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Kinetic Theory Test - 53

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Kinetic Theory Test - 53
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  • Question 1
    1 / -0
    If the total number of $$H_2$$ molecules is double of the $$O_2$$ molecules then the ratio of total kinetic energies pf $$H_2$$ to that of $$O_2$$ at 300 K is :-
    Solution
    Kinetic energy for a single molecule is given as $$E=\dfrac{3}{2}kT$$
    where $$k$$ is Boltzman's constant so for $$2n $$ molecules of Hydrogen the kinetic energy will be $$E_{H_2}=2n(1.5kT)$$
    And that for $$n$$ molecules of $$O_2 $$ will be $$E_{O_2}=n(1.5kT)$$
    ratio of energi$$es$$ will be $$E_{H_2} :E_{O_2}=2:1$$
    Option C is correct.
  • Question 2
    1 / -0
    The relation between the ratio of specific heats $$(\gamma)$$ of gas and degree of freedom  $$'f'$$ will be :
    Solution

  • Question 3
    1 / -0
    $$3$$ mole of gas ''X"  and $$2$$ moles of gas "Y" enters from end "P" and "Q" of the cylinder respectively. The cylinder has the area of cross section , shown as
    under 
    The length of the cylinder is $$150cm$$. The gas "X" intermixes with gas "Y" at the point . If the molecular weight of the gases X and Y is $$20$$ and $$80$$ respectively, then what will be the distance of point A from Q?
    Solution
    $$\begin{array}{l} \frac { { rx } }{ { ry } } =\frac { { { w_{ x } } } }{ { { n_{ y } } } } \sqrt { \frac { { { M_{ y } } } }{ { { M_{ x } } } }  }  \\ =\frac { 3 }{ 2 } \sqrt { \frac { { 80 } }{ { 20 } }  } =\frac { 3 }{ 1 } =3:1 \\ \therefore \frac { { dis\tan  ce\, \, travelled\, \, by\, \, gas\, \, X } }{ { dis\tan  ce\, \, travelled\, \, by\, \, gas\, \, Y } } =3:1 \\ \therefore dis\tan  ce\, \, of\, \, A\, \, from\, \, Q=\frac { { 150 } }{ 3 } =50\, \, cms \end{array}$$
    Hence, OPtion $$B$$ is correct.
  • Question 4
    1 / -0
    The total Kinetic energy of $$1\ mole$$ of $${N}^{}_{2}$$ at $$27^{o}_{}{C}$$ will be approximately :-
    Solution
    For $$n$$ mole of any gas the total  kinetic energy is given as $$E=\dfrac{3}{2}nRT$$
    Where $$R$$ is gas constant having value $$8.31J/mole-K$$ or $$8.31\times 4.18 cal /mole-K=34.74\text{Cal per mole per Kelvin}$$
    $$T$$ is temperature in Kelvin which is $$T=27+273=300K$$
    So putting all values we get $$E=1.5\times 1 \times 34.74\times 300=15633Calorie$$
  • Question 5
    1 / -0
    The total kinetic energy of 1 mole of $${ N }_{ 2 }$$ at $$27^oC$$ will be approximately:
    Solution
    According to law of Equipartition of energy, a molecule can have $$\dfrac{nRT}{2}$$energy per degree of freedom.

    Here given molecule $$N_2$$ is diatomic molecule and we know, there are five degree of freedom of a diatomic molecule ( three translational and two rotational ).

    So, total kinetic energy = $$\dfrac{5nRT}{2}$$

    Here, $$n$$ is number of mole. given, $$n = 1$$

    $$R$$ is universal gas constant i.e., $$R = 2 Cal/mol.K$$

    and T is temperature in Kelvin. given, $$T = 27°C = (27 + 273)K = 300K$$

    now, total kinetic energy = $$5/2 × 1 × 2 × 300$$

    =$$ 5 × 300 = 1500 Cal$$
  • Question 6
    1 / -0
    How many degrees of freedom the gas molecules have if under STP the gas density $$\rho = 1.3 kg/m^3$$ and the velocity of sound propagation in it is $$330 ms^{-1}$$?
    Solution

  • Question 7
    1 / -0
    Mean kinetic energy(or average energy) per gm molecule of a monoatomic gas is given by:
    Solution
    One gram molecule is the no.of molecules present in one mole, i.e, $$6.022 \times 10^23$$ molecules.
    Therefore mean kinetic energy per gram molecule of a monoatomic gas:
    $$=\dfrac{3}{2} Nk_B T$$
    $$=\dfrac{3}{2} RT$$ since $$R=Nk_B$$
  • Question 8
    1 / -0
    The de-Broglie wavelength of a particle accelerated with $$150\ volt$$ potential is $$10^{-10}\ m$$. If it accelerated by $$600\ volts$$ p.d. its wavelength will be
    Solution

    Given,

    $$\lambda =\dfrac{hc}{eV}\ \ \ \ where,\ V=potential$$

    $$\lambda \ \alpha \ \dfrac{1}{V}$$

    $${{10}^{-10}}\ \alpha \ \dfrac{1}{150}\ ......\ (1)$$

    $$\lambda \ \alpha \ \dfrac{1}{600}\ ......\ (2)$$

    Divide (2) by (1)

    $$ \dfrac{\lambda }{{{10}^{-10}}}=\dfrac{150}{600}=\dfrac{1}{4} $$

    $$ \Rightarrow \lambda =0.25\times {{10}^{-10}}m\ =0.25\ {{A}^{o}} $$ 

  • Question 9
    1 / -0
    Three perfect gasses at absolute temperatures $$T_1$$, $$T_2$$ and $$T_3$$ are mixed. If number of molecules of the gasses are $$n_1$$, $$n_2$$ and $$n_3$$ respectively then temperature of mixture will be (assume no loss of energy)
    Solution

    For perfect gas,

    Kinetic Energy of n molecule, $$K.E=n\left( \dfrac{1}{2}{{K}_{B}}T \right)$$

    Where, $${{K}_{B}}$$ is Boltzmann constant 

    If there is no loss of energy.

    Total kinetic energy of mixture is sum of each gas kinetic energy.

    $$ {{n}_{total}}K.{{E}_{total}}={{n}_{1}}K.{{E}_{1}}+{{n}_{2}}K.{{E}_{2}}+{{n}_{3}}K.{{E}_{3}} $$

    $$ \left( {{n}_{1}}+{{n}_{2}}+{{n}_{3}} \right)\left( \dfrac{1}{2}{{K}_{B}}T \right)={{n}_{1}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{1}} \right)+{{n}_{2}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{2}} \right)+{{n}_{3}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{3}} \right) $$

     $$ T=\dfrac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}+{{n}_{3}}{{T}_{3}}}{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}} $$ 

  • Question 10
    1 / -0
    The total kinetic energy of $$1$$ mole of $$N_2$$ at $$27$$C will be approximately
    Solution
    The kinetic enrgy of one mole is given by:
    KE=$$\dfrac{3}{2} K_BT$$
    The kinetic enrgy of 1 mole of $$N_2$$ atoms is:
    KE=$$\dfrac{3}{2}K_B T$$ where $$N$$ is Avogadro's number,$$K_B$$ is Boltzmann's constant and $$T$$ is temperature
    KE=$$\dfrac{3}{2} \times (6.022 \times 10^{23})\times (1.38 \times 10^{-23}) \times 300$$
    $$=3739.662 J$$

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