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A gas can be liquefied by pressure alone...

A gas can be liquefied by pressure alone when its temperature

A

higher than its critical temperature

B

lower than its critical temperature

C

either of these

D

none

Text Solution

Verified by Experts

The correct Answer is:
B
A gas cab be liquefied only if its temperature is lower than its critical temperature.
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A gas can be liquefied most suitably at .

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Knowledge Check

  • A gas can be liquefied if:

    A
    forces of attraction are low under ordinary conditions
    B
    forces of attraction are high under ordinary conditions
    C
    forces of attraction are zero under ordinary conditions
    D
    forces of attraction either high or low under ordinary conditions

  • A real gas can be liquefied:

    A
    under adiabatic expansion
    B
    above critical temperature
    C
    when cooled below critical temperature under applied pressure
    D
    at temperature lower than critical temperature and pressure higher than critical pressure

  • The temperature below which a gas should be cooled before it can be liquefied by pressure only is termed as

    A
    critical temperature
    B
    triple point
    C
    dew point
    D
    freezing point

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    A gas is liquefied

    A gas is liquefied

    A real gas can be liquefied by the application of pressure when its temperature is

    An ideal gas can't be liquefied because

    The essential conditions for liquefaction of gases were discovered by Andrews in 1869 as a result of his study of pressure-volume-temperature relationship for CO_(2) . If was found that above a certain temperature, it was impossible to liquefy a gas whatever the pressure was applied. The temperature below which the gas can be liquefied by the application of pressure alone is called critical temperature (T_(c)) . The pressure required to liquefy a gas at this temperature is called the critical pressure (P_(c)) . The volume occupied by one mole of the substance at the critical temperature and pressure is called critcal volume. Critical constants are related with van der Waals' constant as follows: V_(c) = 3b, P_(c) = (a)/(27b^(2)), T_(c) = (8a)/(27 Rb) Gas A and can be liquefied at room temperature by applying pressure but gas B cannot. This reflects:

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