Ask Question
9 May, 18:53

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 1.00*105 Pa to 4.00*105 Pa. The second process is a compression to a volume of 5.00*10-2 m3 at a constant pressure of 4.00*105 Pa. Find the total work done by the gas during both processes.

+3
Answers (1)
  1. 9 May, 21:36
    0
    The work done = - 2 x 10⁴ J

    Explanation:

    In the first case, the volume is kept constant and pressure varies.

    In isothermal process, the work done

    W₁ = V x ΔP

    here V is the volume of gas and ΔP is the change in pressure

    Thus W₁ = 0

    Because there is no change in volume, therefore displacement is zero.

    In second case pressure is constant, but volume changes

    Thus W₂ = P x ΔV

    here P is the pressure and ΔV is the change in volume

    Therefore W₂ = 4 x 10⁵ x 5 x 10⁻² = 2 x 10⁴ J

    The total work done W = - 2 x 10⁴ J

    Because the work done in compression is negative.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 1.00*105 Pa to ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers