One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 litre, 95 k) $\to$ (4.0 atm, 5.0 litre, 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

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42.3

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44.0

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Not defined because the pressure is not constant

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Solution

The correct option is C
44.0

we clearly know that the enthalpy can be represented as:

$H = U + P V$

$H_{1} = U_{1} + (2 \times 3) = U_{1} + 6$

$H_{2} = U_{2} + (4 \times 5) = U_{2} + 20$

$H_{2} - H_{1} = U_{2} - U_{1} + (20 - 6)$

$Δ H = Δ U + 14 = 30 + 14 = 44 L a t m$

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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 ?

Q. One mole of a non- ideal gas undergoes a change of state

(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

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with a change in internal energy,

Δ U = 30.0 Latm

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Q. One mole of a non- ideal gas undergoes a change of state

(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

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One mole of an ideal gas undergoes a change of state (2.0atm, 30L, 95K) $\to$ (4.0atm, 5.0L, 245K) with change in internal energy $Δ E = 30.0 L a t m$ .The change in enthalpy of the process in L atm is

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One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 litre, 95 k) → (4.0 atm, 5.0 litre, 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

The correct option is C 44.0 we clearly know that the enthalpy can be represented as: H = U + PV H1 = U1 + (2 × 3) = U1 + 6 H2 = U2 + (4 × 5) = U2 + 20 H2 − H1 = U2 − U1 + (20 − 6) Δ H = Δ U + 14 = 30 + 14 = 44 L atm

One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 litre, 95 k) $\to$ (4.0 atm, 5.0 litre, 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

The correct option is C
44.0

we clearly know that the enthalpy can be represented as:

$H = U + P V$

$H_{1} = U_{1} + (2 \times 3) = U_{1} + 6$

$H_{2} = U_{2} + (4 \times 5) = U_{2} + 20$

$H_{2} - H_{1} = U_{2} - U_{1} + (20 - 6)$

$Δ H = Δ U + 14 = 30 + 14 = 44 L a t m$