50 cc of oxygen is collected in an inverted gas jar over water. The atmospheric pressure is 99.4 kPa and the room temperature is 27°C. The water level in the jar is same as the level outside. The saturation vapour pressure at 27°C is 3.4 kPa. Calculate the number of moles of oxygen collected in the jar.
Answers
Answer:
Given
Volume of oxygen = 50cc=50cm3=50
Atmospheric pressure Po=99.4kPa=99.4Pa
Temperature =27=300.15K
Saturation vapor pressure Ps=3.4kPa=3.4Pa
According to question water level in the jar is same as level outside. So
Pressure inside the jar = pressure outside the jar
Pressure outside the jar =atmospheric pressure Po
Pressure inside the jar=Po …. (1)
But
Pressure inside the jar is also = vapor pressure of oxygen+ Saturation vapor pressure
Pressure inside the jar=P+Ps …. (2)
From equation (1) and (2), we can write
Po=P+Ps
P=Po-Ps=99.4-3.4=96Pa
Applying ideal gas equation
PV=nRT
Where V= volume of gas
R=gas constant =8.3JK-1mol-1
T=temperature
n=number of moles of gas
P=pressure of gas.
The number of moles of oxygen collected in the jar is 1.9310-3.
Answer:
Pressure inside the tube
= Atmospheric pressure
= 99.4 K Pa
pressure exerted by O2 vapour
= Atmospheric pressure V.P.
= 99.4 K Pa - 3.4 K Pa
=96 K Pa
Ifνmberofmo≤of O_2=n
Then using gas equation
96×(103)×50×(10−6)
=n×8.3×300
rArr n= ( 96 xx 50 xx (10^3)/8.3 xx 300) =1.9277 xx(10^-3)