One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) arrow (4.0 atm, 5.0 L, 245 K) with a change in internal energy, Δ U = 30.0 L atm. The change in enthalpy (.Δ H.) of the process in L atm is

Question

One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) arrow (4.0 atm, 5.0 L, 245 K) with a change in internal energy, Δ U = 30.0 L atm. The change in enthalpy (.Δ H.) of the process in L atm is (A) 40.0 (B) 22.0 (C) 44.0 (D) 48.0

Tardigrade Admin · Accepted Answer

Relation between enthalpy change and internal energy change for non-ideal gas: Δ H=Δ U+Δ (.PV.) =Δ U+(.P2V2-P1V1.) =30+(.4× 5-2× 3.) =30+14=44 L.atm

Q. One mole of a 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, $Δ U$ = 30.0 L atm. The change in enthalpy $(Δ H)$ of the process in L atm is

40.0

22.0

44.0

48.0

Relation between enthalpy change and internal energy change for non-ideal gas:
$Δ H = Δ U + Δ (P V)$
$= Δ U + (P_{2} V_{2} - P_{1} V_{1})$
$= 30 + (4 \times 5 - 2 \times 3)$
$= 30 + 14 = 44$ L.atm

Q. One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) → (4.0 atm, 5.0 L, 245 K) with a change in internal energy, ΔU = 30.0 L atm. The change in enthalpy (ΔH) of the process in L atm is