Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. Molecular mass of B will be
A
50.00 u
B
12.25 u
C
6.50 u
D
25.00 u
Explanation
We know that $${{{r_A}} \over {{r_B}}} = {{v/{t_A}} \over {v/{t_B}}} = \sqrt {{{{M_B}} \over {{M_A}}}} $$
A bubble of air is underwater at temperature 15oC and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25oC and the pressure is 1.0 bar, what will happen to the volume of the bubble?
$$ \therefore $$ Volume of bubble will be almost
1.6 time to initial volume of bubble.
3
AIPMT 2011 Prelims
MCQ (Single Correct Answer)
A gaseous mixture was prepared by taking equal mole of CO and N2. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N2) in the mixture is
A
0.5 atm
B
0.8 atm
C
0.9 atm
D
1 atm
Explanation
Number of moles $$ \Rightarrow $$ nCO = nN2
Volume of container is same. So, VCO = VN2
.
Also,
temperature is same for both the gases thus,
TCO = TN2
According to ideal gas equation,
PV = nRT
Now, V, n, R and T for both gases are same. So,
PCO = PN2
Now, total pressure is 1 atm and according to
Dalton’s sum of partial pressure,