99 Percentile Qs Bank for JEE MainChemistryThermodynamics (C)
(6 ~g ) of graphite is burnt in a bomb calorimeter at (25^ C ) and (1 ~atm ) pressure. The temperature of water increased from (25^ C ) to (31^ C ). If ( H ) of this reaction is (-248 ~kJ ) mol, find out ( C _V ) (in ( kJ K ⁻¹ ) ) of bomb calorimeter.
Options
- A20.667
- B41.33
- C1488
- D0.145
Correct answer
A. 20.667
Step-by-step solution
Weight of graphite (=6 ~g ) In bomb calorimeter volume is constant Hence, (W=0 ), (work done) ( H= U=q (Heat) ) Given, ( H=-248 ~kJ / mol ) For, (12 ~g ) of graphite required energy is (248 ~kJ / mol ). For, (6 ~g ) of graphite, it will be (= 248 6 12 ) ( aligned aligned From, q & =C_V T 248 6 12 & =C_V (31-25) C_V & = 248 2 6 =20.6 ~kJ / K aligned aligned )