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AP EAMCET · Chemistry · Thermodynamics (C)

\(6 \mathrm{~g}\) of graphite is burnt in a bomb calorimeter at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. The temperature of water increased from \(25^{\circ} \mathrm{C}\) to \(31^{\circ} \mathrm{C}\). If \(\Delta H\) of this reaction is \(-248 \mathrm{~kJ}\) mol, find out \(\mathrm{C}_V\) (in \(\mathrm{kJ} \mathrm{K}^{-1}\) ) of bomb calorimeter.

  1. A 20.667
  2. B 41.33
  3. C 1488
  4. D 0.145
Verified Solution

Answer & Solution

Correct Answer

(A) 20.667

Step-by-step Solution

Detailed explanation

Weight of graphite \(=6 \mathrm{~g}\) In bomb calorimeter volume is constant Hence, \(W=0\), (work done) \(\Delta H=\Delta U=q \text { (Heat) }\) Given, \(\Delta H=-248 \mathrm{~kJ} / \mathrm{mol}\) For, \(12 \mathrm{~g}\) of graphite required energy is…
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