 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2012

The compressibility factor for a real gas at high pressure is :
A
1 + RT/pb
B
1
C
1 + pb/RT
D
1–pb/RT

Explanation

$\left( {P + {a \over {{V^2}}}} \right)\left( {V - b} \right) = RT\,\,$

at high pressure ${a \over {{V^2}}}$ can be neglected

$PV - Pb = RT\,\,\,$

and $\,\,\,PV = RT + Pb$

${{PV} \over {RT}} = 1 + {{Pb} \over {RT}}$

$z = 1 + {{Pb} \over {RT}};Z > 1\,\,\,$ at high pressure
2

AIEEE 2011

'a’ and `b’ are van der Waals’ constants for gases. Chlorine is more easily liquefied than ethane because
A
a and b for Cl2 < a and b for C2H6
B
a and b for Cl2 > a and b for C2H6
C
a for Cl2 > a for C2H6 and b Cl2 < b for C2H6
D
a for Cl2 < a for C2H6 and b Cl2 > b for C2H6

Explanation

The value of $a$ is a measure of the magnitude of the attractive forces between the molecules of the gas. Greater the value of $'a',$ larger is the attractive inter-molecular force between the gas molecules.

The value of $b$ related to the effective size of the gas molecules. It is also termed as excluded volume.

The gases with higher value of $a$ and lower value of $b$ are more liquefiable, hence for $C{{\rm l}_2}$ $''a'''$ should be greater than for ${C_2}{H_6}$ but for it $b$ should be less than for ${C_2}{H_6}.$
3

AIEEE 2010

If 10–4 dm3 of water is introduced into a 1.0 dm3 flask to 300 K, how many moles of water are in the vapour phase when equilibrium is established ?
(Given : Vapour pressure of H2O at 300 K is 3170 Pa ; R = 8.314 J K–1 mol–1)
A
5.56 x 10–3 mol
B
1.53 x 10–2 mol
C
4.46 x 10–2 mol
D
1.27 x 10–3 mol

Explanation

From the ideal gas equation :

$PV = nRT\,\,$

or $\,\,\,n = {{PV} \over {RT}} = {{3170 \times {{10}^{ - 3}}} \over {8.314 \times 300}}$

$= 1.27 \times {10^{ - 3}}$
4

AIEEE 2007

Equal masses of methane and oxygen are mixed in an empty container at 25oC. The fraction of the total pressure exerted by oxygen is
A
2/3
B
1/2
C
1/3 $\times$ 273/298
D
1/3

Explanation

Let the mass of methane and oxygen $=m$ $gm.$

Mole fraction of ${O_2}$

$= {{Moles\,\,of\,\,{O_2}} \over {Moles\,\,of\,\,{O_2}\, + \,Moles\,\,of\,\,C{H_4}}}$

$= {{m/32} \over {m/32 + m/16}}$

$= {{m/32} \over {3m/32}} = {1 \over 3}$

Partial pressure of ${O_2}$

$=$ Total pressure $\times$ mole fraction of ${O_2},$

${P_{{O_2}}} = P \times {1 \over 3} = {1 \over 3}P$