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Enthalpy

One mole of a...

Question

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, $△ U = 30.0 L$ atm. The change in enthalpy $(△ H)$ of the process is:

40.0 L a t m

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42.3 L a t m

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44.0 L a t m

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Not defined, because pressure is not constant

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Solution

The correct option is C $44.0 L a t m$
The relation between enthalpy $(△ H)$ and internal energy $(△ U)$ is given as:

$△ H = △ U + △ (p V) = △ U + (p_{2} V_{2} - p_{1} V_{1})$
putting values given in question we have
$= 30.0 L a t m + [(4.0 a t m) (5.0 L) - (2.0 a t m) (3.0 L)] = 30.0 L a t m + 14.0 L a t m = 44.0 L a t m$

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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 ?

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Enthalpy

Standard XII Chemistry

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

The correct option is C 44.0 L atmThe relation between enthalpy (△H) and internal energy (△U) is given as: △H=△U+△(pV)=△U+(p2V2−p1V1) putting values given in question we have =30.0L atm+[(4.0 atm)(5.0L)−(2.0 atm)(3.0L)]=30.0 L atm+14.0 L atm=44.0 L atm

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, $△ U = 30.0 L$ atm. The change in enthalpy $(△ H)$ of the process is: