MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Special Integrals - 1

The value of ...

Question

The value of critical temperature in terms of Van der Waals constant a and b is given by:

A

T_{C}

=

\frac{8 a}{27 R b}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B

T_{C}

=

\frac{a}{27 R b}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

T_{C}

=

\frac{a}{2 R b}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

T_{C}

=

\frac{- a}{R b}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $T_{C}$ = $\frac{8 a}{27 R b}$
$Critical Tempenature T_{C} = \frac{8 a}{27 R b}$

Suggest Corrections

thumbs-up

0

Similar questions

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

Q. Match Column I with Column II and select the correct answers.

Column I	Column II
A. Critical temperature	1. $\frac{a}{R b}$
B. Boyle's temperature	2. $\frac{2 a}{R b}$
C. Inversion temperature	3. $\frac{T}{T_{c}}$
D. Reduced temperature	4. $\frac{8 a}{27 R b}$

Q. Assertion :The numerical values of

P_{c}, V_{c}, T_{c}

\frac{a}{27 b^{2}}, 3 b

\frac{8 a}{27 R b}

respectively. Reason: The compressibility factor

Z

at critical conditions is

\frac{3}{8}

.

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^{'}

?

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}

T_{C}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Special Integrals - 1

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Special Integrals - 1

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon