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, ΔU = 30 .0 Latm . The change on enthalpy of proces

Question

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, $ΔU = 30 .0 Latm$ . The change on enthalpy of process (in L-atm) is

Answer:

44.00

Solution:

$Δ H = Δ U + (P_{2} V_{2} - P_{1} V_{1})$
$= 30 + \{(4 \times 5) - (2 \times 3)\}$
$= 44 Latm$

TestHub · Accepted Answer

Δ H = Δ U + P 2 ⁡ V 2 ⁡ - P 1 ⁡ V 1 = 30 + 4 × 5 − 2 × 3 = 44 Latm

Chemistry - Thermodynamics - 1 Question with Solution | TestHub

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