0

Guest

Courses New

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
12 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
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
askIITians Faculty 628 Points
3 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
 

Think You Can Provide A Better Answer ?

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More