Question

At NTP 2.8 litres of oxygen were mixed with 19.6 litres of hydrogen.
Calculate the increase in entropy; assume ideal gas behavior.

Answer 1

Answer:

Let us consider the entropy =2.303nRlog(  

V  

1

​  

 

V  

2

​  

 

​  

)

Putting all given values in equation,

=2.303×1×8.314×log(  

11.2

22.4

​  

)

Since,n=1

=2.303×1×8.314×log  

2

1

​  

 

Implies that

=5.76JK  

−1

Explanation:

Answer 2

Given,

2.8 Litres of Oxygen were mixed with 19.6Litres of Hydrogen.

To Find,

What is the actual increase in entropy.

Solution,

Here n =1 (As Hydrogen)

$V_{1}$ = 19.6L

$V_{2}$=(19.6+2.8)=22.4 L.

Molar gas constant or R-value for any ideal gas = 8.314j$k^{-1} mol^{-1}$.

We know the formula of change in Entropy,

=2.303×n×R×㏒$(\frac{V_{2} }{V_{1}} )$,

Here the change of entropy = 2.303×1×8.314×㏒$(\frac{22.4}{19.6} )$

=2.303×8.314×0.058(Approx)

=1.11(Approx) J$k^{-1}$.

Hence, The increase in entropy for this case will be 1.11 J$k^{-1}$.