Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IDH/DU for reactionHard2 minQB
ChemistryHardnumerical

A volume of 50 ml of a liquid in an isolated container is at 1 bar. The pressure is raised abruptly to 41 bar, which decreases the volume of the liquid by 0.5 ml. The change in enthalpy Δ H of the liquid is (Answer as 'abcd', where a = 1 if Δ H is +ve and a = 2 if Δ H is -ve, and 'bcd' is the magnitude of Δ H in J, nearest integer)

Answer:
1198.00
Solution:

Isolated: Δ U = 0. Δ H = P2V2 - P1V1 = 41×49.5 - 1×50 = 2029.5 - 50 = 1979.5 bar ml = 197.95 J ≈ 198 J (+ve). Using VΔ P alone would give a wrong value. Ans = 1198.

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:DH/DU for reaction
⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...