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Liquefaction of Gases and Critical Constants

The greater t...

Question

The greater the critical temperature of a gas, the easier is the liquefaction of the gas.

A

True

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B

False

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C

Nither

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D

None of the above

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Solution

The correct option is C True
As critical temperature increases, the volume decreases( $V α \frac{1}{T}$ ) hence the liquefaction process becomes easier.

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Similar questions

Q. Which of the following statement(s) is/are true:
I - Slope of isotherm at critical point is maximum.
II - Larger the value of

T_{c}

, easier is the liquefaction of gas.
III - van der Waals' equation of state is applicable below critical temperature at all pressures.

Q. Assertion :Greater the value of van der Waals constant 'a', easier is the liquefaction of a gas. Reason: 'a' indirectly measures the magnitude of attractive forces between the molecules.

Q. Assertion :Greater is critical temperature more easily a gas can be liquified. Reason: Critical temperature depends on the magnitude of intermolecular forces of attraction between the gas.

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

C O_{2}

. It 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 critical volume. Critical constants are related with van der Waals' constant as follows:

V_{c} = 3 b, P_{c} = \frac{a}{27 b^{2}}, T_{c} = \frac{8 a}{27 R b}

.

Which of the following parameters is three times the van der Waals' constant

^{'} b^{'}

?

Which of the following statements on critical constants of gases are correct?

I : Larger the $\frac{T_{c}}{P_{c}}$ value of a gas, larger would be the excluded volume.

II : Critical temperature ( $T_{c}$ ) of a gas is greater than its Boyle's temperature ( $T_{B}$ ).

III : At the critical point in the Van der Walls gas isotherm.

$⟮ \frac{d P}{d V_{m}} ⟯ T_{c} = 0$

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Liquefaction of Gases and Critical Constants

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