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Standard XII

Chemistry

Liquefaction of Gases and Critical Constants

An ideal gas ...

Question

An ideal gas obeying kinetic theory of gases can be liquefied, if :

Its temperature is more than critical temperature

T_{c}

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Its pressure is more than critical pressure

P_{c}

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Its pressure is more than

P_{c}

at a temperature less than

T_{c}

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It cannot be liquefied at any value of

P

and

T

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Solution

The correct option is D It cannot be liquefied at any value of $P$ and $T$
An ideal gas is defined to be a system in which there are no intermolecular/interatomic forces. Such a system can only exist as a gas. Any real system will approach ideal gas behavior in the limit that the pressure is extremely low and the temperature is high enough to overcome attractive intermolecular forces.
PV = nRT
Here, n= no of moles of gas
Ideal gas equation is a relation between four variables and it describes the state of any gas. For this reason, it is also called the Equation of State.
As ideal gas has no force of attraction and has negligible volume. Hence, it cannot be liquefied at any $T$ and $P$ .

So, the correct option is $D$

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Similar questions

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}

The relationship between

P_{C}, V_{C}

and

T_{C}

Q. Assertion (A): Real gases tend to behave as ideal gas at low pressure and high Temperature.
Reason (R): At temperature more than critical temperature, a gas can’t be liquefied just by increasing the pressure. Which of the followings is correct?

The critical temperature $(T_{c})$ and critical pressure $p_{c})$ of $C O_{2}$ are ${30.98}^{\circ} C$ and 73 atm respectively.Can $C O_{2}$ (g) be liquefied at $32^{\circ} C$ and 80 atm pressure?

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}

(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^{'}

$N H_{3}$ can be liquefied at ordinary temperature without the application of pressure, but $O_{2}$ cannot, because

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