two gases a and b having the mole ratio is 3:5 in a container exert a pressure of 8 atm .if a is removed ,what would be the pressure due to B only , at temperature constant
Answers
Answer:
diffuse in a container to fill up the space it is in and does not have any forces of attraction between the molecules. In other words, the different molecules in a mixture of gases are so far apart that they act independently; they do not react with each other. The pressure of an ideal gas is determined by its collisions with the container, not collisions with molecules of other substances since there are no other collisions. A gas will expand to fill the container it is in without affecting the pressure of another gas. So it can be concluded that the pressure of a certain gas is based on the number of moles of that gas and the volume and temperature of the system. Since the gases in a mixture of gases are in one container, the Volume (V) and Temperature (T) for the different gases are the same as well. Each gas exerts its own pressure on the system, which can be added up to find the total pressure of the mixture of gases in a container. This is shown by the equation
Ptotal=PA+PB+...(1)
(1)Ptotal=PA+PB+...
Derivation
We have, from the Ideal Gas Law
PV=nRT(2)
(2)PV=nRT
If we know the molar composition of the gas, we can write
ntotal=na+nb+...(3)
(3)ntotal=na+nb+...
Again, based on the kinetic theory of gases and the ideal gas law, Dalton’s law can also be applied to the number of moles so that the total number of moles equals the sum of the number of moles of the individual gases. Here, the pressure, temperature and volume are held constant in the system. The total volume of a gas can be found the same way, although this is not used as much. This yields the equation,
Answer:
Solution:
A:B=3:5
PT=8 atm
∴PA=3 atm
PB=5 atm
If ‘A’ is removed
PA+PB=PT
3+5=8
PB=5 atm