(a) Calculate the pressure exerted by 10^(23) gas molecules, each of mass 10^(-25) kg in a container of volume 1xx10^(-3) m^(3) and having root mean square velocity of 10^(3) ms^(-1) (b) Also, calculate total kinetic energy and (c ) temperature of the gas (b) What would be the final pressure of O_(2) in following experiment ? A collapsed polethylene bag of 30 litre capacity is partially blown up by the addition of 10 litres of N_(2) at 0.965 atm at 298 K, Subsequently enough O_(2) is pumped into bag so that at 298 K and exetrnal pressure of 0.990 atm, the bag contains 30 litre N_(2).

(a) Calculate the pressure exerted by 10^(23) gas molecules, each of m

1.	(a) Calculate the pressure exerted by 10^(23) gas molecules, each of mass 10^(-25) kg in a container of volume 1xx10^(-3) m^(3) and having root mean square velocity of 10^(3) ms^(-1) (b) Also, calculate total kinetic energy and (c ) temperature of the gas (b) What would be the final pressure of O_(2) in following experiment ? A collapsed polethylene bag of 30 litre capacity is partially blown up by the addition of 10 litres of N_(2) at 0.965 atm at 298 K, Subsequently enough O_(2) is pumped into bag so that at 298 K and exetrnal pressure of 0.990 atm, the bag contains 30 litre N_(2).
Answer» Solution :(a) `P=(1 mNu^(2))/(3V) =(1xx10^(-25)xx10^(23) xx(10^(3))^(2))/(3xx10^(-3))` `=3.33xx10^(6) N m^(-2)` (b) Total `KE=((1)/(2) mu^(2))xx N` `=(1)/(2)xx10^(-25)xx(10^(3))^(2)xx10^(23)` `=(1)/(2)xx10^(4)` `=0.5xx10^(4)` Joule (c ) Also total kinetic energy 3/2 nRT `0.5xx10^(4)=(3)/(2)xx(10^(23))/(6.023xx10^(23))xx8.314xxT` `T=(0.5xx10^(4)xx2xx6.023)/(3xx8.314)` `=2415 K` (b) Given pressure of `N_(2)=0.965` atm Volume of `N_(2)=10` litre. Temperature of `N_(2)=298 K` `therefore` For `N_(2)` when bag is fully expanded. Volume of `N_(2)` (alone) =30 litre at 298 K `because P_(1) V_(1) =P_(2)V_(2)` `0.965xx10=P_(2)xx30 therefore P_(N_(2))` (alone) in 30 litre bag at 298 K =0.322 atm Now `P_(M)=P_(O_(2))+P_(N_(1))` `0.990 =P_(O_(z))+0.332` `therefore P_(O_(z))=0.668` atm