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Physical Chemistry

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

Profile image of Arsh alam
14 Years agoGrade
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3 Answers

Profile image of Rahul Kumar
14 Years ago

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.

Profile image of Shubham Kumar Singh
9 Years ago
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
Profile image of Kushagra Madhukar
5 Years ago
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