Question

An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 40.0 atm and releases 74.8 kJ of heat. Before the reaction, the volume of the system was 7.60 L . After the reaction, the volume of the system was 2.40 L . Calculate the total internal energy change, ΔE , in kilojoules

Answer #1

∆H = - 74.8 kJ

Enthalpy is defined in terms of internal energy U, pressure and
volume:

H = U + P∙V

<=>

U = H + P∙V

So the changes are related as:

∆U = ∆H - ∆(P∙V)

Since P is constant you can rewrite

∆U = ∆H - P∙∆V = ∆H - P∙(V_final _ V_initial)

To compute the P∙V in kilojoules you need to covert pressure to
kilopascals and volume to cubic meters.

Note that 1 kJ = 1 kPa∙m³

Hence,

**∆U** = - 74.8 kJ - 40*101.325 kPa * (2.4×10⁻³ m³ -
7.6×10⁻³ m³) = **-53.72 kJ**

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________________________________________________________________________
An ideal gas (which is is a hypothetical gas that conforms to
the laws...

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Before the reaction, the volume of the system was 9.00 L . After
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