One mole of a non- ideal gas undergoes a change of state 2-0 atm- 3-0 L- 95 K - 4-0 atm- 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-

The relation between enthalpy ( H) and internal energy nbsp; ( U) is given as: nbsp; H = U + (pV ) = U + (p_2V_2 p_1V_1) putting values given in ques ...

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Standard XII

Chemistry

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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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

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 ?

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