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Assuming the system performs only pressu...

Assuming the system performs only pressure-volume work, show that heat absorbed by a closed system at constant volume is equal to the increase in its internal energy.

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Assuming the system performs only pressure-volume work, show that heat evolved by a closed systemm at constant pressure is equal to the decrease in its enthalpy.

Pressure-volume (PV) work done by an ideal gaseous system at constant volume is (where E is internal energy of the system)

Knowledge Check

  • P*V Work done by an ideal gaseous system at constant volume is (E is internal energy of the system)-

    A
    `-(DeltaP)/(P)`
    B
    zero
    C
    `-VDeltaP`
    D
    `-DeltaE`.

  • System performed pressure volume work when in between system and surrounding there is

    A
    difference in temperature
    B
    difference in volume
    C
    difference in pressure
    D
    difference in internal energy

  • Pressure of a gas increases when its temperature is increased at constant volume. This is because, with increase in temperature -

    A
    collision frequency of the gas molecules increases.
    B
    motions of the gas molecules become more random
    C
    gas molecules make more collision with the walls of the container
    D
    compresesibility factor of the gas increases.

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    Pressure of a gas increases when its temperature is increased at constant volume. This is because, with increase in temperature -

    Certain amount of an ideal gas (gamma = 1.4) performs 80J of work while undergoing isobaric expansion. Find the amount of heat absorbed by the gas in the process and the change in its internal energy.

    Does the volume of a closed system remain fixed?

    For a process occurring in a closed system, DeltaU=q . If only pressure-volume work is performed by the system then which type of process is this?

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