201. A fixed amount of an ideal gas has its RMS speed reduced to \(80\%\) of its initial value while its volume becomes \(1.25\) times the initial volume. Which ordered pair correctly gives \(\left(P_2/P_1,\rho_2/\rho_1\right)\)?
ⓐ. \(\left(\dfrac{16}{25},\dfrac45\right)\)
ⓑ. \(\left(\dfrac{64}{125},\dfrac45\right)\)
ⓒ. \(\left(\dfrac{16}{25},\dfrac54\right)\)
ⓓ. \(\left(\dfrac45,\dfrac{64}{125}\right)\)
Correct Answer: \(\left(\dfrac{64}{125},\dfrac45\right)\)
Explanation: \(\textbf{Convert the RMS-speed change into temperature:}\)
For the same gas species,
\[
T\propto v_{\rm rms}^2
\]
Thus,
\[
\frac{T_2}{T_1}
=
(0.80)^2
=
0.64
=
\frac{16}{25}
\]
\(\textbf{Use the fixed-amount gas relation:}\)
\[
\frac{P_2}{P_1}
=
\frac{T_2/T_1}{V_2/V_1}
\]
Therefore,
\[
\frac{P_2}{P_1}
=
\frac{16/25}{5/4}
\]
Hence,
\[
\frac{P_2}{P_1}
=
\frac{16}{25}\times\frac45
=
\frac{64}{125}
\]
\(\textbf{Determine the density change:}\)
The gas mass is fixed, so
\[
\rho\propto\frac1V
\]
Therefore,
\[
\frac{\rho_2}{\rho_1}
=
\frac{V_1}{V_2}
=
\frac45
\]
\(\textbf{Final answer:}\)
\[
\boxed{\left(\frac{P_2}{P_1},\frac{\rho_2}{\rho_1}\right)
=
\left(\frac{64}{125},\frac45\right)}
\]
202. A rigid vessel contains an ideal gas. After some gas escapes, the pressure becomes \(1.80\) times its initial value while the RMS speed of the remaining molecules becomes \(1.50\) times its initial value. What fraction of the original molecules escaped, and what is \(T_2/T_1\)?
ⓐ. \(10\%\) escaped; \(T_2/T_1=9/4\)
ⓑ. \(25\%\) escaped; \(T_2/T_1=3/2\)
ⓒ. \(20\%\) escaped; \(T_2/T_1=9/4\)
ⓓ. \(20\%\) escaped; \(T_2/T_1=3/2\)
Correct Answer: \(20\%\) escaped; \(T_2/T_1=9/4\)
Explanation: \(\textbf{Temperature from the speed change:}\)
For the same molecular species,
\[
\frac{T_2}{T_1}
=
\left(\frac{v_2}{v_1}\right)^2
\]
Therefore,
\[
\frac{T_2}{T_1}
=
(1.50)^2
=
2.25
=
\frac94
\]
\(\textbf{Use the rigid-vessel state equation:}\)
Since volume is unchanged,
\[
P\propto NT
\]
Hence,
\[
\frac{P_2}{P_1}
=
\frac{N_2}{N_1}
\frac{T_2}{T_1}
\]
Therefore,
\[
1.80
=
\frac{N_2}{N_1}\left(\frac94\right)
\]
Thus,
\[
\frac{N_2}{N_1}
=
1.80\times\frac49
=
0.80
\]
So \(80\%\) of the original molecules remain.
The escaped fraction is
\[
1-0.80=0.20
\]
\(\textbf{Final answer:}\) \(20\%\) escaped and \(T_2/T_1=9/4\).
203. A model molecular population contains \(50\%\) of its molecules with speed \(u\), \(25\%\) with speed \(2u\), and \(25\%\) with speed \(3u\). A reference population of the same molecular species has every molecule moving with speed \(u\). Which ordered pair correctly gives the RMS speed of the first population and the ratio of its average translational kinetic energy per molecule to that of the reference population?
ⓐ. \(\left(\dfrac{15}{8}u,\dfrac{15}{4}\right)\)
ⓑ. \(\left(\sqrt3\,u,3\right)\)
ⓒ. \(\left(\dfrac{\sqrt{15}}{2}u,\dfrac{15}{4}\right)\)
ⓓ. \(\left(2u,4\right)\)
Correct Answer: \(\left(\dfrac{\sqrt{15}}{2}u,\dfrac{15}{4}\right)\)
Explanation: \(\textbf{Construct the weighted mean square:}\)
RMS speed satisfies
\[
v_{\rm rms}^2=\langle v^2\rangle
\]
Therefore,
\[
\langle v^2\rangle
=
0.50u^2
+
0.25(2u)^2
+
0.25(3u)^2
\]
Thus,
\[
\langle v^2\rangle
=
0.50u^2+1.00u^2+2.25u^2
\]
Hence,
\[
\langle v^2\rangle
=
3.75u^2
=
\frac{15}{4}u^2
\]
Therefore,
\[
v_{\rm rms}
=
\frac{\sqrt{15}}{2}u
\]
\(\textbf{Compare average translational kinetic energies:}\)
For the same molecular mass,
\[
\overline K=\frac12m\langle v^2\rangle
\]
The reference population has \(\langle v^2\rangle_0=u^2\).
Hence,
\[
\frac{\overline K}{\overline K_0}
=
\frac{15}{4}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\left(v_{\rm rms},\frac{\overline K}{\overline K_0}\right)
=
\left(\frac{\sqrt{15}}2u,\frac{15}{4}\right)}
\]
204. Consider the following statements concerning RMS speed for a set of molecular speeds \(v_1,v_2,\ldots\).
Statement I: If every speed is multiplied by the same positive factor \(\lambda\), the RMS speed is also multiplied by \(\lambda\).
Statement II: If the same positive speed \(c\) is added to every molecular speed, the RMS speed necessarily increases by exactly \(c\).
Which option is correct?
ⓐ. Both Statement I and Statement II are incorrect.
ⓑ. Statement I is correct but Statement II is incorrect.
ⓒ. Both Statement I and Statement II are correct.
ⓓ. Statement I is incorrect but Statement II is correct.
Correct Answer: Statement I is correct but Statement II is incorrect.
Explanation: The RMS speed is defined by
\[
v_{\rm rms}=\sqrt{\langle v^2\rangle}.
\]
If every molecular speed becomes \(\lambda v\), then
\[
v_{\rm rms}'=
\sqrt{\langle \lambda^2v^2\rangle}.
\]
Therefore,
\[
v_{\rm rms}'=\lambda\sqrt{\langle v^2\rangle}
=\lambda v_{\rm rms},
\]
so Statement I is correct.
If instead every speed becomes \(v+c\),
\[
v_{\rm rms}'^{\,2}
=
\langle(v+c)^2\rangle.
\]
Expanding,
\[
v_{\rm rms}'^{\,2}
=
\langle v^2\rangle+2c\langle v\rangle+c^2.
\]
This is not generally equal to
\[
(v_{\rm rms}+c)^2
=
\langle v^2\rangle+2cv_{\rm rms}+c^2.
\]
Since \(\langle v\rangle\) need not equal \(v_{\rm rms}\), Statement II is not generally true.
205. Assertion: If a fixed amount of an ideal gas is compressed isothermally from volume \(V\) to \(V/2\), its pressure doubles but its RMS speed remains unchanged.
Reason: For a fixed amount at constant temperature \(P\propto1/V\), whereas \(v_{\rm rms}=\sqrt{3RT/M}\) depends only on temperature and molar mass for a given gas.
ⓐ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
ⓑ. Both Assertion and Reason are true, but Reason is not the correct explanation of 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 is the correct explanation of Assertion.
Explanation: The amount of gas is fixed and the process is explicitly isothermal.
Therefore the ideal-gas equation gives
\[
PV=\text{constant}.
\]
When the volume changes from \(V\) to \(V/2\),
\[
P_2=2P_1.
\]
Thus the pressure part of the Assertion is correct.
For the same gas species,
\[
v_{\rm rms}=\sqrt{\frac{3RT}{M}}.
\]
Neither \(T\) nor \(M\) changes in the stated compression.
Consequently,
\[
v_{{\rm rms},2}=v_{{\rm rms},1}.
\]
The greater pressure is produced by the greater molecular number density, not by a greater RMS speed.
The Reason correctly identifies both the inverse pressure-volume relation and the temperature dependence of RMS speed.
Therefore both statements are true and the Reason correctly explains the Assertion.
206. An ideal gas has pressure \(1.38\times10^5\,\text{Pa}\), molecular number density \(2.0\times10^{25}\,\text{m}^{-3}\), and mass density \(0.80\,\text{kg m}^{-3}\). Take
\[
k_B=1.38\times10^{-23}\,\text{J K}^{-1},
\qquad
N_A=6.0\times10^{23}\,\text{mol}^{-1}.
\]
Which ordered triple is closest to \((T,M,v_{\rm rms})\)?
ⓐ. \((250\,\text{K},48\,\text{g mol}^{-1},5.1\times10^2\,\text{m s}^{-1})\)
ⓑ. \((500\,\text{K},24\,\text{g mol}^{-1},7.2\times10^2\,\text{m s}^{-1})\)
ⓒ. \((500\,\text{K},48\,\text{g mol}^{-1},5.1\times10^2\,\text{m s}^{-1})\)
ⓓ. \((1000\,\text{K},24\,\text{g mol}^{-1},1.02\times10^3\,\text{m s}^{-1})\)
Correct Answer: \((500\,\text{K},24\,\text{g mol}^{-1},7.2\times10^2\,\text{m s}^{-1})\)
Explanation: \(\textbf{Infer temperature from molecular number density:}\)
\[
P=nk_BT
\]
Therefore,
\[
T=\frac{P}{nk_B}
\]
Substituting,
\[
T=
\frac{1.38\times10^5}
{(2.0\times10^{25})(1.38\times10^{-23})}
\]
The denominator is
\[
2.76\times10^2,
\]
so
\[
T=500\,\text{K}
\]
\(\textbf{Infer the mass of one molecule:}\)
Since
\[
\rho=nm,
\]
\[
m=\frac{\rho}{n}
=
\frac{0.80}{2.0\times10^{25}}
=
4.0\times10^{-26}\,\text{kg}
\]
Thus the molar mass is
\[
M=mN_A
=
(4.0\times10^{-26})(6.0\times10^{23})
=
2.4\times10^{-2}\,\text{kg mol}^{-1}
=
24\,\text{g mol}^{-1}
\]
\(\textbf{Obtain RMS speed from pressure and mass density:}\)
\[
v_{\rm rms}=\sqrt{\frac{3P}{\rho}}
\]
Hence,
\[
v_{\rm rms}
=
\sqrt{\frac{3(1.38\times10^5)}{0.80}}
\approx7.2\times10^2\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{(T,M,v_{\rm rms})
\approx
(500\,\text{K},24\,\text{g mol}^{-1},7.2\times10^2\,\text{m s}^{-1})}
\]
207. Consider the following statements about the two forms of the RMS-speed equation.
Statement I: If \(m\) is the mass of one molecule, \(v_{\rm rms}=\sqrt{3RT/m}\).
Statement II: If \(m\) is the mass of one molecule, \(v_{\rm rms}=\sqrt{3RT/(mN_A)}\).
Statement III: If \(M\) is molar mass, \(v_{\rm rms}=\sqrt{3k_BT/M}\).
Statement IV: The identity \(R=N_Ak_B\) makes the correct molecular and molar forms equivalent.
Which combination is correct?
ⓐ. II and IV only
ⓑ. I and II only
ⓒ. I, III and IV only
ⓓ. II, III and IV only
Correct Answer: II and IV only
Explanation: For the mass \(m\) of one molecule, the molecular form is
\[
v_{\rm rms}=\sqrt{\frac{3k_BT}{m}}.
\]
Statement I incorrectly combines the molar gas constant \(R\) directly with a single-molecule mass.
The molar mass corresponding to molecular mass \(m\) is
\[
M=mN_A.
\]
Thus the molar formula becomes
\[
v_{\rm rms}
=
\sqrt{\frac{3RT}{mN_A}},
\]
so Statement II is correct.
Using
\[
R=N_Ak_B,
\]
this reduces to
\[
\sqrt{\frac{3N_Ak_BT}{mN_A}}
=
\sqrt{\frac{3k_BT}{m}}.
\]
Statement III is incorrect because \(k_B\) must be paired with molecular mass, whereas \(R\) is paired with molar mass.
Statement IV correctly states the bridge between the two descriptions.
Therefore only Statements II and IV are correct.
208. Two ideal gases X and Y have equal translational kinetic-energy densities,
\[
\frac{E_{{\rm tr},X}}{V_X}
=
\frac{E_{{\rm tr},Y}}{V_Y}.
\]
Their mass densities and molar masses satisfy
\[
\rho_X=4\rho_Y,
\qquad
M_X=2M_Y.
\]
Which ordered triple correctly gives
\[
\left(
\frac{P_X}{P_Y},
\frac{v_{{\rm rms},X}}{v_{{\rm rms},Y}},
\frac{T_X}{T_Y}
\right)?
\]
ⓐ. \((4,\frac12,2)\)
ⓑ. \((1,2,2)\)
ⓒ. \((\frac14,2,\frac12)\)
ⓓ. \((1,\frac12,\frac12)\)
Correct Answer: \((1,\frac12,\frac12)\)
Explanation: \(\textbf{Convert energy density into pressure:}\)
For an ideal gas,
\[
\frac{E_{\rm tr}}{V}=\frac32P
\]
Equal translational kinetic-energy densities therefore imply
\[
P_X=P_Y
\]
Hence,
\[
\frac{P_X}{P_Y}=1
\]
\(\textbf{Use pressure and mass density to compare RMS speeds:}\)
\[
P=\frac13\rho v_{\rm rms}^2
\]
At equal pressure,
\[
\rho_Xv_X^2=\rho_Yv_Y^2
\]
Therefore,
\[
\frac{v_X^2}{v_Y^2}
=
\frac{\rho_Y}{\rho_X}
=
\frac14,
\]
so
\[
\frac{v_X}{v_Y}=\frac12
\]
\(\textbf{Convert the speed ratio into a temperature ratio:}\)
Since
\[
v_{\rm rms}^2=\frac{3RT}{M},
\]
\[
\frac{T_X}{T_Y}
=
\frac{M_X}{M_Y}
\frac{v_X^2}{v_Y^2}
\]
Thus,
\[
\frac{T_X}{T_Y}
=
2\left(\frac14\right)
=
\frac12
\]
\(\textbf{Final answer:}\)
\[
\boxed{\left(1,\frac12,\frac12\right)}
\]
209. The same ideal-gas species changes between two states according to
\[
P_2=0.90P_1,\qquad
N_2=0.75N_1,\qquad
v_{{\rm rms},2}=1.20v_{{\rm rms},1}.
\]
Which ordered pair correctly gives \(\left(V_2/V_1,\rho_2/\rho_1\right)\)?
ⓐ. \(\left(\dfrac54,\dfrac35\right)\)
ⓑ. \(\left(\dfrac45,\dfrac{15}{16}\right)\)
ⓒ. \(\left(\dfrac65,\dfrac34\right)\)
ⓓ. \(\left(\dfrac65,\dfrac58\right)\)
Correct Answer: \(\left(\dfrac65,\dfrac58\right)\)
Explanation: \(\textbf{Write pressure directly in terms of molecular speed:}\)
For the same species,
\[
P=\frac13\frac{Nm}{V}v_{\rm rms}^2
\]
Therefore,
\[
\frac{P_2}{P_1}
=
\frac{N_2}{N_1}
\frac{V_1}{V_2}
\left(\frac{v_2}{v_1}\right)^2
\]
Substituting,
\[
0.90
=
0.75\frac{V_1}{V_2}(1.20)^2
\]
Since
\[
(1.20)^2=1.44,
\]
\[
0.90
=
1.08\frac{V_1}{V_2}
\]
Hence,
\[
\frac{V_2}{V_1}
=
\frac{1.08}{0.90}
=
1.20
=
\frac65
\]
\(\textbf{Determine the density ratio:}\)
For the same molecular species,
\[
\rho\propto\frac NV
\]
Thus,
\[
\frac{\rho_2}{\rho_1}
=
\frac{0.75}{1.20}
=
0.625
=
\frac58
\]
\(\textbf{Final answer:}\)
\[
\boxed{\left(\frac{V_2}{V_1},\frac{\rho_2}{\rho_1}\right)
=
\left(\frac65,\frac58\right)}
\]
210. The same ideal-gas species is observed in two states for which
\[
\rho_2=4\rho_1,\qquad
v_{{\rm rms},2}=\frac{v_{{\rm rms},1}}{\sqrt2},
\qquad
V_2=\frac12V_1.
\]
Which ordered triple correctly gives
\[
\left(
\frac{P_2}{P_1},
\frac{T_2}{T_1},
\frac{N_2}{N_1}
\right)?
\]
ⓐ. \((4,\frac12,2)\)
ⓑ. \((2,2,\frac12)\)
ⓒ. \((2,\frac12,2)\)
ⓓ. \((1,\frac12,4)\)
Correct Answer: \((2,\frac12,2)\)
Explanation: \(\textbf{Pressure from mass density and RMS speed:}\)
\[
P=\frac13\rho v_{\rm rms}^2
\]
Thus,
\[
\frac{P_2}{P_1}
=
\frac{\rho_2}{\rho_1}
\left(\frac{v_2}{v_1}\right)^2
\]
Substituting,
\[
\frac{P_2}{P_1}
=
4\left(\frac1{\sqrt2}\right)^2
=
4\left(\frac12\right)
=2
\]
\(\textbf{Temperature from RMS speed:}\)
For the same species,
\[
T\propto v_{\rm rms}^2
\]
Therefore,
\[
\frac{T_2}{T_1}=\frac12
\]
\(\textbf{Molecule number from density and volume:}\)
For fixed molecular mass,
\[
N\propto\rho V
\]
Hence,
\[
\frac{N_2}{N_1}
=
\frac{\rho_2}{\rho_1}
\frac{V_2}{V_1}
\]
Thus,
\[
\frac{N_2}{N_1}
=
4\left(\frac12\right)=2
\]
All three ratios are mutually consistent with \(PV=Nk_BT\).
\(\textbf{Final answer:}\)
\[
\boxed{\left(2,\frac12,2\right)}
\]
211. Consider the following graph statements for a fixed ideal-gas species.
Statement I: A graph of \(v_{\rm rms}^2\) against absolute temperature \(T\) is a straight line through the origin.
Statement II: A graph of \(v_{\rm rms}\) against \(\sqrt T\) is a straight line through the origin.
Statement III: A graph of \(v_{\rm rms}\) against \(T\) is a straight line through the origin.
Statement IV: At fixed temperature, a graph of \(1/v_{\rm rms}^2\) against molar mass \(M\) is a straight line through the origin.
Which combination is correct?
ⓐ. I and III only
ⓑ. I, II, III and IV
ⓒ. I, II and IV only
ⓓ. II and IV only
Correct Answer: I, II and IV only
Explanation: For an ideal gas,
\[
v_{\rm rms}=\sqrt{\frac{3RT}{M}}.
\]
For fixed \(M\),
\[
v_{\rm rms}^2=\frac{3R}{M}T.
\]
Therefore Statement I is correct.
The same equation can be written as
\[
v_{\rm rms}
=
\sqrt{\frac{3R}{M}}\sqrt T,
\]
so Statement II is also correct.
However,
\[
v_{\rm rms}\propto\sqrt T,
\]
not \(T\), so a \(v_{\rm rms}\)-versus-\(T\) graph is not a straight line through the origin.
Statement III is therefore incorrect.
At fixed \(T\),
\[
\frac1{v_{\rm rms}^2}
=
\frac{M}{3RT},
\]
which is directly proportional to \(M\).
Thus Statement IV is correct.
Hence Statements I, II and IV are correct.
212. An unknown ideal gas and a reference ideal gas are maintained at the same absolute temperature. The unknown gas has RMS speed equal to \(3/4\) of that of the reference gas. The reference molar mass is \(18\,\text{g mol}^{-1}\). Equal volumes of the two gases are then taken at the same pressure and the same common temperature. Which ordered pair correctly gives the unknown molar mass and the ratio of the total gas masses \(m_{g,u}/m_{g,r}\)?
ⓐ. \(\left(24\,\text{g mol}^{-1},\dfrac43\right)\)
ⓑ. \(\left(32\,\text{g mol}^{-1},\dfrac{16}{9}\right)\)
ⓒ. \(\left(32\,\text{g mol}^{-1},\dfrac34\right)\)
ⓓ. \(\left(40.5\,\text{g mol}^{-1},\dfrac94\right)\)
Correct Answer: \(\left(32\,\text{g mol}^{-1},\dfrac{16}{9}\right)\)
Explanation: \(\textbf{Infer molar mass from the equal-temperature speed comparison:}\)
At fixed temperature,
\[
v_{\rm rms}\propto\frac1{\sqrt M}
\]
Therefore,
\[
\frac{v_u}{v_r}
=
\sqrt{\frac{M_r}{M_u}}
\]
Given,
\[
\frac{v_u}{v_r}=\frac34
\]
Squaring,
\[
\frac{9}{16}
=
\frac{M_r}{M_u}
\]
Thus,
\[
M_u
=
\frac{16}{9}M_r
=
\frac{16}{9}(18)
=
32\,\text{g mol}^{-1}
\]
\(\textbf{Compare molecule numbers at the common gas state:}\)
From
\[
PV=\mu RT,
\]
equal pressure, volume and temperature imply equal mole numbers.
Therefore the two selected samples contain equal numbers of molecules.
Their total gas masses are consequently proportional to their molar masses:
\[
\frac{m_{g,u}}{m_{g,r}}
=
\frac{M_u}{M_r}
=
\frac{16}{9}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\left(M_u,\frac{m_{g,u}}{m_{g,r}}\right)
=
\left(32\,\text{g mol}^{-1},\frac{16}{9}\right)}
\]
213. Two ideal-gas samples X and Y have equal total gas masses and equal total translational kinetic energies. Their molar masses satisfy
\[
M_X=4M_Y.
\]
Which ordered triple correctly gives
\[
\left(
\frac{T_X}{T_Y},
\frac{N_X}{N_Y},
\frac{v_{{\rm rms},X}}{v_{{\rm rms},Y}}
\right)?
\]
ⓐ. \(\left(4,1,\dfrac12\right)\)
ⓑ. \(\left(1,\dfrac14,2\right)\)
ⓒ. \(\left(\dfrac14,4,1\right)\)
ⓓ. \(\left(4,\dfrac14,1\right)\)
Correct Answer: \(\left(4,\dfrac14,1\right)\)
Explanation: \(\textbf{Use equality of total sample masses:}\)
The total gas mass is
\[
m_g=N\frac{M}{N_A}
\]
Since \(m_{g,X}=m_{g,Y}\),
\[
N_XM_X=N_YM_Y
\]
Given
\[
M_X=4M_Y,
\]
we obtain
\[
\frac{N_X}{N_Y}=\frac14
\]
\(\textbf{Use equality of total translational energies:}\)
For an ideal gas,
\[
E_{\rm tr}=\frac32Nk_BT
\]
Since \(E_{{\rm tr},X}=E_{{\rm tr},Y}\),
\[
N_XT_X=N_YT_Y
\]
Therefore,
\[
\frac{T_X}{T_Y}
=
\frac{N_Y}{N_X}
=4
\]
\(\textbf{Compare RMS speeds:}\)
\[
v_{\rm rms}^2=\frac{3RT}{M}
\]
Hence,
\[
\frac{v_{{\rm rms},X}^2}{v_{{\rm rms},Y}^2}
=
\frac{T_X/T_Y}{M_X/M_Y}
=
\frac44=1
\]
Thus,
\[
v_{{\rm rms},X}=v_{{\rm rms},Y}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\left(4,\frac14,1\right)}
\]
214.
For two sealed ideal-gas samples X and Y, total translational kinetic energy \(E_{\rm tr}\) is plotted against \(v_{\rm rms}^2\) while their states are varied. Both graphs are straight lines through the origin. The slope ratio is \(s_X/s_Y=3\). Their molar masses satisfy \(M_X/M_Y=1/2\).
Which ordered pair correctly gives the ratio of total gas masses and molecule numbers?
ⓐ. \((3,3)\)
ⓑ. \((6,3)\)
ⓒ. \((3,6)\)
ⓓ. \(\left(\dfrac32,6\right)\)
Correct Answer: \((3,6)\)
Explanation: \(\textbf{Interpret the energy-speed graph:}\)
For a complete gas sample of total mass \(m_g\),
\[
E_{\rm tr}=\frac12m_gv_{\rm rms}^2
\]
Thus the slope of an \(E_{\rm tr}\)-versus-\(v_{\rm rms}^2\) graph is
\[
s=\frac{m_g}{2}
\]
Therefore,
\[
\frac{m_{g,X}}{m_{g,Y}}
=
\frac{s_X}{s_Y}
=3
\]
\(\textbf{Relate total mass to molecule number:}\)
For each gas,
\[
m_g=N\frac{M}{N_A}
\]
Hence,
\[
\frac{m_{g,X}}{m_{g,Y}}
=
\frac{N_X}{N_Y}
\frac{M_X}{M_Y}
\]
Substituting,
\[
3
=
\frac{N_X}{N_Y}\left(\frac12\right)
\]
Therefore,
\[
\frac{N_X}{N_Y}=6
\]
The graph reveals total sample mass, while the independent molar-mass ratio converts that information into a molecular-population ratio.
\(\textbf{Final answer:}\)
\[
\boxed{\left(\frac{m_{g,X}}{m_{g,Y}},\frac{N_X}{N_Y}\right)
=
(3,6)}
\]
215.
For ideal gases X and Y, \(P/\rho\) is plotted against absolute temperature \(T\). Both graphs are straight lines through the origin, with \(s_X/s_Y=1/4\). In a particular comparison, \(T_X=9T_Y\).
Which ordered pair correctly gives
\[
\left(\frac{M_X}{M_Y},\frac{v_{{\rm rms},X}}{v_{{\rm rms},Y}}\right)?
\]
ⓐ. \(\left(\dfrac14,6\right)\)
ⓑ. \(\left(4,\dfrac32\right)\)
ⓒ. \(\left(4,\dfrac23\right)\)
ⓓ. \(\left(\dfrac14,\dfrac32\right)\)
Correct Answer: \(\left(4,\dfrac32\right)\)
Explanation: \(\textbf{Read molar mass from the graph slope:}\)
The density form of the ideal-gas equation gives
\[
\frac{P}{\rho}=\frac{RT}{M}
\]
Therefore the slope of the \(P/\rho\)-versus-\(T\) graph is
\[
s=\frac{R}{M}
\]
Hence,
\[
\frac{s_X}{s_Y}
=
\frac{M_Y}{M_X}
\]
Given
\[
\frac{s_X}{s_Y}=\frac14,
\]
we obtain
\[
\frac{M_X}{M_Y}=4
\]
\(\textbf{Use the selected temperatures to compare speeds:}\)
The RMS-speed relation is
\[
v_{\rm rms}^2=\frac{3RT}{M}
\]
Therefore,
\[
\frac{v_{{\rm rms},X}^2}{v_{{\rm rms},Y}^2}
=
\frac{T_X/T_Y}{M_X/M_Y}
\]
Substituting,
\[
\frac{v_{{\rm rms},X}^2}{v_{{\rm rms},Y}^2}
=
\frac94
\]
Thus,
\[
\frac{v_{{\rm rms},X}}{v_{{\rm rms},Y}}
=
\frac32
\]
\(\textbf{Final answer:}\)
\[
\boxed{\left(4,\frac32\right)}
\]
216. Ideal gases X and Y have equal pressures and equal RMS speeds. Their molar masses satisfy
\[
M_X=9M_Y.
\]
Which ordered triple is correct?
\[
\left(
\frac{\rho_X}{\rho_Y},
\frac{T_X}{T_Y},
\frac{n_X}{n_Y}
\right)
\]
ⓐ. \(\left(1,\dfrac19,9\right)\)
ⓑ. \(\left(9,1,\dfrac19\right)\)
ⓒ. \(\left(1,9,\dfrac19\right)\)
ⓓ. \(\left(9,9,1\right)\)
Correct Answer: \(\left(1,9,\dfrac19\right)\)
Explanation: \(\textbf{Use equal pressure and equal RMS speed:}\)
The kinetic pressure relation is
\[
P=\frac13\rho v_{\rm rms}^2
\]
Since both \(P\) and \(v_{\rm rms}\) are equal,
\[
\rho_X=\rho_Y
\]
Thus,
\[
\frac{\rho_X}{\rho_Y}=1
\]
\(\textbf{Compare temperatures:}\)
Using
\[
v_{\rm rms}^2=\frac{3RT}{M},
\]
equal RMS speeds require
\[
\frac{T_X}{M_X}
=
\frac{T_Y}{M_Y}
\]
Therefore,
\[
\frac{T_X}{T_Y}
=
\frac{M_X}{M_Y}
=9
\]
\(\textbf{Compare number densities:}\)
For a gas species,
\[
\rho=n\frac{M}{N_A}
\]
At equal mass density,
\[
\frac{n_X}{n_Y}
=
\frac{M_Y}{M_X}
=
\frac19
\]
Thus equal pressure and equal RMS speed do not require equal temperature or equal molecular number density when the molecular masses differ.
\(\textbf{Final answer:}\)
\[
\boxed{\left(1,9,\frac19\right)}
\]
217. In a model molecular population, a fraction \(f\) of the molecules has speed \(u\), while the remaining fraction \(1-f\) has speed \(3u\). The RMS speed of the population is \(\sqrt5\,u\). Which ordered pair correctly gives \(f\) and the arithmetic mean speed?
ⓐ. \(\left(\dfrac12,2u\right)\)
ⓑ. \(\left(\dfrac34,\dfrac32u\right)\)
ⓒ. \(\left(\dfrac12,\sqrt5u\right)\)
ⓓ. \(\left(\dfrac14,\dfrac52u\right)\)
Correct Answer: \(\left(\dfrac12,2u\right)\)
Explanation: \(\textbf{Use the RMS information to recover the population fraction:}\)
By definition,
\[
v_{\rm rms}^2=\langle v^2\rangle
\]
Therefore,
\[
5u^2
=
fu^2+(1-f)(3u)^2
\]
Dividing by \(u^2\),
\[
5=f+9(1-f)
\]
Thus,
\[
5=9-8f
\]
Hence,
\[
8f=4,
\qquad
f=\frac12
\]
\(\textbf{Now calculate the ordinary mean speed:}\)
\[
\overline v
=
fu+(1-f)(3u)
\]
Substituting \(f=1/2\),
\[
\overline v
=
\frac12u+\frac12(3u)
=2u
\]
This reverse inference also shows why the RMS value cannot generally be replaced by the arithmetic mean speed.
\(\textbf{Final answer:}\)
\[
\boxed{\left(f,\overline v\right)
=
\left(\frac12,2u\right)}
\]
218. Consider two ideal gases for which the ratio \(P/\rho\) is the same.
Statement I: Their RMS speeds must be equal.
Statement II: Their absolute temperatures must be equal even if their molar masses are different.
Which option is correct?
ⓐ. Both Statement I and Statement II are incorrect.
ⓑ. Statement I is correct but Statement II is incorrect.
ⓒ. Both Statement I and Statement II are correct.
ⓓ. Statement I is incorrect but Statement II is correct.
Correct Answer: Statement I is correct but Statement II is incorrect.
Explanation: The kinetic pressure relation gives
\[
P=\frac13\rho v_{\rm rms}^2.
\]
Therefore,
\[
\frac{P}{\rho}
=
\frac13v_{\rm rms}^2.
\]
If two gases have equal \(P/\rho\), they must have equal \(v_{\rm rms}^2\), and hence equal RMS speeds.
Thus Statement I is correct.
However, the ideal-gas density relation gives
\[
\frac{P}{\rho}
=
\frac{RT}{M}.
\]
For a common value of \(P/\rho\),
\[
\frac{T}{M}
\]
must be the same.
If the molar masses differ, the temperatures must differ in the same proportion.
Thus equality of \(P/\rho\) does not generally imply equality of temperature.
Statement II is therefore incorrect.
219. For ideal gases of different molar masses, which transformed graph would make every gas lie on the same straight line through the origin?
ⓐ. Plot \(v_{\rm rms}^2/M\) against \(T\).
ⓑ. Plot \(v_{\rm rms}\sqrt M\) against \(T\).
ⓒ. Plot \(v_{\rm rms}/\sqrt M\) against \(\sqrt T\).
ⓓ. Plot \(v_{\rm rms}\sqrt M\) against \(\sqrt T\).
Correct Answer: Plot \(v_{\rm rms}\sqrt M\) against \(\sqrt T\).
Explanation: The RMS-speed relation is
\[
v_{\rm rms}=\sqrt{\frac{3RT}{M}}.
\]
Multiplying by \(\sqrt M\),
\[
v_{\rm rms}\sqrt M
=
\sqrt{3R}\sqrt T.
\]
Thus a plot of \(v_{\rm rms}\sqrt M\) against \(\sqrt T\) is a straight line through the origin.
Its slope is
\[
\sqrt{3R},
\]
which is independent of the gas species.
In option B, the vertical quantity varies as \(\sqrt T\), not as \(T\), so the graph is not linear in the stated axes.
For option C,
\[
\frac{v_{\rm rms}}{\sqrt M}
\propto\frac{\sqrt T}{M},
\]
so the slope still depends on molar mass.
For option A,
\[
\frac{v_{\rm rms}^2}{M}
=
\frac{3RT}{M^2},
\]
whose slope also depends on the particular gas.
Therefore only option D produces a universal straight line.
220. Consider the following relations for an ideal gas.
Statement I:
\[
\frac{Mv_{\rm rms}^2}{T}=3R.
\]
Statement II:
\[
\frac{mv_{\rm rms}^2}{T}=3k_B.
\]
Statement III:
\[
\frac{P}{\rho}=\frac{v_{\rm rms}^2}{3}.
\]
Statement IV:
\[
\frac{E_{\rm tr}}{m_g}=\frac{v_{\rm rms}^2}{3},
\]
where \(m_g\) is the total gas mass.
Which combination is correct?
ⓐ. II, III and IV only
ⓑ. I, II and III only
ⓒ. I and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: For molar mass \(M\),
\[
v_{\rm rms}^2=\frac{3RT}{M}.
\]
Rearranging,
\[
\frac{Mv_{\rm rms}^2}{T}=3R,
\]
so Statement I is correct.
For molecular mass \(m\),
\[
v_{\rm rms}^2=\frac{3k_BT}{m}.
\]
Hence,
\[
\frac{mv_{\rm rms}^2}{T}=3k_B,
\]
so Statement II is also correct.
The kinetic pressure equation is
\[
P=\frac13\rho v_{\rm rms}^2.
\]
Therefore,
\[
\frac{P}{\rho}=\frac{v_{\rm rms}^2}{3},
\]
making Statement III correct.
For the complete gas sample,
\[
E_{\rm tr}=\frac12m_gv_{\rm rms}^2.
\]
Thus,
\[
\frac{E_{\rm tr}}{m_g}
=
\frac12v_{\rm rms}^2,
\]
not \(v_{\rm rms}^2/3\).
Therefore Statement IV is incorrect.
