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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 , 13 Years ago
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anser 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.
 
∆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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