Answer:
V₂=4.57 x 10³ L
Explanation:
Given that
V₁= 2.88 x 10³ L
P₁=722 mm Hg
T₁ = 19°C
T₁ =292 K
P₂=339 mm Hg
T₂= - 55°C
T₂=218 K
Lets take final volume = V₂
We know that ideal gas equation
PV = m R T
By applying mass conservation
V₂=4.57 x 10³ L
Therefore volume will be 4.57 x 10³ L
Answer:
the work done to lift the bucket = 3491 Joules
Explanation:
Given:
Mass of bucket = 10kg
distance the bucket is lifted = height = 11m
Weight of rope= 0.9kg/m
g= 9.8m/s²
initial mass of water = 33kg
x = height in meters above the ground
Let W = work
Using riemann sum:
the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)
=
Work done in lifting the bucket (W) = force × distance
Force (F) = mass × acceleration due to gravity
Force = 9.8 * 10 = 98N
W done by bucket = 98×11 = 1078 Joules
Work done to lift the rope:
At Height of x meters (0≤x≤11)
Mass of rope = weight of rope × change in distance
= 0.8kg/m × (11-x)m
W done = integral of (mass×g ×distance) with upper 11 and lower limit 0
W done =
Note : upper limit is 11 not 1, problem with math editor
W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0
W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]
=7.84(60.5 -0) = 474.32 Joules
Work done in lifting the water
At Height of x meters (0≤x≤11)
Rate of water leakage = 36kg ÷ 11m = kg/m
Mass of rope = weight of rope × change in distance
= kg/m × (11-x)m = 3.27kg/m × (11-x)m
W done = integral of (mass×g ×distance) with upper 11 and lower limit 0
W done =
Note : upper limit is 11 not 1, problem with math editor
W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0
W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]
= 32.046(60.5 -0) = 1938.783 Joules
the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)
the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103
the work done to lift the bucket = 3491 Joules