NEET-XII-Chemistry

Previous Year Paper year:2017

with Solutions - page 3
  • #33
    A gas is allowed to expand in a well insulated
    container against a constant external pressure of
    2.5 atm from an initial volume of 2.50 L to a final
    volume of 4.50 L. The change in internal energy ΔU
    of the gas in joules will be -
    (1) +505 J
    (2) 1136.25 J
    (3) - 500 J
    (4) - 505 J
    digAnsr:   4
    Ans : (4)
    Sol. ΔU = q + w
    Insulated container So, q = 0
    ΔU = -PdV
    = - 2.5 [4.50 - 2.50]
    = - 2.5 × 2 litre - atm = - 5 l atm [1 -atm = 101.3 ≈ 101J]
    = - 5 × 101
    ⇒ - 505 J