401. In a highly evacuated tube, the mean free path becomes comparable with or larger than a characteristic dimension of the tube.
Statement I: This does not mean that every molecule must reach the wall without any intermolecular collision.
Statement II: Further reduction of number density increases the mean free path and makes intermolecular collisions less frequent.
Which option is correct?
ⓐ. Both Statement I and Statement II are correct.
ⓑ. Statement I is incorrect but Statement II is correct.
ⓒ. Statement I is correct but Statement II is incorrect.
ⓓ. Both Statement I and Statement II are incorrect.
Correct Answer: Both Statement I and Statement II are correct.
Explanation: Mean free path is an average distance between successive molecular collisions.
It is not a deterministic distance that every molecule must travel before colliding.
Therefore even when
\[
\lambda
\]
exceeds a characteristic vessel dimension, some molecules may still undergo intermolecular collisions before reaching a wall.
Other molecules may travel unusually long distances without such a collision.
Thus Statement I is correct.
The molecular formula
\[
\lambda=\frac{1}{\sqrt2\,\pi nd^2}
\]
shows that at fixed molecular diameter,
\[
\lambda\propto\frac1n.
\]
Reducing number density therefore increases the average free-flight distance.
A longer mean free path corresponds qualitatively to less frequent intermolecular encounters.
Hence Statement II is also correct.
402. Two samples of the same ideal gas start at the same temperature. Sample X is heated to \(2T\) in a rigid vessel, while sample Y is heated to \(2T\) at constant pressure. Molecular diameter is unchanged. Which ordered pair correctly gives
\[
\left(\frac{\lambda_{X,f}}{\lambda_{X,i}},\frac{\lambda_{Y,f}}{\lambda_{Y,i}}\right)?
\]
ⓐ. \((\frac12,2)\)
ⓑ. \((2,1)\)
ⓒ. \((2,2)\)
ⓓ. \((1,2)\)
Correct Answer: \((1,2)\)
Explanation: \(\textbf{Sample X — rigid vessel:}\)
The molecule number and volume are fixed.
Therefore,
\[
n=\frac NV
\]
remains unchanged.
Since the molecular diameter also remains unchanged,
\[
\lambda=\frac{1}{\sqrt2\,\pi nd^2}
\]
does not change.
Thus,
\[
\frac{\lambda_{X,f}}{\lambda_{X,i}}=1
\]
\(\textbf{Sample Y — constant pressure:}\)
For fixed molecular diameter,
\[
\lambda
=
\frac{k_BT}{\sqrt2\,\pi d^2P}
\]
At constant pressure,
\[
\lambda\propto T
\]
The temperature doubles, so
\[
\frac{\lambda_{Y,f}}{\lambda_{Y,i}}=2
\]
Temperature alone therefore does not determine how mean free path changes; the accompanying density or pressure condition is essential.
\(\textbf{Final answer:}\)
\[
\boxed{(1,2)}
\]
403. Match each change in Column I with the corresponding factor by which the mean free path changes in Column II.
| Column I | Change | Column II | \(\lambda_f/\lambda_i\) |
|---|
| 1 | \(n\to2n,\;d\) unchanged | p | \(\frac12\) |
| 2 | \(d\to2d,\;n\) unchanged | q | \(\frac14\) |
| 3 | \(P\to3P,\;T\to6T,\;d\) unchanged | r | \(2\) |
| 4 | Fixed \(N,d\), \(V\to V/3\) | s | \(\frac13\) |
Choose the correct matching.
ⓐ. \(1-s,\;2-q,\;3-r,\;4-p\)
ⓑ. \(1-p,\;2-r,\;3-q,\;4-s\)
ⓒ. \(1-p,\;2-q,\;3-r,\;4-s\)
ⓓ. \(1-q,\;2-p,\;3-s,\;4-r\)
Correct Answer: \(1-p,\;2-q,\;3-r,\;4-s\)
Explanation: The basic molecular relation is
\[
\lambda\propto\frac1{nd^2}.
\]
Thus doubling number density gives
\[
\lambda_f/\lambda_i=\frac12,
\]
so \(1-p\).
Doubling molecular diameter gives
\[
\lambda_f/\lambda_i=\frac1{2^2}=\frac14,
\]
so \(2-q\).
Using
\[
\lambda\propto\frac{T}{P d^2},
\]
the third change gives
\[
\frac{\lambda_f}{\lambda_i}
=
\frac63
=2,
\]
so \(3-r\).
For fixed \(N\) and \(d\),
\[
n=\frac NV
\]
implies
\[
\lambda\propto V.
\]
Therefore reducing the volume to \(V/3\) gives
\[
\lambda_f/\lambda_i=\frac13,
\]
so \(4-s\).
Hence the matching is
\[
1-p,\;2-q,\;3-r,\;4-s.
\]
404. An ideal gas at \(300\,\text{K}\) has pressure
\[
1.38\times10^5\,\text{Pa}
\]
and measured mean free path
\[
1.69\times10^{-7}\,\text{m}.
\]
Take \(k_B=1.38\times10^{-23}\,\text{J K}^{-1}\). Using the NCERT mean-free-path relation, which ordered pair is closest to its number density and molecular diameter?
ⓐ. \((3.33\times10^{25}\,\text{m}^{-3},2.0\times10^{-10}\,\text{m})\)
ⓑ. \((3.33\times10^{24}\,\text{m}^{-3},4.0\times10^{-10}\,\text{m})\)
ⓒ. \((6.67\times10^{25}\,\text{m}^{-3},1.0\times10^{-10}\,\text{m})\)
ⓓ. \((1.00\times10^{25}\,\text{m}^{-3},2.0\times10^{-9}\,\text{m})\)
Correct Answer: \((3.33\times10^{25}\,\text{m}^{-3},2.0\times10^{-10}\,\text{m})\)
Explanation: \(\textbf{Infer number density from the ideal-gas equation:}\)
\[
P=nk_BT
\]
Therefore,
\[
n=\frac{P}{k_BT}
\]
Substituting,
\[
n
=
\frac{1.38\times10^5}
{(1.38\times10^{-23})(300)}
\]
Hence,
\[
n=3.33\times10^{25}\,\text{m}^{-3}
\]
\(\textbf{Use the measured collision length to infer molecular diameter:}\)
\[
\lambda=\frac{1}{\sqrt2\,\pi nd^2}
\]
Thus,
\[
d^2
=
\frac{1}{\sqrt2\,\pi n\lambda}
\]
Substituting,
\[
d^2
=
\frac{1}
{(4.44)(3.33\times10^{25})(1.69\times10^{-7})}
\]
This gives approximately
\[
d^2\approx4.0\times10^{-20}\,\text{m}^2
\]
Therefore,
\[
d\approx2.0\times10^{-10}\,\text{m}
\]
\(\textbf{Final answer:}\)
\[
\boxed{(n,d)
\approx
(3.33\times10^{25}\,\text{m}^{-3},2.0\times10^{-10}\,\text{m})}
\]
405. Ideal gases X and Y have equal pressures and equal mean free paths. Their absolute temperatures satisfy
\[
T_X=4T_Y.
\]
What ordered pair is required?
\[
\left(\frac{d_X}{d_Y},\frac{n_X}{n_Y}\right)
\]
ⓐ. \((2,\frac14)\)
ⓑ. \((4,\frac12)\)
ⓒ. \((2,4)\)
ⓓ. \((\frac12,4)\)
Correct Answer: \((2,\frac14)\)
Explanation: \(\textbf{Compare number densities from the common pressure:}\)
For an ideal gas,
\[
n=\frac{P}{k_BT}
\]
Since the pressures are equal,
\[
\frac{n_X}{n_Y}
=
\frac{T_Y}{T_X}
\]
Given
\[
T_X=4T_Y,
\]
\[
\frac{n_X}{n_Y}=\frac14
\]
\(\textbf{Now impose equal mean free paths:}\)
\[
\lambda=\frac{1}{\sqrt2\,\pi nd^2}
\]
Equal values of \(\lambda\) require
\[
n_Xd_X^2=n_Yd_Y^2
\]
Therefore,
\[
\frac{d_X^2}{d_Y^2}
=
\frac{n_Y}{n_X}
=4
\]
Hence,
\[
\frac{d_X}{d_Y}=2
\]
The hotter gas has lower number density at the same pressure, so its molecular diameter must be larger to keep the mean free path unchanged.
\(\textbf{Final answer:}\)
\[
\boxed{\left(2,\frac14\right)}
\]
406. Ideal gases X and Y have the same molecular diameter and the same mean free path. Their molar masses satisfy
\[
M_X=4M_Y.
\]
Which ordered pair must be correct?
\[
\left(\frac{n_X}{n_Y},\frac{\rho_X}{\rho_Y}\right)
\]
ⓐ. \((\frac14,4)\)
ⓑ. \((1,4)\)
ⓒ. \((1,\frac14)\)
ⓓ. \((4,1)\)
Correct Answer: \((1,4)\)
Explanation: \(\textbf{Use equal collision lengths first:}\)
Mean free path is
\[
\lambda=\frac{1}{\sqrt2\,\pi nd^2}
\]
The molecular diameters and mean free paths are equal.
Therefore,
\[
n_X=n_Y
\]
Hence,
\[
\frac{n_X}{n_Y}=1
\]
\(\textbf{Convert number density into mass density:}\)
Mass density is
\[
\rho=nm,
\]
where the mass of one molecule is proportional to molar mass:
\[
m=\frac{M}{N_A}
\]
Therefore, at equal number density,
\[
\frac{\rho_X}{\rho_Y}
=
\frac{M_X}{M_Y}
=4
\]
The equal mean free paths constrain molecular number density, not mass density.
\(\textbf{Final answer:}\)
\[
\boxed{(1,4)}
\]
407. Gases X and Y are at the same temperature and have equal mean free paths. Their molecular diameters and molar masses satisfy
\[
d_X=2d_Y,
\qquad
M_X=4M_Y.
\]
Which ordered pair correctly gives
\[
\left(\frac{P_X}{P_Y},\frac{\rho_X}{\rho_Y}\right)?
\]
ⓐ. \((\frac14,\frac14)\)
ⓑ. \((4,1)\)
ⓒ. \((1,4)\)
ⓓ. \((\frac14,1)\)
Correct Answer: \((\frac14,1)\)
Explanation: \(\textbf{Use equal mean free paths to compare number densities:}\)
Equal \(\lambda\) requires
\[
n_Xd_X^2=n_Yd_Y^2
\]
With
\[
d_X=2d_Y,
\]
\[
n_X(4d_Y^2)=n_Yd_Y^2
\]
Hence,
\[
\frac{n_X}{n_Y}=\frac14
\]
\(\textbf{Compare pressures at the common temperature:}\)
\[
P=nk_BT
\]
Thus,
\[
\frac{P_X}{P_Y}
=
\frac{n_X}{n_Y}
=
\frac14
\]
\(\textbf{Compare mass densities:}\)
\[
\rho=nm
\]
Since molecular mass is proportional to molar mass,
\[
\frac{\rho_X}{\rho_Y}
=
\frac{n_X}{n_Y}\frac{M_X}{M_Y}
\]
Therefore,
\[
\frac{\rho_X}{\rho_Y}
=
\frac14(4)=1
\]
Thus the gases can have equal mass density while having different number densities, pressures and molecular diameters.
\(\textbf{Final answer:}\)
\[
\boxed{\left(\frac14,1\right)}
\]
408. A molecule has mean free path
\[
\lambda=2.0\times10^{-7}\,\text{m}
\]
and a supplied representative speed
\[
v=5.0\times10^2\,\text{m s}^{-1}.
\]
During a total molecular path length of \(1.0\,\text{mm}\), which ordered pair is closest to the expected collision count and corresponding average collision frequency?
ⓐ. \((5.0\times10^3,\;2.5\times10^9\,\text{s}^{-1})\)
ⓑ. \((2.0\times10^3,\;1.0\times10^9\,\text{s}^{-1})\)
ⓒ. \((5.0\times10^3,\;5.0\times10^8\,\text{s}^{-1})\)
ⓓ. \((5.0\times10^2,\;2.5\times10^8\,\text{s}^{-1})\)
Correct Answer: \((5.0\times10^3,\;2.5\times10^9\,\text{s}^{-1})\)
Explanation: \(\textbf{Estimate the number of collisions along the travelled path:}\)
\[
N_c\approx\frac{L}{\lambda}
\]
Here,
\[
L=1.0\,\text{mm}=1.0\times10^{-3}\,\text{m}
\]
Therefore,
\[
N_c
=
\frac{1.0\times10^{-3}}
{2.0\times10^{-7}}
=
5.0\times10^3
\]
\(\textbf{Find the time associated with that total path length:}\)
Using the supplied representative speed,
\[
t=\frac{L}{v}
\]
Thus,
\[
t
=
\frac{1.0\times10^{-3}}
{5.0\times10^2}
=
2.0\times10^{-6}\,\text{s}
\]
\(\textbf{Convert collision count into collision frequency:}\)
\[
f_c\approx\frac{N_c}{t}
\]
Hence,
\[
f_c
=
\frac{5.0\times10^3}{2.0\times10^{-6}}
=
2.5\times10^9\,\text{s}^{-1}
\]
Equivalently,
\[
f_c\approx\frac{v}{\lambda}
\]
\(\textbf{Final answer:}\)
\[
\boxed{(5.0\times10^3,\;2.5\times10^9\,\text{s}^{-1})}
\]
409. For an ideal gas of fixed molecular diameter, the state is changed continuously while its mean free path is maintained constant. Which statement is correct?
ⓐ. \(P\propto T\), and the molecular number density remains constant.
ⓑ. \(P\propto1/T\), and the molecular number density increases with temperature.
ⓒ. \(P\) must remain constant, while number density decreases as \(1/T\).
ⓓ. \(T\) must remain constant, while pressure can vary arbitrarily.
Correct Answer: \(P\propto T\), and the molecular number density remains constant.
Explanation: For fixed molecular diameter,
\[
\lambda
=
\frac{k_BT}{\sqrt2\,\pi d^2P}.
\]
Maintaining \(\lambda\) constant requires
\[
\frac{T}{P}=\text{constant}.
\]
Therefore,
\[
P\propto T.
\]
Now use the molecular ideal-gas relation:
\[
n=\frac{P}{k_BT}.
\]
If
\[
\frac{P}{T}
\]
is constant, then
\[
n
\]
is also constant.
This is also immediate from
\[
\lambda=\frac1{\sqrt2\,\pi nd^2},
\]
because both \(\lambda\) and \(d\) are fixed.
Thus a constant-mean-free-path path in the \(P-T\) plane is a straight-line proportionality through the origin for the ideal-gas model.
410. Between two states of a gas,
\[
\lambda_2=\frac16\lambda_1
\]
and
\[
n_2=3n_1.
\]
What is the required ratio of molecular diameters \(d_2/d_1\)?
ⓐ. \(1\)
ⓑ. \(2\)
ⓒ. \(\dfrac1{\sqrt2}\)
ⓓ. \(\sqrt2\)
Correct Answer: \(\sqrt2\)
Explanation: The mean-free-path ratio is
\[
\frac{\lambda_2}{\lambda_1}
=
\frac{n_1d_1^2}{n_2d_2^2}
\]
Substitute
\[
\frac{\lambda_2}{\lambda_1}=\frac16
\]
and
\[
\frac{n_2}{n_1}=3
\]
Thus,
\[
\frac16
=
\frac{1}{3(d_2^2/d_1^2)}
\]
Rearranging,
\[
3\frac{d_2^2}{d_1^2}=6
\]
Hence,
\[
\frac{d_2^2}{d_1^2}=2
\]
Taking the positive square root,
\[
\frac{d_2}{d_1}=\sqrt2
\]
The square root is essential because molecular diameter enters the collision cross-sectional dependence as \(d^2\).
\(\textbf{Final answer:}\)
\[
\boxed{\frac{d_2}{d_1}=\sqrt2}
\]
411. Which dimensional argument correctly supports the NCERT expression
\[
\lambda=\frac{1}{\sqrt2\,\pi nd^2}?
\]
ⓐ. \([n]=L^0\), \([d^2]=L^2\), hence \([\lambda]=L^{-2}\).
ⓑ. \([n]=L^{-2}\), \([d^2]=L^2\), hence \([\lambda]=L^0\).
ⓒ. \([n]=L^{-3}\), \([d^2]=L^2\), hence \([\lambda]=L\).
ⓓ. \([n]=L^{-3}\), \([d^2]=L\), hence \([\lambda]=L^2\).
Correct Answer: \([n]=L^{-3}\), \([d^2]=L^2\), hence \([\lambda]=L\).
Explanation: Number density is molecule number per unit volume.
The molecule count is dimensionless as a count, but number density is not.
Its dimension is
\[
[n]=L^{-3}.
\]
Molecular diameter has dimension
\[
[d]=L.
\]
Therefore,
\[
[d^2]=L^2.
\]
Hence,
\[
[nd^2]
=
L^{-3}L^2
=
L^{-1}.
\]
The constants
\[
\sqrt2
\]
and
\[
\pi
\]
are dimensionless.
Thus,
\[
\left[\frac1{\sqrt2\,\pi nd^2}\right]
=
L,
\]
which is exactly the required dimension for a mean free path.
Therefore option C provides the correct dimensional check.
412. Two vessels contain the same fixed number of molecules of the same gas, and the molecular diameter is unchanged. Their measured mean free paths are equal, but their pressures satisfy
\[
P_2=2P_1.
\]
Which ordered pair must be correct?
\[
\left(\frac{V_2}{V_1},\frac{T_2}{T_1}\right)
\]
ⓐ. \((\frac12,1)\)
ⓑ. \((1,2)\)
ⓒ. \((2,1)\)
ⓓ. \((1,\frac12)\)
Correct Answer: \((1,2)\)
Explanation: \(\textbf{Use equal mean free paths as a density constraint:}\)
For the same molecular diameter,
\[
\lambda=\frac1{\sqrt2\,\pi nd^2}
\]
Equal mean free paths therefore require
\[
n_2=n_1
\]
Since both vessels contain the same fixed number \(N\),
\[
n=\frac NV
\]
Thus,
\[
V_2=V_1
\]
Hence,
\[
\frac{V_2}{V_1}=1
\]
\(\textbf{Now use the pressure ratio:}\)
For the same amount of gas,
\[
PV=Nk_BT
\]
With equal volumes,
\[
P\propto T
\]
Therefore,
\[
\frac{T_2}{T_1}
=
\frac{P_2}{P_1}
=2
\]
Thus equal mean free path here fixes the number density and therefore the volume, allowing the pressure ratio to determine the temperature ratio.
\(\textbf{Final answer:}\)
\[
\boxed{\left(1,2\right)}
\]
413. A molecular ensemble satisfies the usual kinetic-theory assumptions except that its velocity distribution is not isotropic. Let the directional pressures be \(P_x,P_y,P_z\). Consider the following statements.
Statement I: \(P_x=nm\langle v_x^2\rangle\) can still be used for a wall normal to the \(x\)-axis.
Statement II: The scalar relation \(P=\frac13nm\langle v^2\rangle\) follows without any further assumption.
Statement III: The total translational kinetic energy can be written as
\[
E_{\rm tr}=\frac V2(P_x+P_y+P_z).
\]
Statement IV: If the molecular directions are subsequently randomized without changing total translational energy, the final isotropic pressure is
\[
P_f=\frac{P_x+P_y+P_z}{3}.
\]
Which combination is correct?
ⓐ. I, III and IV only
ⓑ. II and IV only
ⓒ. I and II only
ⓓ. I, II, III and IV
Correct Answer: I, III and IV only
Explanation: The wall-collision argument for one particular direction does not require isotropy. For a wall normal to the \(x\)-axis,
\[
P_x=nm\langle v_x^2\rangle,
\]
so Statement I remains valid. The corresponding relations
\[
P_y=nm\langle v_y^2\rangle,\qquad P_z=nm\langle v_z^2\rangle
\]
also remain valid. The scalar factor \(1/3\), however, requires
\[
\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle,
\]
which is precisely the isotropy condition that has been removed. Hence Statement II is not generally valid. Adding the three directional-pressure relations gives
\[
P_x+P_y+P_z=nm\langle v^2\rangle.
\]
Since
\[
E_{\rm tr}=\frac12Nm\langle v^2\rangle,
\]
we obtain
\[
E_{\rm tr}=\frac V2(P_x+P_y+P_z),
\]
so Statement III is correct. If randomization preserves this energy, the sum of the final three equal directional pressures must remain the same. Therefore
\[
3P_f=P_x+P_y+P_z,
\]
making Statement IV correct. Thus I, III and IV are correct.
414. Two anisotropic molecular ensembles A and B occupy equal volumes and have the same molecular species and number density. Their directional pressures are
\[
(P_x,P_y,P_z)_A=(6P_0,2P_0,P_0)
\]
and
\[
(P_x,P_y,P_z)_B=(4P_0,3P_0,2P_0).
\]
The \(x\)-normal walls have equal areas. Each ensemble is then independently randomized without changing its translational kinetic energy. Which ordered triple is correct?
\[
\left(
\frac{E_A}{E_B},
\frac{F_{x,A}}{F_{x,B}},
\frac{P_{f,A}}{P_{f,B}}
\right)
\]
ⓐ. \(\left(1,\dfrac32,1\right)\)
ⓑ. \(\left(\dfrac32,1,\dfrac32\right)\)
ⓒ. \(\left(\dfrac98,\dfrac32,\dfrac98\right)\)
ⓓ. \(\left(1,\dfrac23,1\right)\)
Correct Answer: \(\left(1,\dfrac32,1\right)\)
Explanation: \(\textbf{Compare the translational energies:}\)
For an anisotropic ensemble,
\[
E_{\rm tr}=\frac V2(P_x+P_y+P_z)
\]
For A,
\[
P_{x,A}+P_{y,A}+P_{z,A}
=
(6+2+1)P_0
=
9P_0
\]
For B,
\[
P_{x,B}+P_{y,B}+P_{z,B}
=
(4+3+2)P_0
=
9P_0
\]
The volumes are equal, so
\[
\frac{E_A}{E_B}=1
\]
\(\textbf{Compare the original forces on x-normal walls:}\)
Since the wall areas are equal,
\[
F_x=P_xA
\]
Therefore,
\[
\frac{F_{x,A}}{F_{x,B}}
=
\frac{6P_0}{4P_0}
=
\frac32
\]
\(\textbf{Randomize while conserving energy:}\)
For A,
\[
P_{f,A}
=
\frac{9P_0}{3}
=
3P_0
\]
For B,
\[
P_{f,B}
=
\frac{9P_0}{3}
=
3P_0
\]
Hence,
\[
\frac{P_{f,A}}{P_{f,B}}=1
\]
\(\textbf{Final answer:}\)
\[
\boxed{\left(1,\frac32,1\right)}
\]
415. Two states of the same ideal gas have the same effective molecular diameter. Their mass densities and pressures satisfy
\[
\rho_2=4\rho_1,\qquad P_2=2P_1.
\]
Which ordered pair correctly gives
\[
\left(\frac{\lambda_2}{\lambda_1},\frac{T_2}{T_1}\right)?
\]
ⓐ. \(\left(\dfrac12,\dfrac14\right)\)
ⓑ. \(\left(\dfrac14,2\right)\)
ⓒ. \(\left(4,2\right)\)
ⓓ. \(\left(\dfrac14,\dfrac12\right)\)
Correct Answer: \(\left(\dfrac14,\dfrac12\right)\)
Explanation: \(\textbf{Convert mass density into number-density scaling:}\)
For the same molecular species,
\[
\rho=nm,
\]
and the molecular mass \(m\) is fixed.
Therefore,
\[
n\propto\rho
\]
Hence,
\[
\frac{n_2}{n_1}=4
\]
The molecular diameter is unchanged, so
\[
\lambda=\frac{1}{\sqrt2\,\pi nd^2}
\]
gives
\[
\frac{\lambda_2}{\lambda_1}
=
\frac{n_1}{n_2}
=
\frac14
\]
\(\textbf{Use the ideal-gas density form for temperature:}\)
For a fixed species,
\[
P=\frac{\rho RT}{M}
\]
Thus,
\[
T\propto\frac{P}{\rho}
\]
Therefore,
\[
\frac{T_2}{T_1}
=
\frac{P_2/P_1}{\rho_2/\rho_1}
=
\frac{2}{4}
=
\frac12
\]
The pressure increases even though the temperature falls because the density increases by the larger factor.
\(\textbf{Final answer:}\)
\[
\boxed{\left(\frac14,\frac12\right)}
\]
416. For molecular gases X and Y, a graph of mean free path \(\lambda\) against \(1/\rho\) is a straight line through the origin. The two graph slopes are equal. The molecular masses satisfy
\[
m_X=4m_Y.
\]
Using
\[
\lambda=\frac{1}{\sqrt2\,\pi nd^2}
\]
and \(\rho=nm\), which ordered pair is correct?
\[
\left(
\frac{d_X}{d_Y},
\frac{\lambda_X}{\lambda_Y}\text{ at equal number density}
\right)
\]
ⓐ. \((2,1)\)
ⓑ. \((2,\frac14)\)
ⓒ. \((\frac12,4)\)
ⓓ. \((4,\frac12)\)
Correct Answer: \((2,\frac14)\)
Explanation: \(\textbf{Rewrite mean free path using mass density:}\)
Since
\[
n=\frac{\rho}{m},
\]
\[
\lambda
=
\frac{m}{\sqrt2\,\pi\rho d^2}
\]
Thus, for a graph of \(\lambda\) against \(1/\rho\), the slope is
\[
s=\frac{m}{\sqrt2\,\pi d^2}
\]
The slopes are equal, so
\[
\frac{m_X}{d_X^2}
=
\frac{m_Y}{d_Y^2}
\]
Given
\[
m_X=4m_Y,
\]
we obtain
\[
\frac{4m_Y}{d_X^2}
=
\frac{m_Y}{d_Y^2}
\]
Therefore,
\[
d_X^2=4d_Y^2,
\]
and hence
\[
\frac{d_X}{d_Y}=2
\]
\(\textbf{Now compare at equal number density:}\)
At equal \(n\),
\[
\lambda\propto\frac1{d^2}
\]
Thus,
\[
\frac{\lambda_X}{\lambda_Y}
=
\frac{d_Y^2}{d_X^2}
=
\frac14
\]
\(\textbf{Final answer:}\)
\[
\boxed{\left(2,\frac14\right)}
\]
417. Consider the following statements involving Avogadro's number.
Statement I: \(N_A\approx6.02\times10^{23}\,\text{mol}^{-1}\).
Statement II: \(0.25\) mol of a molecular gas contains approximately \(1.505\times10^{23}\) molecules.
Statement III: The quantity \(N/N_A\) has the unit mole when \(N\) denotes a molecular count.
Statement IV: \(N_A\) is dimensionless because it represents a number of entities.
Which combination is correct?
ⓐ. I, II and III only
ⓑ. II and III only
ⓒ. I and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Avogadro's number gives the number of specified entities per mole, so its numerical value is approximately
\[
N_A=6.02\times10^{23}\,\text{mol}^{-1}.
\]
Thus Statement I is correct. For \(0.25\) mol,
\[
N=\mu N_A.
\]
Hence,
\[
N=(0.25)(6.02\times10^{23})
=
1.505\times10^{23},
\]
so Statement II is correct. Conversely,
\[
\mu=\frac{N}{N_A}.
\]
A molecule count is a count, while division by a quantity with unit \(\text{mol}^{-1}\) produces moles, making Statement III correct. Statement IV is incorrect because \(N_A\) is not merely the bare count \(6.02\times10^{23}\); it carries the unit \(\text{mol}^{-1}\). That unit is essential when converting between amount of substance and entity number. Therefore I, II and III only are correct.
418. An \(8.0\,\text{g}\) sample of oxygen gas has molar mass \(32\,\text{g mol}^{-1}\). Take
\[
N_A=6.02\times10^{23}\,\text{mol}^{-1}.
\]
Which ordered pair correctly gives the number of molecules in the sample and the mass of one oxygen molecule?
ⓐ. \(\left(3.01\times10^{23},2.66\times10^{-26}\,\text{kg}\right)\)
ⓑ. \(\left(1.505\times10^{23},5.32\times10^{-26}\,\text{kg}\right)\)
ⓒ. \(\left(6.02\times10^{23},1.33\times10^{-26}\,\text{kg}\right)\)
ⓓ. \(\left(1.505\times10^{23},5.32\times10^{-23}\,\text{kg}\right)\)
Correct Answer: \(\left(1.505\times10^{23},5.32\times10^{-26}\,\text{kg}\right)\)
Explanation: \(\textbf{Convert sample mass into amount of gas:}\)
\[
\mu=\frac{m_g}{M}
\]
Therefore,
\[
\mu=\frac{8.0}{32}=0.25\,\text{mol}
\]
\(\textbf{Convert moles into molecules:}\)
\[
N=\mu N_A
\]
Thus,
\[
N=(0.25)(6.02\times10^{23})
=
1.505\times10^{23}
\]
\(\textbf{Convert molar mass into mass per molecule:}\)
First express the molar mass in SI units:
\[
M=32\,\text{g mol}^{-1}
=
0.032\,\text{kg mol}^{-1}
\]
Then,
\[
m=\frac{M}{N_A}
\]
Hence,
\[
m
=
\frac{0.032}{6.02\times10^{23}}
\approx5.32\times10^{-26}\,\text{kg}
\]
The molar mass and the mass of one molecule differ by the Avogadro factor.
\(\textbf{Final answer:}\)
\[
\boxed{\left(1.505\times10^{23},5.32\times10^{-26}\,\text{kg}\right)}
\]
419. Given
\[
R=8.28\,\text{J mol}^{-1}\text{K}^{-1}
\]
and
\[
k_B=1.38\times10^{-23}\,\text{J K}^{-1},
\]
which ordered pair is obtained from \(R=N_Ak_B\) for Avogadro's number and the number of molecules in \(0.50\) mol?
ⓐ. \(\left(3.0\times10^{23}\,\text{mol}^{-1},6.0\times10^{23}\right)\)
ⓑ. \(\left(6.0\times10^{-23}\,\text{mol}^{-1},3.0\times10^{-23}\right)\)
ⓒ. \(\left(6.0\times10^{23}\,\text{mol}^{-1},3.0\times10^{23}\right)\)
ⓓ. \(\left(1.2\times10^{24}\,\text{mol}^{-1},6.0\times10^{23}\right)\)
Correct Answer: \(\left(6.0\times10^{23}\,\text{mol}^{-1},3.0\times10^{23}\right)\)
Explanation: \(\textbf{Infer Avogadro's number from the two gas constants:}\)
\[
R=N_Ak_B
\]
Therefore,
\[
N_A=\frac{R}{k_B}
\]
Substituting,
\[
N_A
=
\frac{8.28}{1.38\times10^{-23}}
\]
Since
\[
\frac{8.28}{1.38}=6.0,
\]
\[
N_A=6.0\times10^{23}\,\text{mol}^{-1}
\]
\(\textbf{Apply the mole-to-molecule conversion:}\)
\[
N=\mu N_A
\]
For
\[
\mu=0.50\,\text{mol},
\]
\[
N
=
(0.50)(6.0\times10^{23})
=
3.0\times10^{23}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\left(6.0\times10^{23}\,\text{mol}^{-1},3.0\times10^{23}\right)}
\]
420. A student begins with the macroscopic ideal-gas equation
\[
PV=\mu RT
\]
and wants to obtain its molecular form. Which substitution chain is correct?
ⓐ. \(R=N_Ak_B,\;N=\mu N_A\Rightarrow PV=Nk_BT\)
ⓑ. \(R=N_A/k_B,\;N=\mu N_A\Rightarrow PV=Nk_BT\)
ⓒ. \(R=N_Ak_B,\;N=\mu/N_A\Rightarrow PV=Nk_BT\)
ⓓ. \(R=k_B/N_A,\;N=\mu/N_A\Rightarrow PV=Nk_BT\)
Correct Answer: \(R=N_Ak_B,\;N=\mu N_A\Rightarrow PV=Nk_BT\)
Explanation: The mole-based state equation is
\[
PV=\mu RT.
\]
The universal gas constant and Boltzmann constant are related by
\[
R=N_Ak_B.
\]
Therefore,
\[
PV=\mu N_Ak_BT.
\]
The number of molecules in \(\mu\) moles is
\[
N=\mu N_A.
\]
Substituting this identification gives
\[
PV=Nk_BT.
\]
No additional factor of \(N_A\) is required after \(N\) has been introduced. Using both \(N\) and an extra Avogadro factor would double-count the mole-to-molecule conversion. Thus the microscopic and macroscopic equations are exactly equivalent descriptions of the same ideal-gas state.
