Question

a student forgot to add the reaction mixture to the round bottom flask at 27°c but istead he placed the flask on the flame. after a lapse pf time he realized his mistake and using a pyrometer he found the temp. of flask was 477°c what fraction pf air would have been expelled out ?​

Answer 1

Answer:

Explanation:

Let the volume of the round-bottomed flask be V.

Then, the mass of air inside the flask at 27° C is V.

Now,

V1 = V

T1 = 27°C = 300 K

V2 =?

T2 = 477° C = 750 K

According to Charles’s law,

V1/T1 = V2/T2

=> V2 =V1T2/T1

         = V(750/300)

        = 2.5 V

So, volume of air expelled out = 2.5 V – V = 1.5 V

Hence, fraction of air expelled out = (1.5 V)/2.5 V) = 3/5

Answer 2

Explanation:

For an open vessel containing gas inside, pressure and volume corresponding to the gas remains the same.

So nT= constant for the gas inside.

Therefore n1T1=n2T2

⇒n1(273+27)=n2(227+23)→300n1=500n2→n2=35n1

So, 60% of the original amount of air is still there. Therefore the fraction of air expelled is two-fifths, 25.