One mole of a non - ideal gas undergoes a change of state 2-0 atm- 3-0 L- 95K - 4-0 atm- 5-0L- 245 K with a change in internal energy - U - 30-0L atm- The change in enthalpy of the process in L atm is -

Δ H = Δ U + (P_2V_2 P_1 V_1) = 30 + (4 × 5 2 × 3) = 44 L atm ...

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Standard XI

Chemistry

Enthalpy

One mole of a...

Question

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

40.0

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42.3

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44.0

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Not defined, because it is not constant,

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Solution

The correct option is C
44.0

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

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Q. One mole of a non-ideal gas undergoes change of state (2.0 atm, 3.0 L, 95 K)

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with a change in internal energy,

△ U = 30.0 L

atm. The change in enthalpy

(△ H)

of the process is:

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

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(2.0 a t m, 3.0 L, 95 K) \to (4.0 a t m, 5.0 L, 245 K)

with a change in internal energy,

△ U = 30.0 L

atm. The change in enthalpy

(△ H)

of the process is:

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

The correct option is C 44.0 ΔH=ΔU+(P2V2−P1V1)=30+(4×5−2×3)=−44L atm

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

The correct option is C
44.0

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