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Kinetic Theory Test - 54

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Kinetic Theory Test - 54
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Weekly Quiz Competition
  • Question 1
    1 / -0
    If $$\triangle E$$ is the heat of reaction for $${ C }_{ 2 }{ H }_{ 3 }{ OH }_{ \left( 1 \right)  }+{ 3O }_{ 2\left( g \right)  }\rightarrow { 2CO }_{ 2\left( s \right)  }+3{ H }_{ 2 }{ 0 }_{ \left( 1 \right)  }$$ at constant volume, the $$\triangle H$$ ( Heat of reaction at constant pressure) at constant temperature is:
    Solution
    $$DH=DE+Dn_{gas} RT$$
    $$\Delta n_{brq}=3-1$$ (as a question)
    $$\therefore \ DH=DE+2RT$$
  • Question 2
    1 / -0
    Ideal monoatomic gas is taken through a process $$dQ  =2dU$$. What is the molar heat capacity for the process? ( where dQ is heat  supplied and dU is change in internal energy)
    Solution
    $$d Q= nCp dT$$
    $$Cp-$$ heat capacity of constant pressure 
    $$dU=nCvdT$$
    $$Cv-$$ heat capacity at constant volume
    $$n-$$ number of moles 
    $$dQ= 2d U$$
    $$nCpdT= 2nCvdT$$
    $$Cp= 2Cv$$
    $$= 2 \left( \dfrac{R}{y-1} \right)$$
    (For mono atomic $$y= 5/3$$ )
    $$=\dfrac{2R}{\dfrac{5}{3}-1}$$
    $$Cp= 3R$$
  • Question 3
    1 / -0
    A vessel has $$6g$$ of oxegen at pressure $$P$$ and temperature $$400\ K$$. A small hole is made in it so that oxygen leaks out. How much oxygen leaks out if the final pressure is $$P/2$$ and temperature is $$300\ K$$
    Solution

    From ideas gas equation

      $$ PV=nRT $$

     $$ PV=\dfrac{m}{{{M}_{o}}}RT $$

     $$ m=\dfrac{PV{{M}_{o}}}{RT}\  $$

    In first event

      $$ {{m}_{1}}=\dfrac{{{P}_{1}}V{{M}_{o}}}{R{{T}_{1}}} $$

    $$ 6\ gram=\dfrac{PV{{M}_{o}}}{R\times 400}\  $$

    $$ \dfrac{{{\operatorname{PVM}}_{o}}}{R}=400\times 6\ .\ .....\ (1) $$

    In the second event when hole is made:

     $$ {{m}_{2}}=\dfrac{{{P}_{2}}V{{M}_{o}}}{R{{T}_{2}}} $$

    $$ {{m}_{2}}=\dfrac{\dfrac{P}{2}V{{M}_{o}}}{R\times 300}\ =\dfrac{1}{300\times 2}\dfrac{PV{{M}_{o}}}{R} $$

     $$ {{m}_{2}}=\dfrac{1}{300\times 2}\times \left( 400\times 6 \right)=4\ gram $$

    Leak out oxygen is $$\left( 6gram-4\,gram \right)\ =\ 2\,gram$$ 

  • Question 4
    1 / -0
    Three particles are situated on a light and rigid rod along Y-axis as shown in the figure. If the system is rotating with angular velocity of $$2 rad/sec$$ about X axis, then the total kinetic energy of the system is :
  • Question 5
    1 / -0
    We have a jar A filled with gas characterized by parameters P, V and T and another jar B filed with gas with parameters 2P, Vl4 and 2T, where the symbols have their usual meanings. The ratio of the number of molecules of jar A to those of jar B is 
    Solution

  • Question 6
    1 / -0
    The degrees of freedom of a triatomic gas is? (consider moderate temperature)
    Solution
    The general epression for degree of freedom is $$DOF=3N-n$$
    here, DOF means degree of freedom, N is number of particle, and n is the number of holonomic constraints.
    for a triatomic molecule, the number of particle is 3 and since the separation between three atoms are fixed so, the number of constraints is 3.
    hence, $$DOF=(3\times 3)-3$$
    $$DOF=9-3$$
    $$DOF=6$$
  • Question 7
    1 / -0
    Change in momentum of the gas molecules as they strike the walls
    One mole of an ideal gas (mono-atomic) at temperature $$T_{0}$$ expands slowly according to law $$P^{2} = cT$$ ($$c$$ is constant). If final temperature is $$2T_{0}$$, heat supplied to gas is
  • Question 8
    1 / -0
    At what temperature does the average translational kinetic energy of a molecule in a gas become equal to kinetic energy of an electron accelerated from rest through a potential difference of $$1 \ volt$$?$$(k=1.38\times10^{23} \ J/k)$$
    Solution

  • Question 9
    1 / -0
    The effect of temperature on Maxwell's speed distribution is correctly shown by 
  • Question 10
    1 / -0
    Calculate $$\gamma $$ (ratio of $$C_p$$ and $$C_v$$ ) for triatomic linear gas at high temperature. Assume that the contribution of the vibrational degree of freedom is 75%?
    Solution
    For Triatomic molecule $$N=3$$
    Degree of freedom $$=3N=9$$

    3$$\longrightarrow$$Translation degree of freedom

    For linear, 2$$\longrightarrow$$Rotational degree of freedom

    $$9-(3+2)=4\longrightarrow$$Vibrational degree of freedom

    $$U=3\times \cfrac { 1 }{ 2 } kT+2\times \cfrac { 1 }{ 2 } kT+4\times \cfrac { 75 }{ 100 } kT$$

    as vibrational contributes only 75%

    $$U=\cfrac { 3 }{ 2 } kT+kT+3kT=\cfrac { 11 }{ 2 } kT$$ 

    $$ { C }_{ V }=\cfrac { dU }{ dT } =\cfrac { 11 }{ 2 } k$$ or for 1 mole $$\cfrac{11}{2}R$$

    As for ideal gas, 

    $${ C }_{ P }-{ C }_{ V }=R\\ { C }_{ P }={ C }_{ V }+R=\cfrac { 11 }{ 2 } R+R=\cfrac { 13 }{ 2 } R\\ g=\cfrac { { C }_{ P } }{ { C }_{ V } } =\cfrac { 13 }{ 2 } R\times \cfrac { 2 }{ 11R } =1.18$$

    Thus, the ratio of $$C_{p}$$ and $$C_{v}$$ is 1.18.

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