201. An ideal gas initially occupies state \(A:(P_0,V_0,T_0)\). It expands reversibly and adiabatically to \(B\) with volume \(4V_0\). For the gas, \(\gamma=3/2\). It is then heated at constant volume until its temperature again becomes \(T_0\), reaching state \(C\). Which statement about state \(C\) is correct?
ⓐ. \(P_C=P_0/4\), and state \(C\) is different from state \(A\).
ⓑ. \(P_C=P_0\) because the temperature has returned to \(T_0\).
ⓒ. \(P_C=P_0/8\), because pressure does not change during heating.
ⓓ. \(P_C=2P_0\), because heat has been supplied after the expansion.
Correct Answer: \(P_C=P_0/4\), and state \(C\) is different from state \(A\).
Explanation: \(\textbf{Adiabatic stage A to B:}\)
For \(\gamma=3/2\),
\[
T_AV_A^{1/2}=T_BV_B^{1/2}
\]
Thus,
\[
T_B=T_0\left(\frac{V_0}{4V_0}\right)^{1/2}
=\frac{T_0}{2}
\]
Also,
\[
P_B=P_0\left(\frac14\right)^{3/2}
=\frac{P_0}{8}.
\]
\(\textbf{Constant-volume heating B to C:}\)
The volume remains
\[
V_C=4V_0
\]
At fixed volume for an ideal gas,
\[
P\propto T
\]
The temperature doubles from \(T_0/2\) to \(T_0\), so
\[
P_C=2P_B=\frac{P_0}{4}
\]
Although the temperature has returned to its original value, the volume remains four times the original volume. Therefore state \(C\) is not the same thermodynamic state as \(A\).
\(\textbf{Final answer:}\)
\[
P_C=\frac{P_0}{4}
\]
202. An ideal gas expands adiabatically and does \(600\,\text{J}\) of work. It is then kept at the expanded volume while \(600\,\text{J}\) of heat is supplied. Which conclusion is correct?
ⓐ. The gas necessarily returns to its original thermodynamic state.
ⓑ. Its internal energy increases by \(600\,\text{J}\) because only the second stage matters.
ⓒ. Its internal energy decreases by \(1200\,\text{J}\).
ⓓ. Its final internal energy equals its initial value, but the two thermodynamic states can differ.
Correct Answer: Its final internal energy equals its initial value, but the two thermodynamic states can differ.
Explanation: \(\textbf{Adiabatic expansion:}\)
For Stage I,
\[
Q_1=0
\]
and
\[
W_1=+600\,\text{J}
\]
Hence,
\[
\Delta U_1=Q_1-W_1=-600\,\text{J}.
\]
\(\textbf{Constant-volume heating:}\)
At fixed volume,
\[
W_2=0.
\]
Heat supplied is
\[
Q_2=+600\,\text{J}
\]
Therefore,
\[
\Delta U_2=+600\,\text{J}
\]
Thus,
\[
\Delta U_{\text{total}}=-600+600=0
\]
The final internal energy therefore equals the initial internal energy. For a fixed amount of ideal gas this also restores the original temperature. However, the gas remains at the expanded volume, so its pressure and volume generally differ from their initial values. Equal internal energy does not by itself mean identical thermodynamic state.
203. Four processes involving the same ideal gas are described below.
| Process | Given condition | Claim |
|---|
| I | Reversible adiabatic expansion | \(Q=0,\ W\gt 0,\ \Delta U\lt 0\) |
| II | Reversible adiabatic compression | \(Q=0,\ W\lt 0,\ \Delta U\gt 0\) |
| III | Isothermal expansion | \(\Delta U=0,\ Q=W\gt 0\) |
| IV | Adiabatic expansion | \(Q=0,\ \Delta U=0,\ W\gt 0\) |
Which process contains an inconsistent claim?
ⓐ. Process I
ⓑ. Process II
ⓒ. Process III
ⓓ. Process IV
Correct Answer: Process IV
Explanation: Process I is consistent because an adiabatic expansion has
\[
Q=0
\]
and positive work done by the gas. Hence,
\[
\Delta U=-W\lt 0
\]
Process II is also consistent. Compression gives \(W\lt 0\), so with \(Q=0\),
\[
\Delta U=-W\gt 0
\]
Process III is correct for an ideal gas because isothermal behaviour gives
\[
\Delta U=0,
\]
and therefore
\[
Q=W.
\]
For expansion, both are positive. Process IV is inconsistent because it simultaneously states
\[
Q=0,\qquad W\gt 0
\]
and
\[
\Delta U=0.
\]
The first law instead requires
\[
\Delta U=-W\lt 0
\]
Thus Process IV violates the energy balance.
204. Match the processes in List I with the most appropriate result in List II.
| List I | List II |
|---|
| P. Ideal-gas reversible adiabatic expansion | 1. \(\Delta U=0,\ Q=W\gt 0\) |
| Q. Ideal-gas isothermal expansion | 2. \(Q=0,\ \Delta U\lt 0\) |
| R. Ideal-gas reversible adiabatic compression | 3. \(Q=0,\ \Delta U\gt 0\) |
| S. Heating in a rigid vessel with no other work mode | 4. \(W=0,\ \Delta U=Q\gt 0\) |
ⓐ. P-2, Q-1, R-3, S-4
ⓑ. P-1, Q-2, R-4, S-3
ⓒ. P-3, Q-4, R-2, S-1
ⓓ. P-4, Q-3, R-1, S-2
Correct Answer: P-2, Q-1, R-3, S-4
Explanation: For P, reversible adiabatic expansion has
\[
Q=0,\qquad W\gt 0,
\]
so
\[
\Delta U\lt 0
\]
Thus P-2. For Q, an isothermal ideal-gas expansion has
\[
\Delta U=0
\]
and therefore
\[
Q=W\gt 0,
\]
so Q-1. For R, adiabatic compression has \(Q=0\) and \(W\lt 0\), which gives
\[
\Delta U\gt 0,
\]
so R-3. For S, the rigid vessel has zero pressure-volume work. If heat is supplied and there is no other work mode,
\[
W=0
\]
and therefore
\[
\Delta U=Q\gt 0
\]
Hence S-4, giving the matching P-2, Q-1, R-3, S-4
205. The same ideal gas initially occupies the same state and is compressed quasi-statically to the same smaller final volume by two paths: one isothermal and one reversible adiabatic. Which comparison is correct?
ⓐ. The adiabatic path has lower pressure throughout the compression and requires less work on the gas.
ⓑ. The adiabatic path has higher pressure after leaving the initial point and requires a larger magnitude of work on the gas.
ⓒ. Both paths have the same pressure at each volume because the volume endpoints are identical.
ⓓ. The isothermal path has higher pressure because heat leaves the gas during compression.
Correct Answer: The adiabatic path has higher pressure after leaving the initial point and requires a larger magnitude of work on the gas.
Explanation: Starting from the same initial state, an isothermal path obeys
\[
PV=\text{constant}
\]
A reversible adiabatic path obeys
\[
PV^\gamma=\text{constant}
\]
with
\[
\gamma\gt 1
\]
During compression, the volume decreases. Because pressure varies as \(V^{-\gamma}\) rather than \(V^{-1}\), the adiabatic pressure rises more rapidly. Therefore the adiabatic curve lies above the isothermal curve after they leave their common initial point. Compression work done by the gas is negative, and its magnitude equals the relevant area under the path. Since the adiabatic pressure is higher throughout the common compression interval, the magnitude of the negative work done by the gas is larger. Equivalently, the surroundings must do more positive work on the gas during the adiabatic compression. The associated adiabatic temperature rise is another consequence of this larger mechanical energy input without heat removal.
206. In a reversible adiabatic expansion of an ideal gas, the volume increases by a factor of \(8\) while the pressure decreases by a factor of \(32\). What is the value of \(\gamma\)?
ⓐ. \(4/3\)
ⓑ. \(7/5\)
ⓒ. \(3/2\)
ⓓ. \(5/3\)
Correct Answer: \(5/3\)
Explanation: \(\textbf{Use the adiabatic relation:}\)
\[
P_1V_1^\gamma=P_2V_2^\gamma
\]
The data give
\[
\frac{V_2}{V_1}=8
\]
and
\[
\frac{P_1}{P_2}=32
\]
Therefore,
\[
32=8^\gamma
\]
Writing both sides as powers of \(2\),
\[
32=2^5
\]
and
\[
8^\gamma=(2^3)^\gamma=2^{3\gamma}
\]
Hence,
\[
3\gamma=5
\]
Therefore,
\[
\gamma=\frac53
\]
The pressure-volume data thus allow the adiabatic exponent to be inferred directly from the process.
\(\textbf{Final answer:}\)
\[
\boxed{\gamma=\frac53}
\]
207. During a reversible adiabatic expansion of an ideal gas, its volume increases from \(V\) to \(4V\) while its temperature falls from \(600\,\text{K}\) to \(300\,\text{K}\). What is \(\gamma\)?
ⓐ. \(4/3\)
ⓑ. \(3/2\)
ⓒ. \(5/3\)
ⓓ. \(7/5\)
Correct Answer: \(3/2\)
Explanation: \(\textbf{Adiabatic temperature-volume relation:}\)
\[
T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}
\]
Therefore,
\[
\frac{T_1}{T_2}
=\left(\frac{V_2}{V_1}\right)^{\gamma-1}
\]
Substituting the given ratios,
\[
\frac{600}{300}=4^{\gamma-1}
\]
Hence,
\[
2=4^{\gamma-1}
\]
Writing \(4=2^2\),
\[
2=(2^2)^{\gamma-1}
=2^{2(\gamma-1)}
\]
Therefore,
\[
2(\gamma-1)=1
\]
Thus,
\[
\gamma-1=\frac12
\]
and
\[
\gamma=\frac32
\]
\(\textbf{Final answer:}\)
\[
\boxed{\gamma=\frac32}
\]
208. Two ideal gases \(A\) and \(B\) start from identical values of \(P,V,T\) and each undergoes a reversible adiabatic expansion to the same larger final volume. Their adiabatic indices satisfy \(\gamma_A\gt \gamma_B\gt 1\). Which comparison is correct?
ⓐ. \(T_A\gt T_B\) and \(P_A\gt P_B\) at the final volume.
ⓑ. \(P_A=P_B\) but \(T_A\lt T_B\).
ⓒ. \(T_A=T_B\) but \(P_A\lt P_B\).
ⓓ. \(T_A\lt T_B\) and \(P_A\lt P_B\) at the final volume.
Correct Answer: \(T_A\lt T_B\) and \(P_A\lt P_B\) at the final volume.
Explanation: For each gas,
\[
PV^\gamma=\text{constant}
\]
At the same final volume ratio \(V_f/V_i\gt 1\),
\[
P_f=P_i\left(\frac{V_i}{V_f}\right)^\gamma
\]
Because the base \(V_i/V_f\) is less than \(1\), the larger exponent \(\gamma_A\) gives the smaller final pressure. Hence,
\[
P_A\lt P_B
\]
For temperature,
\[
TV^{\gamma-1}=\text{constant},
\]
so
\[
T_f=T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1}
\]
Again the base is less than \(1\). Since
\[
\gamma_A-1\gt \gamma_B-1,
\]
gas \(A\) undergoes the larger temperature fall:
\[
T_A\lt T_B.
\]
Thus the gas with the larger adiabatic index cools more strongly and has the lower pressure after the same expansion ratio.
209. Two curves originate from the same point \(A\) on a \(P-V\) diagram and both move toward larger volume.
Curve X falls gradually with increasing volume. Curve Y falls more steeply and remains below X after leaving the common initial point. Both curves describe quasi-static processes of the same ideal gas starting from the same state.
If one curve is isothermal and the other is reversible adiabatic, which identification is correct?
ⓐ. X is adiabatic and Y is isothermal because adiabatic expansion maintains greater pressure.
ⓑ. X is isothermal and Y is adiabatic because the adiabatic pressure falls faster with volume.
ⓒ. Both are isothermal because they start from the same temperature.
ⓓ. Both are adiabatic because both pressures decrease during expansion.
Correct Answer: X is isothermal and Y is adiabatic because the adiabatic pressure falls faster with volume.
Explanation: For the isothermal ideal-gas path,
\[
P\propto V^{-1}.
\]
For the reversible adiabatic path,
\[
P\propto V^{-\gamma},
\]
with
\[
\gamma\gt 1.
\]
Starting from the same point, the larger negative exponent causes the adiabatic pressure to fall more rapidly as volume increases. The adiabatic curve therefore lies below the isothermal curve during expansion. Curve Y is described as the more steeply falling curve, so Y must be the adiabatic path. Curve X, which retains the greater pressure at a given expanded volume, is the isothermal path. The additional heat absorbed along the isothermal path helps maintain the temperature and therefore sustains a higher pressure. Hence the correct identification is X as isothermal and Y as adiabatic.
210. An ideal gas initially occupies state \(A:(P_0,V_0,T_0)\). It expands reversibly and adiabatically to volume \(2V_0\), reaching state \(B\). It is then compressed isothermally at the lower temperature \(T_B\) back to volume \(V_0\), reaching state \(C\). If the adiabatic index is \(\gamma\gt 1\), which relation is correct?
ⓐ. \(P_C=P_0\) and \(T_C=T_0\), so the gas returns to state \(A\).
ⓑ. \(P_C=P_0/2\) and \(T_C=T_0\).
ⓒ. \(P_C=P_0/2^{\gamma-1}\) and \(T_C=T_0/2^{\gamma-1}\), so \(C\neq A\).
ⓓ. \(P_C=2^{\gamma-1}P_0\) and \(T_C=T_0/2^{\gamma-1}\).
Correct Answer: \(P_C=P_0/2^{\gamma-1}\) and \(T_C=T_0/2^{\gamma-1}\), so \(C\neq A\).
Explanation: \(\textbf{Adiabatic expansion A to B:}\)
Using
\[
TV^{\gamma-1}=\text{constant},
\]
we obtain
\[
T_B=T_0\left(\frac{V_0}{2V_0}\right)^{\gamma-1}
\]
Therefore,
\[
T_B=\frac{T_0}{2^{\gamma-1}}
\]
The adiabatic pressure relation gives
\[
P_B=P_0\left(\frac{V_0}{2V_0}\right)^\gamma
=\frac{P_0}{2^\gamma}.
\]
\(\textbf{Isothermal compression B to C:}\)
The second process occurs at constant temperature \(T_B\), so
\[
T_C=T_B=\frac{T_0}{2^{\gamma-1}}
\]
Using
\[
P_BV_B=P_CV_C,
\]
with
\[
V_B=2V_0,\qquad V_C=V_0,
\]
we get
\[
P_C=2P_B.
\]
Thus,
\[
P_C=\frac{2P_0}{2^\gamma}
=\frac{P_0}{2^{\gamma-1}}
\]
Although the original volume is restored, both pressure and temperature remain below their original values. Therefore state \(C\) is not state \(A\).
\(\textbf{Final answer:}\)
\[
P_C=\frac{P_0}{2^{\gamma-1}},\qquad T_C=\frac{T_0}{2^{\gamma-1}}.
\]
211. A fixed amount of ideal gas undergoes a reversible adiabatic expansion in which its pressure falls from \(P_1\) to \(P_1/8\). The gas has \(\gamma=3/2\). What are the corresponding ratios \(T_2/T_1\) and \(V_2/V_1\)?
ⓐ. \(T_2/T_1=1/2,\ V_2/V_1=4\)
ⓑ. \(T_2/T_1=1/4,\ V_2/V_1=2\)
ⓒ. \(T_2/T_1=1/2,\ V_2/V_1=8\)
ⓓ. \(T_2/T_1=1/8,\ V_2/V_1=4\)
Correct Answer: \(T_2/T_1=1/2,\ V_2/V_1=4\)
Explanation: \(\textbf{Use the pressure-temperature adiabatic relation:}\)
For a reversible adiabatic ideal-gas change,
\[
T^\gamma P^{1-\gamma}=\text{constant}
\]
Hence,
\[
\frac{T_2}{T_1}
=\left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}
\]
With
\[
\gamma=\frac32,
\]
the exponent is
\[
\frac{\gamma-1}{\gamma}
=\frac{1/2}{3/2}
=\frac13
\]
Therefore,
\[
\frac{T_2}{T_1}
=\left(\frac18\right)^{1/3}
=\frac12
\]
\(\textbf{Infer the volume change:}\)
For the same amount of ideal gas,
\[
\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}
\]
Thus,
\[
\frac{V_2}{V_1}
=\frac{T_2/T_1}{P_2/P_1}
=\frac{1/2}{1/8}
=4
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{T_2}{T_1}=\frac12,\qquad \frac{V_2}{V_1}=4}
\]
212. An ideal gas with \(\gamma=3/2\) expands reversibly and adiabatically from \(P_1=4.0\times10^5\,\text{Pa},\ V_1=2.0\,\text{L}\) to \(V_2=8.0\,\text{L}\). Which pair correctly gives the work done by the gas and the change in its internal energy?
ⓐ. \(W=+400\,\text{J},\ \Delta U=-400\,\text{J}\)
ⓑ. \(W=+1200\,\text{J},\ \Delta U=-1200\,\text{J}\)
ⓒ. \(W=-800\,\text{J},\ \Delta U=+800\,\text{J}\)
ⓓ. \(W=+800\,\text{J},\ \Delta U=-800\,\text{J}\)
Correct Answer: \(W=+800\,\text{J},\ \Delta U=-800\,\text{J}\)
Explanation: \(\textbf{Determine the final pressure:}\)
For a reversible adiabatic process,
\[
P_1V_1^\gamma=P_2V_2^\gamma
\]
Therefore,
\[
P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma
\]
Here,
\[
\frac{V_2}{V_1}=4,
\]
so
\[
P_2=(4.0\times10^5)\left(\frac14\right)^{3/2}
\]
Since
\[
4^{3/2}=8,
\]
we obtain
\[
P_2=5.0\times10^4\,\text{Pa}
\]
\(\textbf{Adiabatic work:}\)
\[
W=\frac{P_1V_1-P_2V_2}{\gamma-1}
\]
Now,
\[
P_1V_1=(4.0\times10^5)(2.0\times10^{-3})=800\,\text{J},
\]
and
\[
P_2V_2=(5.0\times10^4)(8.0\times10^{-3})=400\,\text{J}
\]
Therefore,
\[
W=\frac{800-400}{1/2}=800\,\text{J}
\]
Because the process is adiabatic,
\[
Q=0
\]
and hence
\[
\Delta U=-W=-800\,\text{J}
\]
\(\textbf{Final answer:}\)
\[
\boxed{W=+800\,\text{J},\qquad \Delta U=-800\,\text{J}}
\]
213. An ideal gas with \(\gamma=5/3\) has \(P_1V_1=800\,\text{J}\). It is reversibly compressed adiabatically until its volume becomes \(V_1/8\). What is the work done by the gas during the compression?
ⓐ. \(-1200\,\text{J}\)
ⓑ. \(-2400\,\text{J}\)
ⓒ. \(-3600\,\text{J}\)
ⓓ. \(+3600\,\text{J}\)
Correct Answer: \(-3600\,\text{J}\)
Explanation: \(\textbf{Find the temperature ratio first:}\)
For a reversible adiabatic process,
\[
TV^{\gamma-1}=\text{constant}
\]
Therefore,
\[
\frac{T_2}{T_1}
=\left(\frac{V_1}{V_2}\right)^{\gamma-1}
\]
Here,
\[
\frac{V_1}{V_2}=8
\]
and
\[
\gamma-1=\frac23
\]
Thus,
\[
\frac{T_2}{T_1}=8^{2/3}=4
\]
\(\textbf{Use the initial ideal-gas state relation:}\)
For an ideal gas,
\[
nRT_1=P_1V_1=800\,\text{J}
\]
Hence,
\[
nR(T_1-T_2)
=nRT_1(1-4)
=800(-3)
\]
\(\textbf{Adiabatic work:}\)
\[
W=\frac{nR(T_1-T_2)}{\gamma-1}
\]
Therefore,
\[
W=\frac{-2400}{2/3}
=-3600\,\text{J}
\]
The negative sign is consistent with compression: work is done on the gas.
\(\textbf{Final answer:}\)
\[
\boxed{W=-3600\,\text{J}}
\]
214. Two ideal gases \(A\) and \(B\) begin with the same \(P_0,V_0,T_0\) and separately undergo reversible adiabatic expansion to \(8V_0\). Their adiabatic indices are
\[
\gamma_A=\frac53,\qquad \gamma_B=\frac75.
\]
Which comparison of their final states and works is correct?
ⓐ. \(T_A\gt T_B,\ P_A\gt P_B,\ W_A\gt W_B\)
ⓑ. \(T_A\lt T_B,\ P_A\lt P_B,\ W_A\lt W_B\)
ⓒ. \(T_A\lt T_B,\ P_A\gt P_B,\ W_A\gt W_B\)
ⓓ. \(T_A\gt T_B,\ P_A\lt P_B,\ W_A=W_B\)
Correct Answer: \(T_A\lt T_B,\ P_A\lt P_B,\ W_A\lt W_B\)
Explanation: Both gases undergo the same expansion ratio, but gas \(A\) has the larger value of \(\gamma\). For temperature,
\[
T_f=T_0\left(\frac{V_0}{V_f}\right)^{\gamma-1}.
\]
Because \(V_0/V_f\lt 1\), the larger exponent for gas \(A\) produces the greater temperature fall. Therefore,
\[
T_A\lt T_B
\]
Similarly,
\[
P_f=P_0\left(\frac{V_0}{V_f}\right)^\gamma,
\]
so the larger \(\gamma_A\) gives
\[
P_A\lt P_B
\]
For the work,
\[
W=\frac{P_0V_0-P_fV_f}{\gamma-1}
\]
For gas \(A\),
\[
\frac{P_AV_f}{P_0V_0}=8^{1-5/3}=\frac14,
\]
giving
\[
W_A=\frac{P_0V_0(1-1/4)}{2/3}
=\frac98P_0V_0
\]
For gas \(B\),
\[
\frac{P_BV_f}{P_0V_0}=8^{1-7/5}=8^{-2/5},
\]
which leads to a larger work than \(9P_0V_0/8\). Thus gas \(B\), having the smaller \(\gamma\), retains greater pressure and temperature and performs more work during the same expansion ratio.
\(\textbf{Final answer:}\)
\[
T_A\lt T_B,\qquad P_A\lt P_B,\qquad W_A\lt W_B
\]
215. At one common state of an ideal gas, the magnitude of the slope of its reversible adiabatic curve on a \(P-V\) diagram is observed to be \(3/2\) times the magnitude of the slope of the isotherm through the same state. The gas is then expanded reversibly and adiabatically until its volume becomes four times the initial volume. What is the final pressure in terms of the initial pressure \(P_1\)?
ⓐ. \(P_1/4\)
ⓑ. \(P_1/6\)
ⓒ. \(P_1/8\)
ⓓ. \(P_1/16\)
Correct Answer: \(P_1/8\)
Explanation: \(\textbf{Infer the adiabatic index from the slopes:}\)
For an isothermal ideal-gas curve,
\[
\left|\frac{dP}{dV}\right|_{\text{iso}}=\frac{P}{V}
\]
For a reversible adiabatic curve,
\[
\left|\frac{dP}{dV}\right|_{\text{ad}}=\gamma\frac{P}{V}
\]
Thus the ratio of slope magnitudes at the common state is
\[
\frac{|dP/dV|_{\text{ad}}}{|dP/dV|_{\text{iso}}}=\gamma
\]
The given ratio therefore gives
\[
\gamma=\frac32
\]
\(\textbf{Apply the adiabatic state relation:}\)
\[
P_1V_1^\gamma=P_2(4V_1)^\gamma
\]
Hence,
\[
P_2=P_1\left(\frac14\right)^{3/2}
\]
Since
\[
4^{3/2}=8,
\]
we obtain
\[
P_2=\frac{P_1}{8}
\]
\(\textbf{Final answer:}\)
\[
\boxed{P_2=\frac{P_1}{8}}
\]
216. From the same initial state, an ideal gas expands reversibly to the same final volume by an isothermal path and by an adiabatic path.
The isothermal curve lies above the adiabatic curve throughout the expansion after the common initial point. The area under the isothermal path is \(1000\,\text{J}\), and the area between the two curves is \(250\,\text{J}\).
Which pair correctly describes the adiabatic process?
ⓐ. \(W_{\text{ad}}=750\,\text{J},\ \Delta U_{\text{ad}}=-750\,\text{J}\)
ⓑ. \(W_{\text{ad}}=1250\,\text{J},\ \Delta U_{\text{ad}}=-1250\,\text{J}\)
ⓒ. \(W_{\text{ad}}=750\,\text{J},\ \Delta U_{\text{ad}}=0\)
ⓓ. \(W_{\text{ad}}=250\,\text{J},\ \Delta U_{\text{ad}}=-750\,\text{J}\)
Correct Answer: \(W_{\text{ad}}=750\,\text{J},\ \Delta U_{\text{ad}}=-750\,\text{J}\)
Explanation: \(\textbf{Read the work from the graph:}\)
The work done during quasi-static expansion equals the area under the corresponding \(P-V\) path. The isothermal work is
\[
W_{\text{iso}}=1000\,\text{J}
\]
The isothermal curve lies above the adiabatic curve, and the area between them is \(250\,\text{J}\). Therefore,
\[
W_{\text{ad}}=1000-250
\]
Hence,
\[
W_{\text{ad}}=750\,\text{J}
\]
\(\textbf{Apply the adiabatic energy balance:}\)
For the adiabatic process,
\[
Q_{\text{ad}}=0
\]
Thus,
\[
\Delta U_{\text{ad}}=Q_{\text{ad}}-W_{\text{ad}}
\]
Therefore,
\[
\Delta U_{\text{ad}}=-750\,\text{J}
\]
The internal energy decreases because the gas performs work without receiving compensating heat.
\(\textbf{Final answer:}\)
\[
\boxed{W_{\text{ad}}=+750\,\text{J},\qquad \Delta U_{\text{ad}}=-750\,\text{J}}
\]
217. An ideal gas undergoes two consecutive processes. In Stage I it expands adiabatically and does \(600\,\text{J}\) of work. In Stage II it is compressed isothermally at the lower temperature reached after Stage I, and the work done by the gas is \(-250\,\text{J}\). Which set gives the net heat, net work and net internal-energy change for the complete process?
ⓐ. \(Q_{\text{net}}=0,\ W_{\text{net}}=+350\,\text{J},\ \Delta U_{\text{net}}=-350\,\text{J}\)
ⓑ. \(Q_{\text{net}}=+250\,\text{J},\ W_{\text{net}}=+350\,\text{J},\ \Delta U_{\text{net}}=-100\,\text{J}\)
ⓒ. \(Q_{\text{net}}=-250\,\text{J},\ W_{\text{net}}=+850\,\text{J},\ \Delta U_{\text{net}}=-1100\,\text{J}\)
ⓓ. \(Q_{\text{net}}=-250\,\text{J},\ W_{\text{net}}=+350\,\text{J},\ \Delta U_{\text{net}}=-600\,\text{J}\)
Correct Answer: \(Q_{\text{net}}=-250\,\text{J},\ W_{\text{net}}=+350\,\text{J},\ \Delta U_{\text{net}}=-600\,\text{J}\)
Explanation: \(\textbf{Stage I — adiabatic expansion:}\)
\[
Q_1=0,\qquad W_1=+600\,\text{J}
\]
Therefore,
\[
\Delta U_1=Q_1-W_1=-600\,\text{J}
\]
\(\textbf{Stage II — isothermal compression:}\)
For an ideal gas in an isothermal process,
\[
\Delta U_2=0
\]
Hence,
\[
Q_2=W_2=-250\,\text{J}
\]
The negative heat means that heat is released during the compression.
\(\textbf{Combine the stages:}\)
\[
Q_{\text{net}}=0-250=-250\,\text{J},
\]
\[
W_{\text{net}}=600-250=350\,\text{J},
\]
and
\[
\Delta U_{\text{net}}=-600+0=-600\,\text{J}
\]
A direct check gives
\[
Q_{\text{net}}-W_{\text{net}}
=-250-350=-600\,\text{J}
\]
\(\textbf{Final answer:}\)
\[
\boxed{Q_{\text{net}}=-250\,\text{J},\quad W_{\text{net}}=+350\,\text{J},\quad \Delta U_{\text{net}}=-600\,\text{J}}
\]
218. An ideal gas is first compressed adiabatically so that the work done by the gas is \(-500\,\text{J}\). It is then cooled at constant volume until its temperature returns exactly to its initial value. What are the heat transfer during the second stage and the net internal-energy change for the complete two-stage process?
ⓐ. \(Q_2=-500\,\text{J},\ \Delta U_{\text{net}}=0\)
ⓑ. \(Q_2=+500\,\text{J},\ \Delta U_{\text{net}}=0\)
ⓒ. \(Q_2=-500\,\text{J},\ \Delta U_{\text{net}}=-500\,\text{J}\)
ⓓ. \(Q_2=0,\ \Delta U_{\text{net}}=+500\,\text{J}\)
Correct Answer: \(Q_2=-500\,\text{J},\ \Delta U_{\text{net}}=0\)
Explanation: \(\textbf{Adiabatic compression:}\)
For Stage I,
\[
Q_1=0
\]
The work done by the gas is
\[
W_1=-500\,\text{J}
\]
Hence,
\[
\Delta U_1=Q_1-W_1
=+500\,\text{J}
\]
The ideal-gas temperature therefore rises.
\(\textbf{Return to the initial temperature:}\)
The second stage brings the same ideal gas back to its original temperature. Since ideal-gas internal energy depends only on temperature, the final internal energy must equal the original internal energy. Therefore,
\[
\Delta U_2=-500\,\text{J}
\]
At constant volume,
\[
W_2=0
\]
Thus,
\[
Q_2=\Delta U_2=-500\,\text{J}
\]
The complete process has
\[
\Delta U_{\text{net}}=500-500=0
\]
\(\textbf{Final answer:}\)
\[
\boxed{Q_2=-500\,\text{J},\qquad \Delta U_{\text{net}}=0}
\]
219. An ideal gas initially occupies volume \(V_0\) at pressure \(P_0\) and temperature \(T_0\). It first undergoes free expansion in an insulated enclosure into vacuum until its volume becomes \(4V_0\). It is then compressed reversibly and adiabatically back to volume \(V_0\). If \(\gamma=3/2\), what are the final temperature and pressure after the second stage?
ⓐ. \(T_0,\ P_0\)
ⓑ. \(T_0/2,\ 2P_0\)
ⓒ. \(2T_0,\ P_0/2\)
ⓓ. \(2T_0,\ 2P_0\)
Correct Answer: \(2T_0,\ 2P_0\)
Explanation: \(\textbf{Stage I — insulated free expansion:}\)
The gas expands into vacuum, so
\[
W_1=0
\]
The enclosure is insulated, so
\[
Q_1=0
\]
Hence,
\[
\Delta U_1=0
\]
For an ideal gas,
\[
T_{\text{after free expansion}}=T_0
\]
Its volume is now \(4V_0\).
\(\textbf{Stage II — reversible adiabatic compression:}\)
Using
\[
TV^{\gamma-1}=\text{constant},
\]
we get
\[
T_f( V_0)^{1/2}
=T_0(4V_0)^{1/2}
\]
Therefore,
\[
T_f=2T_0
\]
At the final volume \(V_0\), the ideal-gas relation gives
\[
\frac{P_f}{P_0}=\frac{T_f}{T_0}=2
\]
Thus,
\[
P_f=2P_0
\]
The two-stage process does not restore the original state even though the original volume is recovered.
\(\textbf{Final answer:}\)
\[
\boxed{T_f=2T_0,\qquad P_f=2P_0}
\]
220. Starting from the same equilibrium state, the same ideal gas expands to \(4V_0\) by two different adiabatic processes.
Process P: Free expansion into vacuum inside an insulated enclosure.
Process Q: Reversible adiabatic expansion with \(\gamma=3/2\).
Which comparison of the final states is correct?
ⓐ. \(T_P=T_Q=T_0\) and \(P_P=P_Q=P_0/4\)
ⓑ. \(T_P=T_0,\ T_Q=T_0/2\), and \(P_P\gt P_Q\)
ⓒ. \(T_P=T_0/2,\ T_Q=T_0\), and \(P_P\lt P_Q\)
ⓓ. \(T_P=T_Q=T_0/2\), but \(P_P\gt P_Q\)
Correct Answer: \(T_P=T_0,\ T_Q=T_0/2\), and \(P_P\gt P_Q\)
Explanation: Both processes are adiabatic, so each has
\[
Q=0
\]
For free expansion into vacuum,
\[
W_P=0
\]
Therefore,
\[
\Delta U_P=0,
\]
and for an ideal gas,
\[
T_P=T_0
\]
At volume \(4V_0\),
\[
P_P=\frac{P_0}{4}
\]
For the reversible adiabatic expansion,
\[
T_Q=T_0\left(\frac{V_0}{4V_0}\right)^{\gamma-1}
\]
With \(\gamma=3/2\),
\[
T_Q=\frac{T_0}{2}
\]
Also,
\[
P_Q=P_0\left(\frac14\right)^{3/2}
=\frac{P_0}{8}
\]
Thus,
\[
P_P=\frac{P_0}{4}\gt \frac{P_0}{8}=P_Q
\]
The comparison demonstrates why “adiabatic” alone does not determine the temperature change; the work interaction matters.
