Answer:
The volume of the gas is 11.2 L.
Explanation:
Initially, we have:
V₁ = 700.0 L
P₁ = 760.0 mmHg = 1 atm
T₁ = 100.0 °C
When the gas is in the thank we have:
V₂ =?
P₂ = 20.0 atm
T₂ = 32.0 °C
Now, we can find the volume of the gas in the thank by using the Ideal Gas Law:

(1)
Where R is the gas constant
With the initials conditions we can find the number of moles:
(2)
By entering equation (2) into (1) we have:

Therefore, When the gas is placed into a tank the volume of the gas is 11.2 L.
I hope it helps you!
Answer:
The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.
Solution
Given:
Length = 48 m
Width = 12 m
Depth = 3m
Flow rate = 4 m 3 /s
Water density = 10 3 kg/m 3
Dynamic viscosity = 1.30710 -3 N.sec/m
Now,
At the minimum particular diameter it is stated as follows:
The Reynolds number= 0.1
Thus,
0.1 =ρVTD/μ
VT = Dp² ( ρp- ρ) g/ 10μ²
Where
gn = The case/issue of sedimentation
VT = Terminal velocity
So,
0.1 = Dp³ ( ρp- ρ) g/ 10μ²
This becomes,
0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²
= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)
dp³=3.1343 * 10 ^⁻12
Dp minimum= 1.474 * 10 ^⁻4 meters.