There are around muscles in the human body 640 250 100
1 answer:
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To solve the question we will assume that the gas behaves like an ideal gas, that is to say, that there is no interaction between the molecules. Assuming ideal gas we can apply the following equation:
Where,
P is the pressure of the gas
V is the volume of the gas
n is the number of moles
R is a constant
T is the temperature
Now, we have two states, an initial state, and a final state. The conditions for each state will be.
Initial state (1)
P1=975Torr=1.28atm
V1=3.8L
T1=-18°C=255.15K
Final state(2), STP conditions
P2=1atm
T2=273.15K
V2=?
We will assume that the number of moles remains constant, so the nR term of the first equation will be constant. For each state, we will have:
Since nR is the same for both states, we can equate the equations and solve for V2:
We replace the known values:
At STP conditions the gas would occupy 5.2L. First option
Explanation:
The given data is as follows.
Mass flow rate of mixture = 1368 kg/hr
in feed = 40 mole%
This means that in feed = (100 - 40)% = 60%
We assume that there are 100 total moles/hr of gas in feed stream.
Hence, calculate the total mass flow rate as follows.
40 moles/hr of N_{2}/hr (28 g/mol of ) + 60 moles/hr of
(2 g/mol of
)
= 1120 g/hr + 120 g/hr
= 1240 g/hr
= (as 1 kg = 1000 g)
= 1.240 kg/hr
Now, we will calculate mol/hr in the actual feed stream as follows.
= 110322.58 moles/hr
It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of into the reactor as follows.
= 44129.03 mol/hr
As 1 mole of nitrogen has 28 g/mol of mass or 0.028 kg.
Therefore, calculate the rate flow of into the reactor as follows.
= 1235.612 kg/hr
Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.