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One mole of a non-ideal gas undergoes a change of state (2.0 atm. 3.0 L, 95 K ) à (4.0 atm. 5.0 L, 245 K) with a change in internal energy , DU = 30.0L atm. The change in enthalpy (DH) of the process in L atm is
(A) 40.0
(B) 42.0
(C) 44.0
(D) Not defined, because pressure is not constant

Arsh alam , 14 Years ago
Grade
anser 3 Answers
Rahul Kumar

We define enthalpy in terms of constant pressure and internal enerfy in terms of constant volume.Hence in this case we cannot define change in enthalpy.

Last Activity: 14 Years ago
Shubham Kumar Singh
As we know that , ∆H=U2-U1 +( P2V2-P1V1)And given ∆U=30 L ATM ,P1=2atm ,P2=4atm , V1 =3L , V2 = 5L∆H = ∆U + (4*5 - 3*2) ∆H= 30 + 14∆H= 44 L atm
Last Activity: 9 Years ago
Kushagra Madhukar
Dear student,
Please find the attached solution to your problem below.
 
∆H = U2 - U1 + (P2V2-P1V1)
given, ∆U = 30 L atm ,P1 = 2atm, P2 = 4atm, V1 =3L, V2 = 5L∆H = ∆U + (4*5 - 3*2)
∆H= 30 + 14∆H = 44 L atm
 
Thanks and regards,
Kushagra
 
Last Activity: 5 Years ago
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