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can anyone solve these two sums step by step apply Boyle's law ( p1 v1 = p2 v2 ) no irrelevant answers |
| Answer» 1. Given :INITIAL volume (V1) = 500 dm³Initial Pressure(P1) = 1 Bar FINAL volume(V2) = 200 dm³To find :Minimum pressure for the given change of volume Solution :Let final Pressure be P2 .By Boyle's LAW -P1 x V1 = P2 x V2 . . . . (at const. temp.)=> 1 x 500 = P2 x 200=> P2 = 500/ 200=> P2 = 2.5 Hence, the minimum pressure required is 2.5 bar. (Ans 2. We know that, 1 L = 1 dm³ Given :Initial volume (V1) = 2 LInitial Pressure(P1) = 760 mm Hg Final volume(V2) = 4 dm³(since, 1 dm³ = 1 L 4 dm³ = 4L)=> V2 = 4LTo find :Pressure when volume changes Solution :Let final Pressure be P2 .By Boyle's Law -P1 x V1 = P2 x V2 . . . . (at const. temp.)=> 760 x 2 = P2 x 4=> P2 = (760 x 2)/ 4=> P2 = 760/ 2=> P2 = 380 Hence, the pressure when volume changes to 4 dm³ is 380 mm Hg. (Ans | |