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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 ( D H) of the process in L atm is (A) 40.0 (B) 42.0 (C) 44.0 (D) Not defined, because pressure is not constant

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 

Grade:

3 Answers

Rahul Kumar
131 Points
10 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.

Shubham Kumar Singh
19 Points
5 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
Kushagra Madhukar
askIITians Faculty 629 Points
2 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
 

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