Q44 / 180 · AIIMS 2007 · Chemistry

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

  • A
    \(-42 \mathrm{~kJ}\)
  • Correct answer
    B
    \(-88 \mathrm{~kJ}\)
  • C
    \(+88 \mathrm{~kJ}\)
  • D
    \(+42 \mathrm{~kJ}\).

Solution

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

Hence, the correct answer is option: 'B'.

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