. Consider the following statements.
Statement I: Internal energy \(U\) is a thermodynamic state variable.
Statement II: Heat \(Q\) and work \(W\) are process quantities and may depend on the path.
Statement III: For two fixed equilibrium states \(A\) and \(B\), \(\Delta U_{A\to B}\) is the same for every allowed path joining them.
Statement IV: If two paths have the same \(\Delta U\), they must separately have the same heat transfer and the same work.
Which combination is correct?
ⓐ. I and IV only
ⓑ. II and III only
ⓒ. I, II and III only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Statement I is correct because the value of internal energy is determined by the thermodynamic state of the system. Statement II is also correct because heat and work describe energy transferred during a particular process, so their values can vary when the route between states changes. Statement III follows directly from the state-function character of \(U\). If the initial and final states are fixed, then
\[
\Delta U=U_B-U_A
\]
has one definite value regardless of the intermediate path. Statement IV is incorrect. Two paths can have different amounts of heat and different amounts of work while still producing the same difference \(Q-W\) and hence the same \(\Delta U\). This is one of the central contrasts between a state quantity and path-dependent transfer quantities. Therefore Statements I, II and III only are correct.
117. A system goes from state \(A\) to state \(B\) by three different paths. Using \(Q\gt 0\) for heat supplied to the system and \(W\gt 0\) for work done by the system, the data are:
| Path | \(Q\) | \(W\) |
|---|
| P | \(+900\,\text{J}\) | \(+300\,\text{J}\) |
| Q | \(+500\,\text{J}\) | \(-100\,\text{J}\) |
| R | \(+200\,\text{J}\) | \(W_R\) |
What value of \(W_R\) is required if all three paths genuinely connect the same states?
ⓐ. \(-400\,\text{J}\)
ⓑ. \(+400\,\text{J}\)
ⓒ. \(-200\,\text{J}\)
ⓓ. \(+800\,\text{J}\)
Correct Answer: \(-400\,\text{J}\)
Explanation: \(\textbf{Determine the endpoint internal-energy change:}\)
For Path P,
\[
\Delta U=Q_P-W_P
\]
Thus,
\[
\Delta U=900-300=600\,\text{J}
\]
Path Q confirms the same endpoint value:
\[
\Delta U=500-(-100)=600\,\text{J}
\]
This agreement is required because \(U\) is a state function.
\(\textbf{Apply the same internal-energy change to Path R:}\)
\[
600=Q_R-W_R
\]
Substituting \(Q_R=200\,\text{J}\),
\[
600=200-W_R
\]
Hence,
\[
-W_R=400
\]
and therefore
\[
W_R=-400\,\text{J}
\]
The negative sign means that the net work is done on the system along Path R.
\(\textbf{Final answer:}\)
\[
\boxed{W_R=-400\,\text{J}}
\]
118. Two quasi-static paths \(P\) and \(Q\) connect the same equilibrium states \(A\) and \(B\). Along \(P\), the gas does \(800\,\text{J}\) of work and absorbs \(1100\,\text{J}\) of heat. Along \(Q\), the \(P-V\) area shows that the gas does only \(300\,\text{J}\) of work. How much heat is absorbed along \(Q\)?
ⓐ. \(300\,\text{J}\)
ⓑ. \(600\,\text{J}\)
ⓒ. \(800\,\text{J}\)
ⓓ. \(1100\,\text{J}\)
Correct Answer: \(600\,\text{J}\)
Explanation: \(\textbf{Use Path P to determine the state-function change:}\)
\[
\Delta U=Q_P-W_P
\]
Substituting,
\[
\Delta U=1100-800=300\,\text{J}
\]
Because \(A\) and \(B\) are the same endpoints for both processes, this value of \(\Delta U\) must also apply to Path Q.
\(\textbf{Apply the endpoint constraint to Path Q:}\)
\[
\Delta U=Q_Q-W_Q
\]
Therefore,
\[
300=Q_Q-300
\]
Hence,
\[
Q_Q=600\,\text{J}
\]
The two paths have different work and therefore different heat transfers, but their internal-energy changes are identical. In fact,
\[
Q_P-Q_Q=W_P-W_Q=500\,\text{J},
\]
which is another way to see the required compensation between the path quantities.
\(\textbf{Final answer:}\)
\[
\boxed{Q_Q=600\,\text{J}}
\]
119. A thermodynamic system initially has internal energy \(U_i\). During a process it receives \(300\,\text{J}\) as heat, while the surroundings simultaneously do \(500\,\text{J}\) of work on the system. Using the convention that work done by the system is positive, which result is correct?
ⓐ. \(\Delta U=-800\,\text{J}\)
ⓑ. \(\Delta U=-200\,\text{J}\)
ⓒ. \(\Delta U=+200\,\text{J}\)
ⓓ. \(\Delta U=+800\,\text{J}\)
Correct Answer: \(\Delta U=+800\,\text{J}\)
Explanation: \(\textbf{Translate the two transfer directions into signs:}\)
Heat enters the system, so
\[
Q=+300\,\text{J}
\]
The surroundings do \(500\,\text{J}\) of work on the system. Therefore work done by the system is
\[
W=-500\,\text{J}
\]
\(\textbf{Determine the internal-energy change:}\)
Using
\[
\Delta U=Q-W,
\]
we get
\[
\Delta U=300-(-500)
\]
Thus,
\[
\Delta U=800\,\text{J}
\]
Both transfers increase the energy retained by the system in this process, even though their thermodynamic signs are different because the work convention refers to work done by the system. Internal energy can therefore change through heat transfer, work transfer, or both.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta U=+800\,\text{J}}
\]
120. Along one path from equilibrium state \(A\) to equilibrium state \(B\), a system absorbs \(700\,\text{J}\) of heat and does \(200\,\text{J}\) of work. Which alternative heat-work pair could describe another path between the same two states?
ⓐ. \(Q=+300\,\text{J},\ W=-200\,\text{J}\)
ⓑ. \(Q=+300\,\text{J},\ W=+200\,\text{J}\)
ⓒ. \(Q=-300\,\text{J},\ W=-200\,\text{J}\)
ⓓ. \(Q=+800\,\text{J},\ W=+100\,\text{J}\)
Correct Answer: \(Q=+300\,\text{J},\ W=-200\,\text{J}\)
Explanation: \(\textbf{Fix the state-function change from the known path:}\)
\[
\Delta U=Q-W
\]
For the given path,
\[
\Delta U=700-200=500\,\text{J}
\]
Any second path connecting exactly the same states must satisfy
\[
Q-W=500\,\text{J}
\]
For option A,
\[
Q-W=300-(-200)=500\,\text{J},
\]
so it is compatible with the same endpoints. For option B,
\[
300-200=100\,\text{J}
\]
For option C,
\[
-300-(-200)=-100\,\text{J},
\]
and for option D,
\[
800-100=700\,\text{J}
\]
Only option A preserves the required internal-energy change while allowing both heat and work individually to differ from the first path.
\(\textbf{Final answer:}\)
\[
\boxed{Q=+300\,\text{J},\qquad W=-200\,\text{J}}
\]
