1
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

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,

PCO + PN2 = 1 atm

$$ \Rightarrow $$ 2PN2 = 1 atm {$$ \because $$ PCO = PN2 }

$$ \Rightarrow $$ PN2 = 0.5 atm
2
MCQ (Single Correct Answer)

AIPMT 2011 Mains

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?
A
Volume will become greater by a factor of 1.6.
B
Volume will become greater by a factor of 1.1
C
Volume will become smaller by a factor of 0.70
D
Volume will become greater by a factor of 2.5

Explanation

Given

P1 = 1.5 bar, T1 = 273 + 15 = 288 K, V1 = V

P2 = 1.0 bar, T1 = 273 + 25 = 298K, V2 = ?

$${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$$

$$ \Rightarrow $$ $${{1.5 \times V} \over {288}} = {{1 \times {V_2}} \over {298}}$$

$$ \Rightarrow $$ V2 = 1.55V

$$ \therefore $$ Volume of bubble will be almost 1.6 time to initial volume of bubble.
3
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled ?
A
2.0
B
2.8
C
4.0
D
1.4

Explanation

Average velocity (vAV) = $$\sqrt {{{8RT} \over {\pi M}}} $$

$$ \Rightarrow $$ vAV $$ \propto $$ $$\sqrt T $$

$$ \therefore $$ $${{{{\left( {{v_{AV}}} \right)}_2}} \over {{{\left( {{v_{AV}}} \right)}_1}}} = \sqrt {{{2T} \over T}} $$ = 1.4
4
MCQ (Single Correct Answer)

AIPMT 2010 Mains

The pressure exerted by 6.0 g of methane gas in a 0.03 m3 vessel at 129oC is (Atomic masses : C = 12.01, H = 1.01 and R = 8.314 J K$$-$$1 mol$$-$$1 )
A
215216 Pa
B
13409 Pa
C
41648 Pa
D
31684 Pa

Explanation

PV = nRT

$$ \Rightarrow $$ $$P = {{nRT} \over V}$$ = $$ {w \over m}{{RT} \over V}$$

= $${6 \over {16.05}} \times {{8.314 \times 402} \over {0.03}}$$

= 41648 Pa

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