One mole of a real gas at its critical temperature has a relation between its pressure in atm and volume in litres as follows-P-2RTc-V-3X-V2-Y-V3Where Tc is the critical temperature in K and R is the gas constant while X and Y are some constants- Find the correct options about the gas and its given relation-

At critical temperature, we know that, dP/dV=0 ⇒ 2RT/V^2+6X/V^3 3Y/V^4=0 ⇒ 2RT_cV^2 6XV+3Y=0 This quadratic equation should have only one root corresponding to ...

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Standard XI

Chemistry

Real Gases

One mole of a...

Question

One mole of a real gas at its critical temperature has a relation between its pressure (in atm) and volume (in litres) as follows:
$P = \frac{2 R T_{c}}{V} - \frac{3 X}{V^{2}} + \frac{Y}{V^{3}}$
Where $T_{c}$ is the critical temperature (in K) and R is the gas constant while X and Y are some constants. Find the correct option(s) about the gas and its given relation.

The unit of the constant X is

a t m L

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The unit of the constant Y is atm

a t m L^{3}

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T_{c} = \frac{3 X^{2}}{2 R Y}

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At critical volume

\frac{d P}{d V} = 0

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Solution

The correct option is D At critical volume $\frac{d P}{d V} = 0$
At critical temperature, we know that, $\frac{d P}{d V} = 0$
$\Rightarrow - \frac{2 R T}{V^{2}} + \frac{6 X}{V^{3}} - \frac{3 Y}{V^{4}} = 0$
$\Rightarrow 2 R T_{c} V^{2} - 6 X V + 3 Y = 0$
This quadratic equation should have only one root corresponding to $V_{c}$
So, the discriminant $b^{2} - 4 a c = 0$ $(6 X)^{2} - 4 \times 2 R T_{c} \times 3 Y = 0$
$\Rightarrow T_{c} = \frac{3 X^{2}}{2 R Y}$

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Similar questions

Q. One mole of a real gas at its critical temperature has a relation between its pressure (in atm) and volume (in litres) as follows:

P = \frac{2 R T_{c}}{V} - \frac{3 X}{V^{2}} + \frac{Y}{V^{3}}

Where

T_{c}

is the critical temperature (in K) and R is the gas constant while X and Y are some constants. Find the correct option(s) about the gas and its given relation.

Q. One mole of a real gas at its critical temperature has a relation between its pressure (in atm) and volume (in litres) as follows:

P = \frac{2 R T_{c}}{V} - \frac{3 X}{V^{2}} + \frac{Y}{V^{3}}

Where

T_{c}

is the critical temperature (in K) and R is the gas constant while X and Y are some constants. Find the correct option(s) about the gas and its given relation.

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}

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^{'}

Q. Critical temperature and pressure of a real gas is

27^{o} C

and 75 atm respectively. The critical volume of 1 mole of real gas is :

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The correct option is D At critical volume dPdV=0At critical temperature, we know that, dPdV=0 ⇒−2RTV2+6XV3−3YV4=0 ⇒2RTcV2−6XV+3Y=0 This quadratic equation should have only one root corresponding to Vc So, the discriminant b2−4ac=0(6X)2−4×2RTc×3Y=0 ⇒Tc=3X22RY