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Kinetic Theory Test - 31

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Kinetic Theory Test - 31
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  • Question 1
    1 / -0

    28gm of $$N_2$$ gas is contained in a flask at a pressure of 10atm and at a temperature of 57$$^{0}$$C. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to 27$$^{0}$$C. The quantity of N$$_{2}$$ gas that leaked out is

    Solution
    Since the volume of the flask remains the same, using Ideal Gas Equation we claim,
    $$n\propto \dfrac{P}{T}$$
    $$\implies \dfrac{n_2}{n_1}=\dfrac{P_2T_1}{P_1T_2}$$
    $$\implies n_2=\dfrac{11}{20}n_1$$
    Therefore, fraction of gas leaked out is $$(1-\dfrac{11}{20})=\dfrac{9}{20}$$
    Thus, amount of gas leaked out=$$\dfrac{9}{20}\times 28gm=\dfrac{63}{5}gm$$
  • Question 2
    1 / -0

    $$Assertion$$: gases are characterised with two coefficients of expansion

    $$Reason$$: when heated both volume and pressure increase with the rise in temperature

    Solution
    The ideal gas equation $$PV=\eta RT$$ tells us that the pressure and volume are directly proportional to the temperature. The coefficients of expansion for pressure is $$\gamma_p$$ and that of volume is $$\alpha_v$$.
  • Question 3
    1 / -0

    Assertion: Gasses obey Boyle's law at high temperature and low pressure only.

    Reason: At low pressure and high temperature, gasses would behave like ideal gases.

    Solution
    Due to non idealities gases obeys vander waals equation instead of ideal equation and at high temperature and low pressure vander waals equation becomes ideal gas equation by taking approximations. therefore, at high temperatur and low pressure gases obeys boyles law and behaves ideally.
    assertion is true reason is true and reason is correct explanation of assertion.
    option(A)
  • Question 4
    1 / -0

    Follwing operation are carried out on a sample of ideal gas initially at pressure P volume V and kelvin temperature T.

    a) At constant volume, the pressure is increased fourfold.

    b) At constant pressure, the volume is doubled

    c) The volume is doubled and pressure halved.

    d) If heated in a vessel open to atmosphere, one-fourth of the gas escapes from the vessel.

    Arrange the above operations in the increasing order of final temperature.

    Solution
    ideal gas equation
    $$PV = nRT$$          eq(1)
    where
    $$P$$ = Pressure
    $$V$$ = Volume
    $$n$$ = number of moles
    $$R$$ = universal gas constant
    $$T$$ = Temperature

    suppose initial values are $${P}_{1},{V}_{1},{n}_{1},{T}_{1}$$ and
    final values are $${P}_{2},{V}_{2},{n}_{2},{T}_{2}$$

    so
    $${P}_{1}{V}_{1}={n}_{1}R{T}_{1}$$          eq(2)
    $${P}_{2}{V}_{2}={n}_{2}R{T}_{2}$$          eq(3)

    (a) At constant volume, the pressure is increased fourfold
    $${V}_{2} = {V}_{1}$$
    $${P}_{2} = 4 {P}_{1}$$
    $${n}_{2} = {n}_{1}$$
    $$\dfrac{eq(3)}{eq(2)}$$
    $$\dfrac{{T}_{2}}{{T}_{1}} = 4$$
    $${T}_{2} = 4 {T}_{1}$$

    (b) At constant pressure, the volume is doubled
    $${V}_{2} = 2 {V}_{1}$$
    $${P}_{2} = {P}_{1}$$
    $${n}_{2} = {n}_{1}$$
    $$\dfrac{eq(3)}{eq(2)}$$
    $$\dfrac{{T}_{2}}{{T}_{1}} = 2$$
    $${T}_{2} = 2 {T}_{1}$$

    (c) The volume is doubled and pressure halved.
    $${V}_{2} = 2 {V}_{1}$$
    $${P}_{2} =\dfrac{1}{2} {P}_{1}$$
    $${n}_{2} = {n}_{1}$$
    $$\dfrac{eq(3)}{eq(2)}$$
    $$\dfrac{{T}_{2}}{{T}_{1}} = 1$$
    $${T}_{2} = {T}_{1}$$

    (d) If heated in a vessel open to atmosphere, onefourth of the gas escapes from the vessel.

    for open vessel pressure and volume will be constant.

    $${V}_{2} = {V}_{1}$$
    $${P}_{2} = {P}_{1}$$

    one-fourth of gas escaped from vessel, so three-fourth gas left
    $${n}_{2} = \dfrac{3}{4}{n}_{1}$$

    $$\dfrac{\dfrac{3}{4} {T}_{2}}{{T}_{1}} = 1$$

    $${T}_{2} = \dfrac{4}{3}{T}_{1}$$

    arranging final temperature in increasing order

    (c),(d),(b),(a)
  • Question 5
    1 / -0

    PV = n RT holds good for :

    a) Isobaric process          b) Isochoric process

    c) Isothermal process     d) Adiabatic process

    Solution
    PV=nRT
    is ideal gas equation it is valid for each and every process neglecting the non idealities.
  • Question 6
    1 / -0

    Assertion: In Joules bulb apparatus, as reservoir is moved up, the mercury level raises into the bulb. 

    Reason: The pressure on the enclosed gas increases.

    Solution
    Due to increment in pressure, to balance forces, mercury level in bulb must rise.Increment in height will necessarily ensure that Joules bulb apparatus has been in equilibrium condition

  • Question 7
    1 / -0

    Assertion:With increase in temperature, the pressure of given gas increases

    Reason:Increase in temperature causes decrease in no. of collision of molecules with walls of container.

    Solution
    With increase in temperature, the pressure of given gas increases. This is because the increase in temperature causes increase in no. of collision of molecules with the walls of the container.
  • Question 8
    1 / -0

    According to kinetic theory of gasses at Zero kelvin

    a) Pressure of ideal gas is zero

    b) Volume of ideal gas is zero

    c) Internal energy of ideal gas is zero

    d) Matter exists in gaseous state only

    Solution
    Since at zero Kelvin it is not true that matter exist in gaseous state only. 
    Where as from PV=nRT
    P=0
    V=0
    KE=3RT/2=0; since T=0 K
    Therefore, option(C) is true.

  • Question 9
    1 / -0

    Which of the following processes will quadruple the pressure

    a) Reduce V to half and double T

    b) Reduce V to 1/8th and reduce T to half

    c) Double V and half T

    d) Increase both V and T to double the values

    Solution
    (a) First we half the volume at constant T pressure will become double now we double our temperature keeping volume same it will again double the pressure and finally pressure will become quadruple.
    (b)When we make volume V/8 keeping temperature constant pressure becomes 8 times now when we half the temperature pressure also gets halved and become 4 times
    (c)When we double V and half T pressure will become 1/4 times
    (d) if we increase both V and T to double then pressure will not change.
    Therefore only (a) & (b) are correct
    Therefore option(B) is correct
  • Question 10
    1 / -0

    Assertion:PV/T=constant for 1 mole of gas. This constant is same for all gases.

    Reason:1 mole of different gases at NTP occupy same volume of 22.4 litres.

    Solution
    For 1 mole 
    PV/T=R
    and universal gas constant R is constant for every gas
    therefore 1 mole of gas at NTP will also occupie 22.4 L of volume
    Assertion is true  reason is true 
    and reason is correct explaination
    option(A)
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