Question

calculate the total number of degrees of freedom possessed by the molecules in 1 cm cube of H2 gas at temperature 273 Kelvin and 1 ATM pressure​

Answer 1

The total number of degrees of freedom possessed by the molecules is

$1.34\times 10^{20}$

The gas is kept at 273 K and 1 atm pressure.

So, we can also say that the gas is kept in STP .

Now, we know at STP :

One mole of gas has volume of $22400 \ cm^3$ .

We can also say, $6.022\times 10^{23}$ molecules of $H_2$ have volume equal to $22400 \ cm^3$.

Therefore ,

Total number of molecules in $1\ cm^3$ is , $n=\dfrac{6.022\times 10^{23}}{22400}=2.68\times 10^{19}$ molecules.

Now,

We know $H_2$ is diatomic gas .

Therefore , degree of freedom of one molecule is 5 [ degree of freedom of diatomic gas is 5 ].

So , degree of freedom of $2.68\times 10^{19}$ molecules is , $2.68\times 10^{19}\times 5=1.34\times 10^{20}.$

Hence , this is the required solution.

Learn More :

Degree of freedom

https://brainly.in/question/10248345

Answer 2

Explanation:

The gas is kept at 273 K and 1 atm pressure.

So, we can also say that the gas is kept in STP .

Now, we know at STP :

One mole of gas has volume of 22400 \ cm^322400 cm

3

.

We can also say, 6.022\times 10^{23}6.022×10

23

molecules of H_2H

2

have volume equal to 22400 \ cm^322400 cm

3

.

Therefore ,

Total number of molecules in 1\ cm^31 cm

3

is , n=\dfrac{6.022\times 10^{23}}{22400}=2.68\times 10^{19}n=

22400

6.022×10

23

=2.68×10

19

molecules.

Now,

We know H_2H

2

is diatomic gas .

Therefore , degree of freedom of one molecule is 5 [ degree of freedom of diatomic gas is 5 ].

So , degree of freedom of 2.68\times 10^{19}2.68×10

19

molecules is , 2.68\times 10^{19}\times 5=1.34\times 10^{20}.2.68×10

19

×5=1.34×10

20

.

Hence , this is the required solution.

Learn More :

Degree of freedom

https://brainly.in/question/1024834