phy mock questions

1 Answer

  • 天同
    Lv 7
    8 years ago
    Favorite Answer

    1. Cold air has a high density and hence sinks to the floor. Hot air rises and is cooled down by the air-conditioner. Thus a convection current is set up.

    2. (a) Use ideal gas equation: PV = nRT

    n = PV/RT

    No. of moles of gas in partition A, n1 = P(20A)/R(300)

    No of moles of gas in partition B, n2 = P(30A)/R(300)

    where A is the cross-sectional area of the container

    Hence, n1/n2 = 20/30 = 2/3

    Since mass of gas is proportional to the no. of moles,

    ratio of gas masses = 2:3

    (b) Use: P = (1/3).d.c^2

    where d is the gas density, and c is the rms speed

    Before the gas is heated, Po = (1/3)d(co)^2

    After the gas temperature is raised to 477'C (= 750 K)

    P = (1/3).d(c)^2

    hence, P/Po = (c/co)^2

    By Pressure Law, Po/300 = P/750

    i.e. P/Po = 750/300

    Thus, (c/co)^2 = 750/300

    c/co = square-root(750/300) = 1.58

    (c) New pressure at partition A, Pa = (n1)R(300)/(L1)A

    New pressure at partition B, Pb = (n2)R(750)/(L2)A

    where L1 and L2 are the new lengths of partitions A and B respectively

    When the piston is in the new equilibrium position, Pa = Pb

    (n1)R(300)/(L1)A = (n2)R(750)/(L2)A

    (L1)/(L2) = (n1/n2).(300/750) = (2/3).(300/750) = 0.2667

    But L1 + L2 = 50

    thus, L1 + L1/0.2667 = 50

    L1 = 10.53 cm

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