One mole of an ideal gas undergoes a change of state 2-0atm- 30L- 95K - 4-0atm- 5-0L- 245K with change in internal energy - E-30-0Latm-The change in enthalpy of the process in L atm is

The correct option is B 10.0 ΔH = ΔE + ΔPV ΔE = 30.0 L.atm ΔPV = (5 × 4 2 × 30) = 40.0 L.atm ΔH = 10.0 L.atm ...

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Standard XII

Chemistry

Heat of Reaction

One mole of a...

Question

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

40.0

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- 10.0

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- 30.3

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Not defined, because it is not constant.

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Solution

The correct option is B $- 10.0$
$Δ H = Δ E + Δ P V$
$Δ E = 30.0 L . a t m$
$Δ P V = (5 \times 4 - 2 \times 30) = - 40.0 L . a t m$
$Δ H = - 10.0 L . a t m$

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One mole of an ideal gas undergoes a change of state (2.0atm, 30L, 95K)→ (4.0atm, 5.0L, 245K) with change in internal energy ΔE=30.0Latm.The change in enthalpy of the process in L atm is

The correct option is B −10.0ΔH=ΔE+ΔPVΔE=30.0 L.atmΔPV=(5×4−2×30)=−40.0 L.atmΔH=−10.0 L.atm

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

The correct option is B $- 10.0$
$Δ H = Δ E + Δ P V$
$Δ E = 30.0 L . a t m$
$Δ P V = (5 \times 4 - 2 \times 30) = - 40.0 L . a t m$
$Δ H = - 10.0 L . a t m$