Question

The relation between critical constants is represented as 8Pc Vc = XRTc ,what is the value of 'x'

Answer 1

Answer:

Solution

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Correct option is

A

RT

c

P

c

V

c

=

8

3

V

c

=3b,P

c

=

27b

a

a

andT

c

=

27Rb

8a

RT

c

P

c

V

c

=

R×

27Rb

8b

27b

2

a

×3b

=

27Rb

8a

9b

a

=

27b

8Ra

9b

a

RT

c

P

c

V

c

=

9b

a

×

8a

27b

=

8

3

=0.375

Answer 2

Answer:

For the given relation of critical constants the value of 'x' is equal to 3.

Explanation:

As the kinetic theory of gases, the gas molecules are constantly moving. When the temperature decreases, the kinetic energy of molecules also decreases, the intermolecular force of attraction increases. So, the liquification of gas takes place.

All the gases can be by lowering the temperature at atmospheric pressure. But it is not possible to liquify all gases at room temperature by increasing the pressure.

Critical temperature $T_{c}$ is the critical temperature above which a gas cannot be liquified.

Critical pressure $P_{c$ is the minimum pressure sufficient to liquify gas at critical temperature.

Critical volume $V_{c}$ is the volume occupied by one mole of gas at critical temperature.

Now, $T_{c}=\frac{8a}{27Rb}$                                                   ................(1)

and, $V_{c}=3b$                                                         ..............(2)

and, $P_{c}=\frac{a}{27b^{2}}$                                                       .............(3)

From equation (2):

$b=\frac{V_{c}}{3}$

Fill the value of 'b' in equation (3);

$P_{c}=\frac{a}{27(\frac{V_{c}}{3} )^{2}}$

$a=3P_{c}V_{c}^{2$

Fill the value of 'a' and 'b' in equation (1);

$T_{c}=\frac{8*3*P_{c}V_{c}^{2}}{27*R*(\frac{V_{c}}{3} )}$

$T_{c}=\frac{8P_{c}V_{c}}{3R}$

$8P_{c}V_{c}=3RT_{c}$

Therefore, the value of x is equal to 3.