IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

jatinpreet

When a bottle of perume is opened, odorous molecules mix with air and slowly diffuse throughout the entire room. which is correct for this process ?

a)

G<0

b)

S>0

c)

H nearly equal to 0

d)

S<0

Please explain this question.

agent007

hi jatinpreet,

i think the answer is (b), and (c)

explanation :

G =

H - T

S

firstly, the process is spontaneous as the liquified perfume is converted into vapour. when liquid is converted to gas, entropy is +ve. so, (b) is correct.

secondly, as the process is purely spontaneous, there is no heat change involved. that means enthalpy is zero. therefore, (c) is also correct.

hope it is clear.

kavincoolzz

i think its 'a' because the process is spontaneous...

agent007

he is right, (a) is also correct i guess

jatinpreet

Thanks for ur replies.

Can u explain how is

G<0.

jatinpreet

Oh got it now. Need to use my brain more often

vinu

A NICE RECOGNITION!

akshaysood121

hello jatinpreet!

I think 'a' and 'b' are correct.

because,

G=

H - T

S

CLEARLY THE LIQUID PERFUME IS CHANGING TO VAPOUR STATE , SO THE ENTROPY IS +VE (favourable). AS IT IS MIXING WITH AIR , SO

S , WILL BE LARGE.SO

G WILL BE -VE. HENCE

G<0.

ac

Hi,
Choices (a), (b) and (c) are correct as explained above.

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