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.
- A 20.667
- B 41.33
- C 1488
- D 0.145
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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