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Asked by vksinghdrdoceo | 04 Nov, 2020, 19:49: PM
Expert Answer
Question - (11)
In isobaric process , pressure P is constant
From perfect gas equation , we get , P = ( ρ R T ) / M ..............(1)
where ρ is density , R is universal gas constant , T is absolute temperature and M is molar mass .
If P is constant , then we get from eqn.(1) , ( ρ T ) is constant
Hence from column-1 , graph (IV) describes the isobaric process
at constant pressure , specific heat for monoatomic gas = (5/2) R per mole that matches with (iii) of column-2
workdone = Pressure × change in volume = P( V2 - V1 ) = nR ( T2 - T1)
workdone matches with (P) of colum-3
Hence (IV) (iii ) (P) matches with isobaric process.
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Question-(12)
All the options are related to graphs (II) and (III) .
Process indicated by graphs (II) and (III) are not realistic for monoatomic gases because
for gas , pressure is directly proportional to temperature . Similarly volume is directly proportional to temperature .
hence none of the option is correct.
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Question-(13)
If in a process , change in internal energy ΔU = 0 , then it is isothermal process
Gaph for isothermal process in colum-1 is (I)
Since internal energy ΔU = m × C × ΔT = 0
where ΔT is change in temperature , ΔT = 0 , hence we get C = ∞ in above equation
Hence (iv) in column-2 is matching for molar heat capacity
for isothermal process , workdone = nRT ln(V2 / V1 ) , matches to the equation (S) given in colum-3
for ΔU = 0 , matching combination is (I) (iv) (S)
Answered by Thiyagarajan K | 05 Nov, 2020, 09:26: AM
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