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Kinetic Theory Test - 36

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Kinetic Theory Test - 36
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  • Question 1
    1 / -0
    The value of $$\gamma$$ of triatomic gas (linear arrangement) molecules is
    Solution
    The number of degrees of freedom for a triatomic gas with linear arrangement is 7.
    We have the relation $$\gamma$$$$=1+\dfrac{2}{f}$$ (where $$f$$ is number of degrees of freedom)
    $$\therefore \gamma=1+\dfrac{2}{7}=\dfrac{9}{7}$$
  • Question 2
    1 / -0
    At what temperature will the linear kinetic energy of a gas molecule be equal to that of an electron accelerated through a potential difference of 10V?
    Solution
    $$The\quad kinetic\quad energy\quad of\quad gas\quad molecule\quad =\quad 3/2\quad kT\\ Energy\quad of\quad the\quad accelerated\quad electron\quad =eV\\ Hence\quad eV=\quad 3/2\quad kT\quad For\quad the\quad given\quad case\\ k={ 1.38 }\times 10^{ -23 }\\ 1.6\times { 10 }^{ -19 } \times 10 = 1.5 kT\\ T=\dfrac { 1.6\times { 10 }^{ -19 }\quad \times 10\quad \times 2 }{ 3k } \\ T={ 11.3\times 10 }^{ 3 } K$$
  • Question 3
    1 / -0
    An enclosure of volume 3 litre contains 16g of oxygen, 7g of nitrogen and 11g of carbondioxide at 27$$^o$$C. The pressure exerted by the mixture is approximately
    Solution
    Total number of moles = number of moles of oxygen + number of moles of nitrogen + number of moles of carbon-di-oxide
    $$\dfrac { 16 }{ 32 } +\dfrac { 7 }{ 14 } +\dfrac { 11 }{ 44 } =1mole$$
    Molecular weight of Carbon-di-oxide = 44g
    Molecular weight of Oxygen =32g
    Molecular weight of  nitrogen=14g
    Using ideal gas equation,
    $$P=\dfrac { nRT }{ V } =\dfrac { 1 \times8.3 \times300 }{ (3\times{ 10 }^{ -3 }) } =8.3atm(nearly)$$
    Hence, Option B is correct.

  • Question 4
    1 / -0
    The law of equipartition of energy is applicable to the system whose constituents are :
    Solution
    The original idea of equipartition  is to assign a average kinetic energy to each degree of freedom if we take transnational case only. So if there would be ordered motion instead of random motion then there is no need of equilibrium theorem  as it would be useless to take about averages as they all are in ordered motion.
    Hence, the answer is in random motion.
  • Question 5
    1 / -0
    The pressure of a gas filled in a closed vessel increases by 0.4%. When temperature is increases by 1$$^o$$C the initial temperature of the gas is
    Solution
    Since volume is constant i.e number of moles is constant.
    $$P_1$$ and $$T_1$$ are the initial pressure and temperature.
    According to the question,
    $${ P }_{ 2 }=1.004{ P }_{ 1 }\\ { T }_{ 2 }={ T }_{ 1 }+1$$
    $$\therefore \dfrac { { P }_{ 1 } }{ { T }_{ 1 } } =\dfrac{P_2}{T_2}=\dfrac { (1+0.004){ P }_{ 1 } }{ { T }_{ 2 } } \\ \Rightarrow { T }_{ 2 }=1.004{ T }_{ 1 }=({ T }_{ 1 }+1)$$ 
    $$0.004T_1 = 1$$
    $$\\ \Rightarrow { T }_{ 1 }=250K$$
    Hence, Option B is correct
  • Question 6
    1 / -0
    The most probable velocity for monoatomic gas is
    Solution
    $$KE_p=\dfrac{mv_p^2}{2}=kT$$ [standard result]
    $$v_p=\sqrt{\dfrac{2kT}{m}}$$
  • Question 7
    1 / -0
    The dimension of universal gas constant R are
    Solution
    using ideal gas equation
    $$R=\dfrac { PV }{ nT } $$
    dimensions of pressure $$M{ L }^{ -1 }{ T }^{ -2 }$$
    dimensions of Volume $${L}^{3}$$
    number of moles is dimensionless
    dimensions of Temperature $$\theta$$
    so, dimensions of R$$ \dfrac { (M{ L }^{ -1 }{ T }^{ -2 })({ L }^{ 3 }) }{ \theta  } =M{ L }^{ 2 }{ T }^{ -2 }{ \theta  }^{ -1 }$$
    Hence,Option B is correct
  • Question 8
    1 / -0
    If the mean kinetic energy per unit volume of a gas is n times its pressure, then the value of n is
    Solution
    Mean kinetic energy of a gas = $$\dfrac{1}{2}MV_{rms}^{2}$$
    $$=\dfrac{1}{2}\times M\times\dfrac{3PV}{M}$$
    $$=\dfrac{3}{2}PV$$
    Mean kinetic energy per unit volume=$$\dfrac{3}{2}P$$

  • Question 9
    1 / -0
    In outer space there are 10 molecules per cm$$^3$$ on an average and the temperature there is 3K. The average pressure of this light gas is
    Solution
    $$P= \displaystyle \frac{nKT}{V} = \frac{10 \times 1.38 \times 10^{-23} \times 3}{10^{-6}}$$
    $$= 4.14 \times 10^{-16} Nm^{-2}$$
  • Question 10
    1 / -0
    A cylinder contains 2kg of air at a pressure of 10$$^5$$Pa. If 2 kg more air is pumped into it, keeping the temperature constant, the pressure will be
    Solution
    Since both volume and Temperature is constant,
    $$\dfrac { P }{ n } =constant=\dfrac { P }{ m } \\ n=no.of\quad moles=\dfrac { mass\quad of\quad air(m) }{ Molecular\quad weight\quad of\quad water(M) } $$
    $$\dfrac { { 10 }^{ 5 } }{ 2 } =\dfrac { P }{ (2+2) } \\ \Rightarrow P=2\times{ 10 }^{ 5 }Pa$$
    Hence, Option B is correct
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