One mole of non`-` ideal gas undergoes a change of state `(1.0 atm, 3.0L, 200 K )` to `(4.0 atm, 5.0L,250 K)` with a change in internal energy `(DeltaU)=40 L-atm`. The change in enthalpy of the process in `L-atm`,
A
43
B
57
C
42
D
None of these
Text Solution
Verified by Experts
The correct Answer is:
B
When both P and V are changing `" "DeltaH=DeltaU +Delta(PV)` `" "DeltaU+(P_(2)V_(2)-P_(1)V_(1))` `" "DeltaH=40+(20-3)` = 57 L-atm
Topper's Solved these Questions
THERMODYNAMICS
NARENDRA AWASTHI|Exercise Level 1 (Q.1 To Q.30)|6 Videos
THERMODYNAMICS
NARENDRA AWASTHI|Exercise Level 1 (Q.31 To Q.60)|2 Videos
One mole of a non-ideal gas undergoes a change of state from (2,0 atm, 3.0 L, 100 K) to (4.0 atm, 5.0 L, 250 K) with a change in internal energy, DeltaU=30.0 Latm. The change on enthalpy of process in (L - atm) is
One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.01, 95K) to (4.0 atm, 5.01, 245K) with a change in internal energy, DeltaU=30.0L atm. The change in enthalpy DeltaH of the process in L atm is.
One mole of a non-ideal gas undergoes a change of state ( 2.0 atm, 3.0L, 95 K) rarr (4.0 atm, 5.0L, 245 K) with a change in internal energy , Delta U= 30.0 L. atm . Calculate change in enthalpy of the process in L. atm .
One mole of non-ideal gas undergoes a change of state ( 2.0 "atm" , 3.0 L , 95 K ) to ( 4.0 "atm" , 5.0 L , 245 K ) with a change in internal energy, DeltaU=30.0 L atm . The change in enthalpy (DeltaH) of the process in L "atm" is
NARENDRA AWASTHI-THERMODYNAMICS-Level 3 - Match The Column