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AIIMS 2007
Exam
AIIMS
Year
2007
Subject
Chemistry
Chapter
Thermodynamics (C)

AIIMS 2007 · Chemistry · Thermodynamics (C)

Calculate change in internal energy if ( Δ H=-92.2 ~kJ, P=40 ~atm )...

Calculate change in internal energy if (Δ\DeltaH=-92.2  kJ,\mathrm{~kJ},P=40  atm\mathrm{~atm}) and (Δ\DeltaV=-1  L\mathrm{~L}).

Options

  1. A(-42 \mathrm{~kJ})
  2. B(-88 \mathrm{~kJ})
  3. C(+88 \mathrm{~kJ})
  4. D(+42 \mathrm{~kJ}).

Correct answer

B. (-88 \mathrm{~kJ})

Step-by-step solution

(\Delta H=\Delta E+P \Delta V) ΔE=ΔH−PΔV=−92.2−40×(−1)×101×10−3=−92.2+4.04=−88.16 kJ≈−88 kJ.\begin{aligned} \Delta E & =\Delta H-P \Delta V=-92.2-40 \times(-1) \times 101 \times 10^{-3} \\ & =-92.2+4.04=-88.16 \mathrm{~kJ} \approx-88 \mathrm{~kJ} . \end{aligned})$
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