can anyone solve these two sums step by step apply Boyle's law ( p1 v1 = p2 v2 ) no irrelevant answers
Answers
Answer:
1.
Given :
Initial volume (V1) = 500 dm³
Initial Pressure(P1) = 1 Bar
Final volume(V2) = 200 dm³
To 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 L
Initial Pressure(P1) = 760 mm Hg
Final volume(V2) = 4 dm³
(since, 1 dm³ = 1 L
4 dm³ = 4L)
=> V2 = 4L
To 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
Hope this answer is correct....
