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In this quesn, how they have assumed thta final pressure of both the vessels will be same. 
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Asked by aaryamanmodern | 17 Jan, 2023, 10:18: PM
answered-by-expert Expert Answer
Initially both vessels of equal volume V is at same pressure Po and absolute temperature  To .
 
Let n be number of moles in each vessel . Hence we write ideal gas equation as
 
begin mathsize 14px style P subscript o space left parenthesis 2 V right parenthesis space equals space left parenthesis 2 n right parenthesis space R space T subscript o end style
 
where R is universal gas constant. Hence we write
 
begin mathsize 14px style left parenthesis 2 n right parenthesis space equals space fraction numerator P subscript o space left parenthesis 2 V right parenthesis over denominator R space T subscript o end fraction end style  ..............................................(1)
If one vessel is maintained at temperature To and other one is maintained at temperature T1  ( T1 > To ) ,
 
then vessel at temperature To will have more number of moles and vessel at temperature T1
 
will have less number of moles because both vessels are connected by a narrow tube
 
and they are at equal pressure.
 
Let n1 be the number of moles in the vessel that is maintained at temperature To .
 
Let n2 be the number of moles in the vessel that is maintained at temperature T1 .
 
Then we have ,   begin mathsize 14px style n subscript 1 space equals space fraction numerator P space V over denominator R space T subscript o end fraction end style  .............................(2)
and   begin mathsize 14px style n subscript 2 space equals space fraction numerator P space V over denominator R space T subscript 1 end fraction end style......................................(3)
 
By adding eqn.(2) and (3) , we get
 
begin mathsize 14px style n subscript 1 space plus space n subscript 2 space equals space fraction numerator P space V over denominator R space end fraction open parentheses 1 over T subscript o space plus space 1 over T subscript 1 close parentheses space end style .............................(4)
Since number of moles in both vessels remain constant, n1 + n2 = ( 2 n )
 
Hence we equate eqn.(1) and (4) to get
 
begin mathsize 14px style space fraction numerator P space V over denominator R end fraction open parentheses 1 over T subscript o plus 1 over T subscript 1 close parentheses space space equals space space fraction numerator P subscript o space left parenthesis 2 V right parenthesis over denominator R space T subscript o end fraction space space end style
 
begin mathsize 14px style P space open parentheses fraction numerator T subscript o plus T subscript 1 over denominator T subscript o space T subscript 1 end fraction close parentheses space equals space fraction numerator 2 space P subscript o over denominator T subscript o end fraction end style
begin mathsize 14px style P space equals space fraction numerator 2 space P subscript o over denominator T subscript o end fraction space cross times space fraction numerator T subscript o space T subscript 1 over denominator T subscript o plus T subscript 1 end fraction space equals space P subscript o space cross times fraction numerator 2 space T subscript 1 over denominator T subscript o plus T subscript 1 end fraction end style
Let us substitute values , Po = 105 kPa ,  To = 300 K  and  T1 = 400 K
 
P = 105 × ( 800/700) = 120 kPa
Answered by Thiyagarajan K | 17 Jan, 2023, 11:40: PM

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