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Kinetic Theory Of Gases Test - 4

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Kinetic Theory Of Gases Test - 4
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  • Question 1
    1 / -0

    At absolute zero temperature may be defined as that temperature at which

    Solution

    At 0K temperature we know that there is no molecular motion, that is the KE of the particles gets 0. Thus we can say the combined KE of a gaseous system is zero, but as there combined mass cant be zero thus the combined of the square of velocities of the particles is zero, which means that the root mean square velocity of the gas is zero.

  • Question 2
    1 / -0

    Pressure exerted by an ideal gas molecule is given by the expression

  • Question 3
    1 / -0

    If the r.m.s speed of oxygen at NTP is x m/s. If the gas is heated at constant pressure till its volume is four fold, what will be its final temperature and r.m.s speed?

    Solution


  • Question 4
    1 / -0

    The perfect gas equation is PV = nRT where n is the

    Solution

    n is the number of moles.

  • Question 5
    1 / -0

    In a mixture of ideal gases at a fixed temperature the heavier molecule has the lower average speed. This is easiest to conclude from the equation

    Solution

    In option A, v is inversely proportional to mass. Hence, the condition of the question is seen accurately in that option.

  • Question 6
    1 / -0

    According to kinetic theory of gases, 0K is that temperature at which for an ideal gas

    Solution

    According to kinetic theory of gases, 0K is that temperature at which for an ideal gas the internal energy is zero because at 0K nearly all molecular motion stops.

  • Question 7
    1 / -0

    The kinetic energy of one mole of an ideal gas is E=3/2 RT. Then Cρ will be

    Solution

    We know that,
    Cp=cv+R
    E=3Rt/2
    also,
    E=Cv
    cp=R+3RT/2
    =5/2 RT
    =2.5R

  • Question 8
    1 / -0

    Value of gas constant, R for one mole of a gas is independent of the

    Solution

    We know that PV=nRT also PM=dRT
    So in the equation The value of R depends on P , V , n , T , d , M
    except atomicity 
     so the ans is A

  • Question 9
    1 / -0

    The pressure exerted by an ideal gas is numerically equal to _________ of the mean kinetic energy of translation per unit volume of the gas.

    Solution

    Translational (KE) per unit volume

  • Question 10
    1 / -0

    The average kinetic energy of a molecule in an ideal gas is

    Solution

    KE = f/2 KT

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