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 , 13 Years ago
Grade
3 Answers
Rahul Kumar
Last Activity: 13 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
Last Activity: 7 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
Last Activity: 4 Years ago
Dear student,
Please find the attached solution to your problem below.