Kinetic Theory NCERT Soln
Class 11 Physics : Chapter 13
Complete NCERT Solutions for Kinetic Theory.
Detailed answers for gas laws, molecular speeds, and kinetic energy calculations.
Question 13.1
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
Answer & Explanation:
Given: Diameter of O2 molecule, \( d = 3 \, \text{Å} = 3 \times 10^{-10} \, \text{m} \)
Volume of one molecule:
\( V_{\text{mol}} = \frac{4}{3} \pi \left( \frac{d}{2} \right)^3 = \frac{4}{3} \pi (1.5 \times 10^{-10})^3 \)
Number of molecules in 1 mole at STP:
\( N_A = 6.02 \times 10^{23} \)
Total molecular volume:
\( V_{\text{mol}} \times N_A \)
Molar volume at STP:
\( 22.4 \, \text{litres} = 22.4 \times 10^{-3} \, \text{m}^3 \)
Fraction:
\( \frac{V_{\text{mol}} \times N_A}{22.4 \times 10^{-3}} \approx 4 \times 10^{-4} \)
Result: The molecular volume is about 0.04% of the actual volume occupied by the gas at STP.
Volume of one molecule:
\( V_{\text{mol}} = \frac{4}{3} \pi \left( \frac{d}{2} \right)^3 = \frac{4}{3} \pi (1.5 \times 10^{-10})^3 \)
Number of molecules in 1 mole at STP:
\( N_A = 6.02 \times 10^{23} \)
Total molecular volume:
\( V_{\text{mol}} \times N_A \)
Molar volume at STP:
\( 22.4 \, \text{litres} = 22.4 \times 10^{-3} \, \text{m}^3 \)
Fraction:
\( \frac{V_{\text{mol}} \times N_A}{22.4 \times 10^{-3}} \approx 4 \times 10^{-4} \)
Result: The molecular volume is about 0.04% of the actual volume occupied by the gas at STP.
Question 13.2
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
Answer & Explanation:
Using ideal gas equation:
\( PV = nRT \)
At STP:
\( P = 1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa} \)
\( T = 0 \, ^\circ\text{C} = 273 \, \text{K} \)
\( n = 1 \, \text{mol} \)
\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \)
\( V = \frac{nRT}{P} = \frac{1 \times 8.314 \times 273}{1.013 \times 10^5} \)
\( V \approx 0.0224 \, \text{m}^3 = 22.4 \, \text{litres} \)
Conclusion: 1 mole of any ideal gas occupies 22.4 litres at STP.
\( PV = nRT \)
At STP:
\( P = 1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa} \)
\( T = 0 \, ^\circ\text{C} = 273 \, \text{K} \)
\( n = 1 \, \text{mol} \)
\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \)
\( V = \frac{nRT}{P} = \frac{1 \times 8.314 \times 273}{1.013 \times 10^5} \)
\( V \approx 0.0224 \, \text{m}^3 = 22.4 \, \text{litres} \)
Conclusion: 1 mole of any ideal gas occupies 22.4 litres at STP.
Question 13.3
Figure 13.8 shows plot of PV/T versus P for 1.00×10-3 kg of oxygen gas at two different temperatures. (a) What does the dotted plot signify? (b) Which is true: T1 > T2 or T1 < T2? (c) What is the value of PV/T where the curves meet on the y-axis? (d) If we obtained similar plots for 1.00×10-3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mol-1 K-1)
Answer & Explanation:
(a) Dotted plot signifies ideal gas behaviour (\( PV/T = \text{constant} \)).
(b) \( T_1 > T_2 \) because at higher temperature, real gas behaves more ideally (less deviation).
(c) At y-axis (\( P \to 0 \)), \( PV/T = \mu R \). For O2:
\( \mu = \frac{1 \times 10^{-3}}{32 \times 10^{-3}} = \frac{1}{32} \, \text{mol} \)
\( PV/T = \frac{1}{32} \times 8.31 \approx 0.26 \, \text{J K}^{-1} \)
(d) For same PV/T, μ must be same. For H2:
\( \mu = \frac{m}{2.02 \times 10^{-3}} = \frac{1}{32} \)
\( m = \frac{2.02 \times 10^{-3}}{32} \approx 6.31 \times 10^{-5} \, \text{kg} \)
So, 0.0631 g of H2 gives same PV/T.
(b) \( T_1 > T_2 \) because at higher temperature, real gas behaves more ideally (less deviation).
(c) At y-axis (\( P \to 0 \)), \( PV/T = \mu R \). For O2:
\( \mu = \frac{1 \times 10^{-3}}{32 \times 10^{-3}} = \frac{1}{32} \, \text{mol} \)
\( PV/T = \frac{1}{32} \times 8.31 \approx 0.26 \, \text{J K}^{-1} \)
(d) For same PV/T, μ must be same. For H2:
\( \mu = \frac{m}{2.02 \times 10^{-3}} = \frac{1}{32} \)
\( m = \frac{2.02 \times 10^{-3}}{32} \approx 6.31 \times 10^{-5} \, \text{kg} \)
So, 0.0631 g of H2 gives same PV/T.
Question 13.4
An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol-1 K-1, molecular mass of O2 = 32 u).
Answer & Explanation:
Using \( PV = \mu RT \)
Initially:
\( P_1 = 15 + 1 = 16 \, \text{atm} = 16 \times 1.013 \times 10^5 \, \text{Pa} \)
\( V = 30 \times 10^{-3} \, \text{m}^3 \)
\( T_1 = 300 \, \text{K} \)
\( \mu_1 = \frac{P_1 V}{R T_1} \)
Finally:
\( P_2 = 11 + 1 = 12 \, \text{atm} \)
\( T_2 = 290 \, \text{K} \)
\( \mu_2 = \frac{P_2 V}{R T_2} \)
Moles withdrawn:
\( \Delta \mu = \mu_1 - \mu_2 \)
Mass withdrawn:
\( \Delta m = \Delta \mu \times 32 \times 10^{-3} \, \text{kg} \)
Calculation gives:
\( \Delta m \approx 0.131 \, \text{kg} = 131 \, \text{g} \)
Answer: About 131 g of oxygen is taken out.
Initially:
\( P_1 = 15 + 1 = 16 \, \text{atm} = 16 \times 1.013 \times 10^5 \, \text{Pa} \)
\( V = 30 \times 10^{-3} \, \text{m}^3 \)
\( T_1 = 300 \, \text{K} \)
\( \mu_1 = \frac{P_1 V}{R T_1} \)
Finally:
\( P_2 = 11 + 1 = 12 \, \text{atm} \)
\( T_2 = 290 \, \text{K} \)
\( \mu_2 = \frac{P_2 V}{R T_2} \)
Moles withdrawn:
\( \Delta \mu = \mu_1 - \mu_2 \)
Mass withdrawn:
\( \Delta m = \Delta \mu \times 32 \times 10^{-3} \, \text{kg} \)
Calculation gives:
\( \Delta m \approx 0.131 \, \text{kg} = 131 \, \text{g} \)
Answer: About 131 g of oxygen is taken out.
Question 13.5
An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?
Answer & Explanation:
At bottom:
\( P_1 = P_{\text{atm}} + \rho g h = 1.013 \times 10^5 + (1000 \times 9.8 \times 40) \approx 4.93 \times 10^5 \, \text{Pa} \)
\( V_1 = 1.0 \times 10^{-6} \, \text{m}^3 \)
\( T_1 = 285 \, \text{K} \)
At surface:
\( P_2 = 1.013 \times 10^5 \, \text{Pa} \)
\( T_2 = 308 \, \text{K} \)
Using combined gas law:
\( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \)
\( V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \)
\( V_2 \approx 5.3 \times 10^{-6} \, \text{m}^3 = 5.3 \, \text{cm}^3 \)
Result: The bubble expands to about 5.3 cm3 at the surface.
\( P_1 = P_{\text{atm}} + \rho g h = 1.013 \times 10^5 + (1000 \times 9.8 \times 40) \approx 4.93 \times 10^5 \, \text{Pa} \)
\( V_1 = 1.0 \times 10^{-6} \, \text{m}^3 \)
\( T_1 = 285 \, \text{K} \)
At surface:
\( P_2 = 1.013 \times 10^5 \, \text{Pa} \)
\( T_2 = 308 \, \text{K} \)
Using combined gas law:
\( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \)
\( V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \)
\( V_2 \approx 5.3 \times 10^{-6} \, \text{m}^3 = 5.3 \, \text{cm}^3 \)
Result: The bubble expands to about 5.3 cm3 at the surface.
Question 13.6
Estimate the total number of air molecules in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure.
Answer & Explanation:
Using ideal gas equation:
\( PV = N k_B T \)
\( P = 1.013 \times 10^5 \, \text{Pa} \)
\( V = 25.0 \, \text{m}^3 \)
\( T = 300 \, \text{K} \)
\( k_B = 1.38 \times 10^{-23} \, \text{J K}^{-1} \)
\( N = \frac{PV}{k_B T} \)
\( N \approx \frac{1.013 \times 10^5 \times 25}{1.38 \times 10^{-23} \times 300} \)
\( N \approx 6.1 \times 10^{26} \, \text{molecules} \)
Answer: About 6.1 × 1026 molecules.
\( PV = N k_B T \)
\( P = 1.013 \times 10^5 \, \text{Pa} \)
\( V = 25.0 \, \text{m}^3 \)
\( T = 300 \, \text{K} \)
\( k_B = 1.38 \times 10^{-23} \, \text{J K}^{-1} \)
\( N = \frac{PV}{k_B T} \)
\( N \approx \frac{1.013 \times 10^5 \times 25}{1.38 \times 10^{-23} \times 300} \)
\( N \approx 6.1 \times 10^{26} \, \text{molecules} \)
Answer: About 6.1 × 1026 molecules.
Question 13.7
Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Answer & Explanation:
Average kinetic energy per molecule:
\( E = \frac{3}{2} k_B T \)
(i) \( T = 300 \, \text{K} \)
\( E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \approx 6.21 \times 10^{-21} \, \text{J} \)
(ii) \( T = 6000 \, \text{K} \)
\( E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 6000 \approx 1.24 \times 10^{-19} \, \text{J} \)
(iii) \( T = 10^7 \, \text{K} \)
\( E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 10^7 \approx 2.07 \times 10^{-16} \, \text{J} \)
Note: Helium is monatomic; for diatomic/polyatomic gases, energy is higher due to rotational/vibrational modes.
\( E = \frac{3}{2} k_B T \)
(i) \( T = 300 \, \text{K} \)
\( E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \approx 6.21 \times 10^{-21} \, \text{J} \)
(ii) \( T = 6000 \, \text{K} \)
\( E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 6000 \approx 1.24 \times 10^{-19} \, \text{J} \)
(iii) \( T = 10^7 \, \text{K} \)
\( E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 10^7 \approx 2.07 \times 10^{-16} \, \text{J} \)
Note: Helium is monatomic; for diatomic/polyatomic gases, energy is higher due to rotational/vibrational modes.
Question 13.8
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?
Answer & Explanation:
Number of molecules: Yes, according to Avogadro's law, equal volumes at same T and P contain equal number of molecules.
vrms:
\( v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} \)
Since T is same, vrms depends on molecular mass m.
• Neon (atomic mass 20 u) – lightest
• Chlorine (71 u) – heavier
• UF6 (352 u) – heaviest
vrms is largest for neon (lightest molecules).
Conclusion: Same number of molecules; vrms highest for neon.
vrms:
\( v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} \)
Since T is same, vrms depends on molecular mass m.
• Neon (atomic mass 20 u) – lightest
• Chlorine (71 u) – heavier
• UF6 (352 u) – heaviest
vrms is largest for neon (lightest molecules).
Conclusion: Same number of molecules; vrms highest for neon.
Question 13.9
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at -20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Answer & Explanation:
\( v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} \)
For same vrms:
\( \frac{T_{\text{Ar}}}{m_{\text{Ar}}} = \frac{T_{\text{He}}}{m_{\text{He}}} \)
\( T_{\text{He}} = 253 \, \text{K} \)
\( m_{\text{He}} = 4.0 \, \text{u} \)
\( m_{\text{Ar}} = 39.9 \, \text{u} \)
\( T_{\text{Ar}} = T_{\text{He}} \times \frac{m_{\text{Ar}}}{m_{\text{He}}} = 253 \times \frac{39.9}{4.0} \)
\( T_{\text{Ar}} \approx 2520 \, \text{K} \)
Answer: Argon must be at about 2520 K.
For same vrms:
\( \frac{T_{\text{Ar}}}{m_{\text{Ar}}} = \frac{T_{\text{He}}}{m_{\text{He}}} \)
\( T_{\text{He}} = 253 \, \text{K} \)
\( m_{\text{He}} = 4.0 \, \text{u} \)
\( m_{\text{Ar}} = 39.9 \, \text{u} \)
\( T_{\text{Ar}} = T_{\text{He}} \times \frac{m_{\text{Ar}}}{m_{\text{He}}} = 253 \times \frac{39.9}{4.0} \)
\( T_{\text{Ar}} \approx 2520 \, \text{K} \)
Answer: Argon must be at about 2520 K.
Question 13.10
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).
Answer & Explanation:
Given:
\( P = 2 \, \text{atm} = 2.026 \times 10^5 \, \text{Pa} \)
\( T = 290 \, \text{K} \)
\( d = 2 \times 10^{-10} \, \text{m} \)
Number density:
\( n = \frac{P}{k_B T} \approx 5.0 \times 10^{25} \, \text{m}^{-3} \)
Mean free path:
\( \ell = \frac{1}{\sqrt{2} \pi d^2 n} \approx 5.6 \times 10^{-8} \, \text{m} \)
Average speed:
\( \langle v \rangle = \sqrt{\frac{8 k_B T}{\pi m}} \approx 470 \, \text{m/s} \) ( \( m = 4.65 \times 10^{-26} \, \text{kg} \) )
Collision frequency:
\( f = \frac{\langle v \rangle}{\ell} \approx 8.4 \times 10^9 \, \text{s}^{-1} \)
Time between collisions:
\( \tau = \frac{1}{f} \approx 1.2 \times 10^{-10} \, \text{s} \)
Comparison: Collision time (\( \sim 10^{-12} \, \text{s} \)) is much smaller than free time (\( \sim 10^{-10} \, \text{s} \)).
\( P = 2 \, \text{atm} = 2.026 \times 10^5 \, \text{Pa} \)
\( T = 290 \, \text{K} \)
\( d = 2 \times 10^{-10} \, \text{m} \)
Number density:
\( n = \frac{P}{k_B T} \approx 5.0 \times 10^{25} \, \text{m}^{-3} \)
Mean free path:
\( \ell = \frac{1}{\sqrt{2} \pi d^2 n} \approx 5.6 \times 10^{-8} \, \text{m} \)
Average speed:
\( \langle v \rangle = \sqrt{\frac{8 k_B T}{\pi m}} \approx 470 \, \text{m/s} \) ( \( m = 4.65 \times 10^{-26} \, \text{kg} \) )
Collision frequency:
\( f = \frac{\langle v \rangle}{\ell} \approx 8.4 \times 10^9 \, \text{s}^{-1} \)
Time between collisions:
\( \tau = \frac{1}{f} \approx 1.2 \times 10^{-10} \, \text{s} \)
Comparison: Collision time (\( \sim 10^{-12} \, \text{s} \)) is much smaller than free time (\( \sim 10^{-10} \, \text{s} \)).
Question 13.11
A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
Answer & Explanation:
Initially (horizontal):
Air column length = 15 cm, \( P = P_{\text{atm}} = 76 \, \text{cm Hg} \)
When vertical (open end down): Mercury thread will move down due to gravity, increasing air column volume.
Let new air column length = \( L \) cm.
Pressure on air = \( P_{\text{atm}} + \) pressure due to mercury column
If mercury thread length remains 76 cm, and it moves down \( x \) cm, then:
Air column length = \( 15 + x \)
Pressure = \( (76 + x) \) cm Hg (approx, considering equilibrium)
Using Boyle’s law:
\( P_1 V_1 = P_2 V_2 \)
\( 76 \times 15 = (76 + x) \times (15 + x) \)
Solving: \( x \approx 4.8 \, \text{cm} \) (approx)
Result: Air column increases to about 19.8 cm; mercury moves down ~4.8 cm.
Air column length = 15 cm, \( P = P_{\text{atm}} = 76 \, \text{cm Hg} \)
When vertical (open end down): Mercury thread will move down due to gravity, increasing air column volume.
Let new air column length = \( L \) cm.
Pressure on air = \( P_{\text{atm}} + \) pressure due to mercury column
If mercury thread length remains 76 cm, and it moves down \( x \) cm, then:
Air column length = \( 15 + x \)
Pressure = \( (76 + x) \) cm Hg (approx, considering equilibrium)
Using Boyle’s law:
\( P_1 V_1 = P_2 V_2 \)
\( 76 \times 15 = (76 + x) \times (15 + x) \)
Solving: \( x \approx 4.8 \, \text{cm} \) (approx)
Result: Air column increases to about 19.8 cm; mercury moves down ~4.8 cm.
Question 13.12
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3 s-1. Identify the gas. [Hint: Use Graham's law of diffusion: R1/R2 = (M2/M1)1/2]
Answer & Explanation:
Graham’s law:
\( \frac{R_{\text{H}}}{R_{\text{gas}}} = \sqrt{\frac{M_{\text{gas}}}{M_{\text{H}}}} \)
\( M_{\text{H}} = 2 \, \text{g/mol} \)
\( R_{\text{H}} = 28.7 \, \text{cm}^3/\text{s} \)
\( R_{\text{gas}} = 7.2 \, \text{cm}^3/\text{s} \)
\( \frac{28.7}{7.2} = \sqrt{\frac{M_{\text{gas}}}{2}} \)
\( \frac{M_{\text{gas}}}{2} = \left( \frac{28.7}{7.2} \right)^2 \approx 15.9 \)
\( M_{\text{gas}} \approx 31.8 \, \text{g/mol} \)
Closest molecular mass: Oxygen (O2 = 32 g/mol)
Identified gas: Oxygen.
\( \frac{R_{\text{H}}}{R_{\text{gas}}} = \sqrt{\frac{M_{\text{gas}}}{M_{\text{H}}}} \)
\( M_{\text{H}} = 2 \, \text{g/mol} \)
\( R_{\text{H}} = 28.7 \, \text{cm}^3/\text{s} \)
\( R_{\text{gas}} = 7.2 \, \text{cm}^3/\text{s} \)
\( \frac{28.7}{7.2} = \sqrt{\frac{M_{\text{gas}}}{2}} \)
\( \frac{M_{\text{gas}}}{2} = \left( \frac{28.7}{7.2} \right)^2 \approx 15.9 \)
\( M_{\text{gas}} \approx 31.8 \, \text{g/mol} \)
Closest molecular mass: Oxygen (O2 = 32 g/mol)
Identified gas: Oxygen.
Question 13.13
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density. Use the law of atmospheres to derive the equation for sedimentation equilibrium of a suspension in a liquid column.
Answer & Explanation:
Law of atmospheres:
\( n_2 = n_1 \exp\left[ -\frac{mg(h_2 - h_1)}{k_B T} \right] \)
For suspended particle in liquid:
Apparent weight = \( mg – \text{buoyant force} = mg – V \rho' g = mg \left( 1 – \frac{\rho'}{\rho} \right) \)
where \( \rho \) = particle density, \( \rho' \) = liquid density.
Replacing \( mg \) with apparent weight:
\( n_2 = n_1 \exp\left[ -\frac{mg \left( 1 – \frac{\rho'}{\rho} \right) (h_2 - h_1)}{k_B T} \right] \)
Using \( k_B = \frac{R}{N_A} \) and \( m = \frac{M}{N_A} \):
\( n_2 = n_1 \exp\left[ -\frac{M g (\rho – \rho') (h_2 - h_1)}{\rho R T} \right] \)
This is the sedimentation equilibrium equation.
\( n_2 = n_1 \exp\left[ -\frac{mg(h_2 - h_1)}{k_B T} \right] \)
For suspended particle in liquid:
Apparent weight = \( mg – \text{buoyant force} = mg – V \rho' g = mg \left( 1 – \frac{\rho'}{\rho} \right) \)
where \( \rho \) = particle density, \( \rho' \) = liquid density.
Replacing \( mg \) with apparent weight:
\( n_2 = n_1 \exp\left[ -\frac{mg \left( 1 – \frac{\rho'}{\rho} \right) (h_2 - h_1)}{k_B T} \right] \)
Using \( k_B = \frac{R}{N_A} \) and \( m = \frac{M}{N_A} \):
\( n_2 = n_1 \exp\left[ -\frac{M g (\rho – \rho') (h_2 - h_1)}{\rho R T} \right] \)
This is the sedimentation equilibrium equation.
Question 13.14
Given densities of some solids and liquids, give rough estimates of the size of their atoms. [Hint: Assume atoms are tightly packed]
Answer & Explanation:
For tightly packed atoms, volume per atom:
\( \frac{M}{\rho N_A} \)
Assuming spherical atom:
\( \frac{4}{3} \pi r^3 = \frac{M}{\rho N_A} \)
\( r = \left[ \frac{3M}{4 \pi \rho N_A} \right]^{1/3} \)
Example for Carbon (diamond):
\( M = 12 \times 10^{-3} \, \text{kg/mol} \), \( \rho = 2220 \, \text{kg/m}^3 \)
\( r \approx 0.77 \times 10^{-10} \, \text{m} = 0.77 \, \text{Å} \)
Similar calculations for others:
• Gold: ~1.4 Å
• Nitrogen (liquid): ~1.8 Å
• Lithium: ~1.5 Å
• Fluorine: ~1.3 Å
Note: These are rough estimates; actual atomic sizes vary.
\( \frac{M}{\rho N_A} \)
Assuming spherical atom:
\( \frac{4}{3} \pi r^3 = \frac{M}{\rho N_A} \)
\( r = \left[ \frac{3M}{4 \pi \rho N_A} \right]^{1/3} \)
Example for Carbon (diamond):
\( M = 12 \times 10^{-3} \, \text{kg/mol} \), \( \rho = 2220 \, \text{kg/m}^3 \)
\( r \approx 0.77 \times 10^{-10} \, \text{m} = 0.77 \, \text{Å} \)
Similar calculations for others:
• Gold: ~1.4 Å
• Nitrogen (liquid): ~1.8 Å
• Lithium: ~1.5 Å
• Fluorine: ~1.3 Å
Note: These are rough estimates; actual atomic sizes vary.
📘 Exam Preparation Tip:
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