One mole of a non-ideal gas changes state from (1.0 atm, 4.0 L, 100 K) to (3.0 atm, 6.0 L, 300 K) with a change in internal energy Δ U = 25.0 L-atm. The enthalpy change Δ H of the process in L-atm is
Answer:
39.00
Solution:
Δ H = Δ U + (P₂V₂ - P₁V₁) = 25.0 + (18.0 - 4.0) = 39 L-atm. Ans = 0039.

Question

One mole of a non-ideal gas changes state from (1.0 atm, 4.0 L, 100 K) to (3.0 atm, 6.0 L, 300 K) with a change in internal energy Δ U = 25.0 L-atm. The enthalpy change Δ H of the process in L-atm is

Answer:

39.00

Solution:

Δ H = Δ U + (P₂V₂ - P₁V₁) = 25.0 + (18.0 - 4.0) = 39 L-atm. Ans = 0039.

TestHub · Accepted Answer

Δ H = Δ U + (P 2 V 2 - P 1 V 1 ) = 25.0 + (18.0 - 4.0) = 39 L-atm. Ans = 0039.

Chemistry - Thermodynamics - I Question with Solution | TestHub

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