One mole of a non ideal gas undergoes a change of state $(2.0 atm, 3.0 L, 98 K) \to (4.0 atm, 5.0 L, 245 K)$ with a change in internal energy, $Δ U = 30.0 Latm$ . The change in enthalpy $Δ H$ of the process in $L atm$ is:

40.0 L atm

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42.3 L atm

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44.0 L atm

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Not defined because pressure is not constant

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Solution

The correct option is C $44.0 L atm$
$H = U + P V$
$∴ Δ H = (U_{2} + P_{2} V_{2}) - (U_{1} + P_{1} V_{1})$
$= (U_{2} - U_{1}) + P_{2} V_{2} - P_{1} V_{1}$
$= 30 + (4 \times 5 - 2 \times 3) = 44 L atm$

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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 L,98 K)→(4.0 atm,5.0 L,245 K) with a change in internal energy, ΔU=30.0 Latm. The change in enthalpy ΔH of the process in L atm is:

The correct option is C 44.0 L atmH=U+PV ∴ΔH=(U2+P2V2)−(U1+P1V1) =(U2−U1)+P2V2−P1V1 =30+(4×5−2×3)=44 L atm

One mole of a non ideal gas undergoes a change of state $(2.0 atm, 3.0 L, 98 K) \to (4.0 atm, 5.0 L, 245 K)$ with a change in internal energy, $Δ U = 30.0 Latm$ . The change in enthalpy $Δ H$ of the process in $L atm$ is:

The correct option is C $44.0 L atm$
$H = U + P V$
$∴ Δ H = (U_{2} + P_{2} V_{2}) - (U_{1} + P_{1} V_{1})$
$= (U_{2} - U_{1}) + P_{2} V_{2} - P_{1} V_{1}$
$= 30 + (4 \times 5 - 2 \times 3) = 44 L atm$