`1 g` mole of oxygen at `27^@ C` and `1 ` atmosphere pressure is enclosed in a vessel.
(a) Assuming the molecules to be moving with `v_(r m s)`, find the number of collisions per second which the molecules make with one square metre area of the vessel wall.
(b) The vessel is next thermally insulated and moves with a constant speed `v_(0)`. It is then suddenly stoppes. The process results in a rise of temperature of the gas by `1^@ C`. Calculate the speed `v_0.[k = 1.38 xx 10^-23 J//K` and `N_(A) = 6.02 xx 10^23 //mol]`.

Answer

A, B, C

Explanation

(i) `F=PxxA=10^5xx1=10^5N`
`F=(DeltaP)/(Deltat) rArr Deltap=FxxDeltat=10^5xx1=10^5…..(i)`
Now, momentum change per second
`(Deltap)=nxx2mv…..(ii)`
Where n is the number of collisions per second per
From (i) and (ii)
`nxx2mv=10^5 :. n=10^5/(2mv)`
Root mean square velocity
`v=sqrt((3RT)/M)=sqrt((3xx8.314xx300)/(32//1000))=483.4 m//s`
According to mole concept `6.023xx10^23` molecules will
have mass 32g
`:. 1 molecule will have mass 32/(6.023xx10^23)g`
`:. n=(10^5xx6.023xx10^23)/(2xx32xx483.4)=1.97xx10^27`
(ii) The kinetic energy of motion of molecules will be
converted into heat energy.
`K.E. of 1gm mole of oxygen=1/2MV_0^2...(i)`
Where `v_0` is the velocity with which the vessel was
moving.
The heat gainded by 1 gm mole of molecules at constat
volume for `1^@C` rise in temperature
`=nC_vDeltaT=1xxC_vxx1=C_v.....(ii)`
From (i) and (ii)
`1/2MV_0^2=C_v But, C_v=R/(gamma-1)`
`1/2MV_0^2=R/(gamma-1)`
`:. v_0=sqrt((2R)/(M(gamma-1)))=sqrt((2xx8.314)/(32/100xx(1.41-1)))=35.6 ms^-1`
`[:' gamma=1.41 for O_2 (diatomic gas)]`