One mole of a non-ideal gas undergoes a change state ( $2 a t m, 3 L, 95 K$ ) to ( $4 a t m, 5 L, 245 K$ ) with a change of internal energy, $Δ U = 30 L$ atm.The change in enthalpy ( $Δ H$ ) of the process in $L$ atm:

40.0

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42.3

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44.0

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not defined, because pressure is not constant

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Solution

The correct option is C $44.0$
As we know
$Δ H = Δ U + Δ (P V)$
$Δ H = 30 + (P_{2} V_{2} - P_{1} V_{1})$
$Δ H = 30 + (20 - 6)$
$Δ H = 30 + 14 = 44$

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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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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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△ U = 30.0 L

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One mole of a non-ideal gas undergoes a change state (2atm,3L,95K) to (4atm,5L,245K) with a change of internal energy, ΔU=30L atm.The change in enthalpy (ΔH) of the process in L atm:

The correct option is C 44.0 As we know ΔH=ΔU+Δ(PV)ΔH=30+(P2V2−P1V1)ΔH=30+(20−6)ΔH=30+14=44

One mole of a non-ideal gas undergoes a change state ( $2 a t m, 3 L, 95 K$ ) to ( $4 a t m, 5 L, 245 K$ ) with a change of internal energy, $Δ U = 30 L$ atm.The change in enthalpy ( $Δ H$ ) of the process in $L$ atm:

The correct option is C $44.0$
As we know
$Δ H = Δ U + Δ (P V)$
$Δ H = 30 + (P_{2} V_{2} - P_{1} V_{1})$
$Δ H = 30 + (20 - 6)$
$Δ H = 30 + 14 = 44$