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Liquefaction of Gases and Critical Constants

Which of the ...

Question

Which of the following statements(s) is true?
I - The slope of isotherm at the critical point is maximum.
II - Larger the value of $T_{c}$ , easier is the liquification of the gas.
III - Van der Waals equation of state is applicable below critical temperature at all pressures.

Only I

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I & II

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II & III

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Only II

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Solution

The correct option is B I & II
I - Slope of isotherm below critical point < 0.
Slope of isotherm above critical point < 0.
Slope of isotherm at critical point = 0.

II - $T_{c} = \frac{8 a}{27 R b}$
$T_{c} \propto a$
Larger value of $T_{c}$ means less decrease in temperature is required to liquify the gas. Gas will liquify at higher temperature. So, easier will be the liquification.

III - When a gas is below the critical temperature at high pressure, it is a liquid. So, Van der Waals equation of state is not valid.

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Similar questions

T_{c}

I :

T_{c} / p_{c}

I I :

(T_{c})

(T_{b})

I I I :

(\frac{δ p}{δ V_{m}}) T_{c} = 0

The value of critical temperature $T_{C}$ for a gas is :

\frac{P}{P_{c}} = P_{r}, \frac{T}{T_{c}} = T_{r}, a n d \frac{V_{m}}{V_{m, c}} = V_{r}

P_{r}

P_{c}

T_{r}

T_{c}

V_{r}

V_{c}

P_{r}, T_{r}

V_{r}

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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