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Kinetic Theory Test - 41

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Kinetic Theory Test - 41
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  • Question 1
    1 / -0
    .

    Solution

  • Question 2
    1 / -0
    A gas behaves more closely like an ideal gas at
    Solution
    Due to low pressure, the size of molecules becomes less significant compared to the empty spaces between them.
    Due to high temperature the potential energy due to inter-molecular forces becomes less significant compared with the particles kinetic energy.
    As temperature increases the kinetic energy of particles.
    Hence, the answer is low pressure and high temperature.
  • Question 3
    1 / -0
    If T represents the absolute temperature of an ideal gas, the volume coefficient of thermal expansion at constant pressure is proportional to:
    Solution
    Coefficient of thermal expansion at constant pressure is:

    $$\implies { \alpha  }_{ V }=\dfrac { 1 }{ V } \dfrac { dV }{ dT } $$

    $$ PV=nRT$$

    $$ V=\dfrac { nRT }{ P } $$

    $$ { \alpha  }_{ V }=\dfrac { P }{ nRT } \dfrac { dV }{ dT } $$

    $$ { \alpha  }_{ V }\rightarrow \dfrac { 1 }{ T } $$
  • Question 4
    1 / -0
    A sample of an ideal gas is contained in a cylinder. The volume of the gas is suddenly decreased. A student makes the following statements to explain the change in pressure of the gas.
    I. The average kinetic energy of the gas atoms increases.
    II. The atoms of the gas hit the walls of the cylinder more frequently.
    III. Temperature of the gas remains unchanged.
    Which of these statements is true?
    Solution
    Since the volume of the 'closed cylinder' decreases, the pressure inside the cylinder increases, which is a measure of frequency of atoms hitting the walls of the cylinder. 
    Also the temperature increases, which is a measure of average kinetic energy of the gas atoms.
  • Question 5
    1 / -0
    A certain mass of an ideal gas undergoes a reversible isothermal compression. Its molecules, compared with the initial state, will then have the same
    (i) root mean square velocity
    (ii) mean momentum
    (iii) mean kinetic energy
    Solution
    Given process is isothermal i.e, $$T$$ is constant so, root mean velocity $$=\sqrt { \dfrac { 3KT }{ M }  } $$ is constant mean momentum is always zero mean kinetic energy is $$\dfrac{1}{2}MV^2_{av}$$ where
    $$\Rightarrow V_{av}=\sqrt { \dfrac { 8KT }{ \pi M }  } $$ is constant at constant $$T.$$
    Hence, $$(i),(ii),(iii)$$ are correct answer.
  • Question 6
    1 / -0
    The number of air molecules in a ( 5m x 5m x 4m ) room at standard temperature and pressure is of the order of :
    Solution
    Volume is $$100m^3$$
    $$PV=NKT$$
    $$10^5\times 100=N\times1.38\times10^{-23}\times 273$$
    $$N=2.69\times10^{27}=3\times10^{27}$$
  • Question 7
    1 / -0
    The mean kinetic energy of a gas molecule is proportional to 
    Solution
    The average kinetic energy of gas molecules is directly proportional to absolute temperature only; this implies that all molecular motion ceases if the temperature is reduced to absolute zero.
    Hence, option C is correct.

  • Question 8
    1 / -0
    Which of the following statement is incorrect for gases?
    Solution
    The mass of a gas can be determined by weighing the container in which the gas is enclosed and again weighing the container after removing the gas. The difference between the two weights gives the mass of the gas. The mass of the gas is related to the number of moles of the gas i.e mass in gm/molar mass.
  • Question 9
    1 / -0
    The internal energy of a gas:
    Solution
    At a given temperature, the pressure of a container is determined by the number of times gas molecules strike the container walls. If the gas is compressed to a smaller volume, then the same number of molecules will strike against a smaller surface area; the number of collisions against the container will increase, and, by extension, the pressure will increase as well.
    Increasing the kinetic energy of the particles will increase the pressure of the gas.
    So, the internal energy of a gas Is the total kinetic energy of randomly moving molecules.
    Hence, option C is correct.
  • Question 10
    1 / -0
    When the temperature is increased from 0$$^o$$C to 273$$^o$$C, in what ratio the average kinetic energy of molecules changes?
    Solution
    Average K.E. $$= \displaystyle \frac{3}{2}$$ RT
    At 0$$^o$$C, average K.E. $$= \displaystyle \frac{3}{2} \times R \times 273$$
                                               $$[T = (0 + 273) K]$$
    At 273$$^o$$C,    
    average K.E. $$= \displaystyle \frac{3}{2} \times R \times (273 + 273)$$
    $$= \displaystyle \frac{3}{2} \times R \times 2 \times 273$$
    $$\therefore $$ Ratio = 2
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