Mass of gas

At constant volume and temperature, the number of moles of a gas is directly proportional to its partial pressure.

CS2(g) + 3O2(g) → CO2(g) + 2SO2(g)
Mole ratio CS2 : O2 : CO2 : SO2 = 1 : 3 : 1 : 3

Let y atm be the partial pressure of CS2 in the mixture.
Decrease in partial pressure of CS2 = y atm
Decrease in partial pressure of O2 = 3y atm
Increase in partial pressure of CO2 = y atm
Increase in partial pressure of SO2= 2y atm

Overall decrease in partial pressure:
(y + 3y) - (y + 2y) = 300 - 240
y = 60
Partial pressure of CS2 in the original mixture = 60 atm


Consider CS2 originally present in the mixture:

Partial pressure of CS2, P = 60 atm
Volume, V = 10 L
Temperature, T = 100 + 273 = 373 K
Molar mass, M = 12 + 32.06x2 = 76.12 g mol-1
Gas constant R = 0.08206 L atm mol-1 K-1

PV = nRT
PV = (m/M)RT

Mass of CS2 in original pressure, m
= MPV/RT
= (76.12)(60)(10)/(0.08206)(373)
= 1492 g
=

what is in the product gas is it pure O2 or mixed with CO2 and CS

2009-05-31 14:27:21 補充：
chemical eqn
CS2+3O2 ---->CO2+2SO2
initial
Pv=nRT
R=82.06cm^3atm/gmolK
T=373.15K
V=10L=10000cm3
P=3atm
3*10000/(82.06*373.15)=n=0.9798gmol of O2 and CS2 mix
final
Pv/RT=n
P=2.4atm R,T,V is unchange
2.4*10000/(82.06*373.15)= 0.7838gmol of SO2,CO2 and O2 mix

from chemical equation 4mole react gives 3moles of product
therefore the change in mole is one 4-3= 1mole also equal to no. of mole of CS2 initially (from equation). it also mean every 4mole react will have 1mole losses e.g. 2mole react 0.5 losses(also 0.5mole of CS2 react)
so therefore the change in mole0.9798-0.7838= 0.196mole is equal to the number of mole of CS2.
0.196*MW=XXXg sorry dont know the MW

2009-06-01 08:21:40 補充：
the man who did the question below method is right u can get the same ans but he uses the wrong partial pressure. CS2 partial pressure is 0.6atm instead of 60atm.

2009-06-01 08:24:10 補充：
but you are better learn of his method is more tranditional. easier for you to understead.

2009-06-01 08:26:36 補充：
0.196*76.12=14.92g

Take a time period of 1 s, the mass of gas emitted = 0.1 kg
Hence, the momentum of the emitted gas = 0.1 x 50 Ns = 5 Ns

Using impulse = change of momentum
F.t = 5
where F is the force given to the exhaust gas, which is equal to the reaction force acting on the rocket, and t is the time period (= 1 s)

thus F = 5 N

using force = mass x acceleration
5 = 2a
this gives a = 5/2 m/s2 = 2.5 m/s2
回應

2009-05-31 19:16:15 補充：
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