Kinetic Theory - CBSE Test Papers

 CBSE Test Paper 01

Chapter 13 Kinetic Theory

---

1. In a 30.0-s interval, 500 hailstones strike a glass window with an area of 0.600 m2${\mathrm{m}}^2$ at an angle of 45.0o to the window surface. Each hailstone has a mass of 5.00 g and a speed of 8.00 m/s. If the collisions are elastic, what are the average force and pressure on the window? 1

   1. 0.913 N
   2. 0.943 N
   3. 0.978 N
   4. 1.043 N

2. Nine particles have speeds of 5.00, 8.00, 12.0, 12.0, 12.0, 14.0, 14.0, 17.0, and 20.0 m/s. What is the rms speed? 1

   1. 9.35 m/s
   2. 13.3 m/s
   3. 12.3 m/s
   4. 14.7 m/s

3. The molar specific heat at constant volume, Cv${\mathrm{C}}_{\mathrm{v}}$ for solids is 1

   1. R
   2. 5R
   3. 3R
   4. 2R

4. Estimate the average thermal energy of a helium atom the temperature of 10 million Kelvin (the typical core temperature in the case of a star). 1

   1. 2.1 ×$\times$ 10−16$10^{-16}$ J
   2. 2.0 ×$\times$ 10−16$10^{-16}$ J
   3. 1.9 ×$\times$ 10−16$10^{-16}$ J
   4. 2.3 ×$\times$ 10−16$10^{-16}$ J

5. If there are n number of molecules per unit volume and m is the mass of each , vx${\mathrm{v}}_{\mathrm{x}}$ is the x-component of velocity, pressure can be written as 1

   1. P=nmv2x¯¯¯¯¯$P=nm\overline{v_x^2}$
   2. P=nv2x¯¯¯¯¯$P=n\overline{v_x^2}$
   3. P=2nmv2x¯¯¯¯¯$P=2nm\overline{v_x^2}$
   4. P=mv2x¯¯¯¯¯$P=m\overline{v_x^2}$

6. Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197g mole-1. 1

7. What happens when an electric fan is switched on in a closed room? 1

8. State Boyle's law in terms of pressure and density. 1

9. At room temperature, a diatomic gas molecule has five degrees of freedom and at high temperature it has seven degrees of freedom, explain. 2

10. A gas is contained in a closed vessel. How pressure due to the gas will be affected if force of attraction between the molecules disappear suddenly? 2

11. Briefly explain Boyle's law on the basis of Kinetic theory of gases. 2

12. Show that average kinetic energy of translation per molecule of a gas is directly proportional to the absolute temperature of gas. 3

13. Show that constant - temperature bulk modulus K of an ideal gas is the pressure P of the gas. 3

14. We have 0.5 g of hydrogen gas in a cubic chamber of size 3cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state? (Hydrogen molecules can be consider as sphere of radius 1A∘$1\stackrel{\circ }{A}$) 3

15. A box of 1.00m3 is filled with nitrogen at 1.5 atm at 300K. The box has a hole of an area 0.010 mm2. How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm. 5

CBSE Test Paper 01
Chapter 13 Kinetic Theory

---

Answer

1. 2. 0.943 N
      Explanation: F=500×ΔpΔt$F=500\times \frac{\mathrm{\Delta }p}{\mathrm{\Delta }t}$
      =500×2mvcos45∘Δt$=500\times \frac{2mv\cos {45}^{\circ }}{\mathrm{\Delta }t}$
      =500×2×5×10−3×8×130×1.41$=\frac{500\times 2\times 5\times {10}^{-3}\times 8\times 1}{30\times 1.41}$
      F = 0.943N

2. 2. 13.3 m/s
      Explanation: vrms=v12+v22+v32+.......vn2n−−−−−−−−−−−−−√$v_{rms}=\sqrt{\frac{{v_1}^2+{v_2}^2+{v_3}^2+.......{v_n}^2}{n}}$
      vrms$v_{rms}$=25+64+144+144+144+196+196+289+4009−−−−−−−−−−−−−−−−−−−−−−−−−√$=\sqrt{\frac{25+64+144+144+144+196+196+289+400}{9}}$=16029−−−−√$=\sqrt{\frac{1602}{9}}$
      vrms = 13.3 m/sec

3. 3. 3R
      Explanation: CV=12Rf$C_V=\frac{1}{2}Rf$
      an atom in a solid though has no degree of freedom for translational and rotational motion. It have 3 x 2 = 6 degree of freedom due to vibration along 3 axis.
      CV=62R=3R$C_V=\frac{6}{2}R=3R$

4. 1. 2.1 ×$\times$ 10−16$10^{-16}$ J
      Explanation: E=32kT$E=\frac{3}{2}kT$=32×1.38×10−23×10×106$=\frac{3}{2}\times 1.38\times {10}^{-23}\times 10\times {10}^6$=2.07×10−16J$=2.07\times {10}^{-16}J$
      k = boltzman constant

5. 1.  

      P=nmv2x¯¯¯¯¯$P=nm\overline{v_x^2}$
      Explanation: Pressure is due to collision of molecule with the wall of container.
      Collision by moleculeto the wall of conntainer assumed to be perfectly elastic.
      By the principle of conservation of momentum, the momentum imparted to the wall in the collision = 2mvx
      the number of molecules hitting the wall in time Δt$\mathrm{\Delta }t$ = nAvxΔt$nA{v}_x\mathrm{\Delta }t$
      n = no. of molecule per unit volume
      A = Area of container
      The total momentum transferred to the wall by these molecules in Δt$\mathrm{\Delta }t$ time = 2mvx(nAvxΔt)$2m{v}_x\left(nA{v}_x\mathrm{\Delta }t\right)$
      Pressure P = F/A = rate of momentum transfer per unit area
      P=2mvx(nAvxΔt)/ΔtA=nmv2x$P=\frac{2m{v}_x\left(nA{v}_x\mathrm{\Delta }t\right)/\mathrm{\Delta }t}{A}=nm{v^2}_x$
      all molecules in a gas do not have the same velocity; there is a distribution in velocities. The above equation therefore, stands for pressure due to the group of molecules with speed vx in the x-direction and n stands for the number density of that group of molecules. The total pressurer due to all groups
      P=nmv2x¯¯¯¯¯$P=nm\overline{v_x^2}$

6. 197 gm gold has number of atoms = 6.023×1023$\times {10}^{23}$
   1 gm gold will have number of atoms=6.023×1023197$=\frac{6.023\times {10}^{23}}{197}$
   39.4 gm gold has number of Au atoms =39.4×6.023×1023197$=\frac{39.4\times 6.023\times {10}^{23}}{197}$
   = 1.2×1023 atoms$1.2\times {10}^{23}atoms$

7. When electric fan is switched on, first electrical energy is converted into mechanical energy and then mechanical energy is converted into heat. The heat energy will increase the Kinetic energy of air molecules; hence temperature of room will increase.

8. Temperature remaining unchanged, density of a given gas is directly proportional to the pressure of the gas.
   Mathematically, if T = constant, then ρ∝P$\rho \propto P$.

9. At low temperature, a diatomic gas has three translational and two rotational degrees of freedom, making total number of degrees of freedom equal to 5.
   But at high temperature, as the gas molecule starts to vibrate which give two additional degrees of freedom. In that case total number of degrees of freedom becomes 5+2 = 7.

10. As force of attraction between molecules disappears, then molecules will hit the wall with more speeds. As a result, rate of change of momentum will also increase (since mass remains same). Again from Newton's 2nd law of motion we know that F=ΔpΔt$F=\frac{\mathrm{\Delta }p}{\mathrm{\Delta }t}$,where F is average force on the wall due to molecules,
    Δ$\mathrm{\Delta }$p is change in momentum and Δ$\mathrm{\Delta }$t is the time duration. Due to increase in Δ$\mathrm{\Delta }$p, force F will also increase. As a result, pressure, p =FA$\frac{F}{A}$ will increase. Here, A is area of one wall, which remains constant throughout the whole process.

11. According to the kinetic theory of gases, pressure exerted by a gas is given by
    P = 13VmNv2$\frac{1}{3V}mN{v}^2$
    where v = rms speed of gas molecules, V is volume of the gas, and mN is total mass of the gas
    ∴$\therefore$ PV = 13mNv2$\frac{1}{3}mN{v}^2$
    Multiplying and dividing by 2 on RHS, we get
    PV = (2N3)(12mv2)$\left(\frac{2N}{3}\right)\left(\frac{1}{2}m{v}^2\right)$ = (2N3)K.E.av$\left(\frac{2N}{3}\right)K.E{.}_{av}$
    The kinetic energy of a gas, as per kinetic theory, is directly proportional to its absolute temperature.
    At a constant temperature, K.E. is constant. Thus,
    PV = constant
    This explains Boyle's law.

12. According to the kinetic theory of gases, the pressure p exerted by one mole of an ideal gas is
    P=13×MV×C2$P=\frac{1}{3}\times \frac{M}{V}\times C^2$ M = Molar mass of the gas
    or PV=13MC2$PV=\frac{1}{3}M{C}^2$ V = Volume of the gas
    Since PV = RT (for 1mole of an ideal gas at absolute temperature T)
    or 13MC2=RT$\frac{1}{3}M{C}^2=RT$ (R = Universal gas contant)
    ∴C2=3RTM$\therefore C^2=\frac{3RT}{M}$ (T = Absolute temperature)
    So, C∝T−−√$C\propto \sqrt{T}$ , [since R and M are constants for the gas]
    Also, 13MC2=RT$\frac{1}{3}M{C}^2=RT$....(i)
    Dividing equation (i) by number of molecules of the gas = N
    1M3NC2=RNT$\frac{1M}{3N}{C}^2=\frac{R}{N}T$ (K = Boltzmann's constant)
    ⇒13mC2=KT$\Rightarrow \frac{1}{3}m{C}^2=KT$[m = M/N = mass of one single molecule of the gas]
    or 12mC2=32KT$\frac{1}{2}m{C}^2=\frac{3}{2}KT$
    Since, 12mC2$\frac{1}{2}m{C}^2$ = average kinetic energy of translation per molecule of the gas
    So, 12mC2∝T$\frac{1}{2}m{C}^2\propto T$ ....(ii)
    as 32K=$\frac{3}{2}K=$ constant
    Hence it is proved from equation (ii) that the average kinetic energy of translation per molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

13. Isothermal bulk modulus is defined as volume times negative partial derivative of pressure with respect to volume at constant temperature:
    K = - V (∂P/∂V)
    An ideal gas satisfies the following equation of state:
    PV = nRT
    So the pressure for an ideal gas is given by:
    P=nRTV$P=\frac{nRT}{V}$
    The partial derivative of pressure w.r.t. volume at constant temperature is:
    (∂P/∂V)T=(∂(nRT/V)/∂P)T$(\mathrm{\partial }P/\mathrm{\partial }V{)}_T=(\mathrm{\partial }(nRT/V)/\mathrm{\partial }P{)}_T$
    =nRTd(1/V)/dV$=nRTd(1/V)/dV$
    =−nRT/V2$-nRT/V^2$
    =−nRT/VV$\frac{-nRT/V}{V}$
    = - P/V
    Hence, K = - V ( - P/V ) = P

14. Volume of 1 molecule of hydrogen
    =43πr3=43×3.14×(10−10)3$=\frac{4}{3}\pi r^3=\frac{4}{3}\times 3.14\times {\left({10}^{-10}\right)}^3$ m3
    r=1A∘=10−10m$r=1\stackrel{\circ }{A}={10}^{-10}\mathrm{m}$ (Given)
    ∴$\therefore$ Volume of 1 molecule of hydrogen=4×1.05×10−30m3=4.20×10−30m3$=4\times 1.05\times {10}^{-30}{\mathrm{m}}^3=4.20\times {10}^{-30}{\mathrm{m}}^3$
    Number of moles in 0.5 g of H2 gas=0.52=0.25 mole$H_{2}gas=\frac{0.5}{2}=0.25mole$ [∴H2has2 mole ]$\left[\therefore H_{2}has2\text{ mole }\right]$
    ∴$\therefore$ Volume of H2$H_2$ molecules of .25 mole
    =0.25×6.023×1023×4.2×10−30m3$=0.25\times 6.023\times {10}^{23}\times 4.2\times {10}^{-30}{m}^3$ (since 1 mole of H2 = 6.023 × 1023 number of molecules)
    =1.05×6.023×10+23−30$=1.05\times 6.023\times {10}^{+23-30}$
    =6.324×10−7$=6.324\times {10}^{-7}$ m3
    Hence net volume 0.5 gm of H2 molecules, V'≈6.3×10−7m3$\approx 6.3\times {10}^{-7}{\mathrm{m}}^3$
    Now for ideal gas at constant temperature, according to Boyle's law,
    PiVi=PfVf$P_i{V}_i=P_f{V}_f$ (Pi, Vi, Pf and Vf are initial pressure, initial volume, final pressure and final volume respectively; Pi = 1 atm, Pf = 100 atm, Vi = volume of the cube)
    ∴Vf=PiViPf=1100×(3×10−2)3$\therefore V_f=\frac{P_i{V}_i}{P_f}=\frac{1}{100}\times {\left(3\times {10}^{-2}\right)}^3$ [∵vol. of cube Vi=(side)3 and Pi=1 atm at NTP]$\left[\because vol.\text{ of cube }{V}_i=(side{)}^3\text{ and }{P}_i=1\text{ atm at }NTP\right]$
    ⇒Vf=27×10−6100=2.7×10−5−2=2.7×10−7m3$\Rightarrow V_f=\frac{27\times {10}^{-6}}{100}=2.7\times {10}^{-5-2}=2.7\times {10}^{-7}{m}^3$
    Hence on compression, the final volume of the gas, Vf becomes of the same order of molecular volume, V'. As the molecules lie very close to each other, so intermolecular forces can not be ignored, i.e. the gas can not be treated as an ideal gas. Therefore one is not justified assuming the ideal gas law.

15. Initial volume ofthe given box =1m3=V1$=1m^3=V_1$
    Initial pressure =1.5atm=P1$=1.5atm=P_1$
    Final pressure = Initial pressure - pressure reduced by what amount = 1.5 – 0.1 =1.4 atm =P'2
    Air pressure out side the box =P2=1atm$=P_2=1atm$
    Initial temperature T1=300K$T_1=300\mathrm{K}$
    Final temperature T2=300K$T_2=300\mathrm{K}$
    A = area of hole =0.01mm2=0.01×10−6m2=10−8m2$=0.01{\mathrm{m}\mathrm{m}}^2=0.01\times {10}^{-6}{\mathrm{m}}^2={10}^{-8}{\mathrm{m}}^2$
    initial pressure difference between the box and the outside atmosphere, ΔP=(1.5−1)$\mathrm{\Delta }P=(1.5-1)$atm = 0.5 atm
    mass of a N2 gas molecule =0.028Kg6.023×1023=46.5×10−27Kg$=\frac{0.028\mathrm{K}\mathrm{g}}{6.023\times {10}^{23}}=46.5\times {10}^{-27}\mathrm{K}\mathrm{g}$ (= molar mass / Avogadro's number)
    Boltzmann's constant, KB=1.38×10−23$K_B=1.38\times {10}^{-23}$kg m2s-2K-1
    Let ρn1${\rho }_{n1}$is the initial number of N2$N_2$ gas molecule per unit volume per unit time. Now, in time Δt$\mathrm{\Delta }t$ let vtx$v_{tx}$ is the speed of the molecules along x axis.
    Number of molecules colliding in time Δt$\mathrm{\Delta }t$ on a wall of the cube
    =12ρn1[(vtx)Δt]A$=\frac{1}{2}{\rho }_{n1}\left[\left(v_{tx}\right)\mathrm{\Delta }t\right]A$ , A being area
    12$\frac{1}{2}$is multiplied as other12$\frac{1}{2}$molecule will strike to the opposite wall of the cube.
    v2rms(N2molecule)$v_{rms}^2\left(N_{2}molecule\right)$ =v2tx+v2ty+v2tz$=v_{tx}^2+v_{ty}^2+v_{tz}^2$
    ∵|vtx|=|vty|=|vtz|$\because \left|v_{tx}\right|=\left|v_{ty}\right|=\left|v_{tz}\right|$
    Then v2rms=3v2tx$v_{rms}^2=3v_{tx}^2$
    KE of the nitrogen gas molecule=32KBT$=\frac{3}{2}{K}_{B}T$
    12mv2rms=32KBT$\frac{1}{2}m{v}_{rms}^2=\frac{3}{2}{K}_{B}T$
    ⇒m×3v2tx=3KBT$\Rightarrow m\times 3v_{tx}^2=3K_{B}T$
    ⇒vtx=KBTm−−−−√$\Rightarrow v_{tx}=\sqrt{\frac{K_{B}T}{m}}$.....(A)
    Number of N2${\mathrm{N}}_2$gas molecules striking to a wall in Δt$\mathrm{\Delta }t$ time =12ρn1KBTm−−−−√Δt×A$=\frac{1}{2}{\rho }_{n1}\sqrt{\frac{K_{B}T}{m}}\mathrm{\Delta }t\times A$ ,outward [putting the value of vtx from equation (A).]
    Temperature inside the box and air are equal to T
    The number of air molecule striking to hole in time Δt$\mathrm{\Delta }t$
    inward =12ρn2KBTm−−−−√Δt×A$=\frac{1}{2}{\rho }_{n2}\sqrt{\frac{K_{B}T}{m}}\mathrm{\Delta }t\times A$
    ρn2=${\rho }_{n2}=$ number of air molecules per unit volume striking the wall in unit time
    Net number of molecules (going outward)
    12ρn1KBTm−−−−√ΔtA−12ρn2KBTm−−−−√ΔtA$\frac{1}{2}{\rho }_{n1}\sqrt{\frac{K_{B}T}{m}}\mathrm{\Delta }tA-\frac{1}{2}{\rho }_{n2}\sqrt{\frac{K_{B}T}{m}}\mathrm{\Delta }tA$.
    Net number of molecules going out from hole inΔt$\mathrm{\Delta }t$time
    =12[ρn1−ρn2]KBTm−−−−√.Δt×A$=\frac{1}{2}\left[{\rho }_{n1}-{\rho }_{n2}\right]\sqrt{\frac{K_{B}T}{m}}.\mathrm{\Delta }t\times A$ ......(i)
    Ideal gas equation for μ$\mu$ moles of the gas,P1V=μRT⇒μ=P1VRT$P_{1}V=\mu RT\Rightarrow \mu =\frac{P_{1}V}{RT}$
    As for box μV=P1RT$\frac{\mu }{V}=\frac{P_1}{RT}$ (μ$\mu$= No. of moles of gas in box)
    ρn1=N( Total number of molecules in box ) volume of box =μNAV${\rho }_{n1}=\frac{N(\text{ Total number of molecules in box })}{\text{ volume of box }}=\frac{\mu N_A}{V}$, NA being Avogadro's number.
    =P1NART$=\frac{P_1{N}_A}{RT}$ .....(ii)(putting the value of μV$\frac{\mu }{V}$) per unit volume
    Let after time T, pressure reduced by 0.1atm and becomes
    P'2 = (1.5−.1)=1.4 atm$(1.5-.1)=1.4atm$
    Then final new density of NA$N_A$moleculeρ′n1${\rho }_{n1}^{\mathrm{\prime }}$ (say)
    Therefore, ρ′n1==P2NART${\rho }_{n1=}^{\mathrm{\prime }}=\frac{P_2{N}_A}{RT}$ per unit volume (iii)
    Net number of molecules going out from volume V
    =(ρn1−ρ′n1)V=P1NARTV−P′2NARTV$=\left({\rho }_{n1}-{\rho }_{n1}^{\prime }\right)V=\frac{P_1{N}_A}{RT}V-\frac{P_2^{\prime }{N}_A}{RT}V$
    =NAVRT[P1−P′2]$=\frac{N_{A}V}{RT}\left[P_1-P_2^{\mathrm{\prime }}\right]$ .....(iv)(using equations ii, iii)
    P′2$P_2^{\mathrm{\prime }}$ = final pressure of box.
    From equation (i), total number of molecules going out in time τ$\tau$ from hole
    =12[ρn1−ρn2]KBTm−−−−√×τ×A$=\frac{1}{2}\left[{\rho }_{n1}-{\rho }_{n2}\right]\sqrt{\frac{K_{B}T}{m}}\times \tau \times A$.....(v)
    Again, ρn1−ρn2=P1NART−P2NART${\rho }_{n1}-{\rho }_{n2}=\frac{P_1{N}_A}{RT}-\frac{P_2{N}_A}{RT}$
    ∴ρn1−ρn2=NART[P1−P2]$\therefore {\rho }_{n1}-{\rho }_{n2}=\frac{N_A}{RT}\left[P_1-P_2\right]$.....(vi) (P2= Pressure of air outside the box)
    Net number of molecules going out in time τ$\tau$ from above(using equations v and vi)
    =12NART[P1−P2]KBTm−−−−√⋅τ.A$=\frac{1}{2}\frac{N_A}{RT}\left[P_1-P_2\right]\sqrt{\frac{K_{B}T}{m}}\cdot \tau .A$ ......(vii)
    From equations (iv) and (vii),
    NAVRT(P1−P′2)=12NART(P1−P2)KBTm−−−−√⋅τ.A$\frac{N_{A}V}{RT}\left(P_1-P_2^{\mathrm{\prime }}\right)=\frac{1}{2}\frac{N_A}{RT}\left(P_1-P_2\right)\sqrt{\frac{K_{B}T}{m}}\cdot \tau .A$
    ∴τ=NAVRT(P1−P′2)×2RTNA1(P1−P2)mKBT−−−−√⋅1A$\therefore \tau =\frac{N_{A}V}{RT}\left(P_1-P_2^{\prime }\right)\times \frac{2RT}{N_A}\frac{1}{\left(P_1-P_2\right)}\sqrt{\frac{m}{K_{B}T}}\cdot \frac{1}{A}$
    ∴τ=2(P1−P′2)(P1−P2)⋅VAmKBT−−−−√$\therefore \tau =\frac{2\left(P_1-P_2^{\prime }\right)}{\left(P_1-P_2\right)}\cdot \frac{V}{A}\sqrt{\frac{m}{K_{B}T}}$
    =2[1.5−1.4](1.5−1)110−846.5×10−271.38×10−23×300−−−−−−−−−−√$=\frac{2[1.5-1.4]}{(1.5-1)}\frac{1}{{10}^{-8}}\sqrt{\frac{46.5\times {10}^{-27}}{1.38\times {10}^{-23}\times 300}}$ (As P2 = atmospheric pressure = 1 atm)
    =2×0.10.5×10−84650×10−27+23−2138×3−−−−−−−−−−√$=\frac{2\times 0.1}{0.5\times {10}^{-8}}\sqrt{\frac{4650\times {10}^{-27+23-2}}{138\times 3}}$
    =0.4×10−8775×10−569−−−−−−√$=0.4\times {10}^{-8}\sqrt{\frac{775\times {10}^{-5}}{69}}$
    =0.4×108×10−3×11.23−−−−√=0.4×105×3.35$=0.4\times {10}^8\times {10}^{-3}\times \sqrt{11.23}=0.4\times {10}^5\times 3.35$
    ∴τ=1.34×105$\therefore \tau =1.34\times {10}^5$ seconds.
    This is the required time in the question.