1

### JEE Main 2017 (Online) 9th April Morning Slot

At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen (N2 ) at 4 bar. The molar mass of gaseous molecule is :
A
28 g mol$-$1
B
56 g mol$-$1
C
112 g mol$-$1
D
224 g mol$-$1

## Explanation

Density = ${{Mass} \over {Volume}}$

PV = RT $\Rightarrow$ V = ${{RT} \over P}$

So, Density(d) = ${{MP} \over {RT}}$

Now, d1 = x, P1 = 4, M1 = 28, d2 = 2x, P2 = 2, M2 = ?

$\therefore$ ${{{d_1}} \over {{d_2}}} = {{{M_1}{P_1}} \over {R{T_1}}} \times {{R{T_2}} \over {{M_2}{P_2}}} = {{{M_1}{P_1}} \over {{M_2}{P_2}}}$ [As T1 = T2 ]

$\Rightarrow$ M2 = ${{{M_1}{P_1}{d_2}} \over {{d_1}{P_2}}}$

= ${{2x \times 28 \times 4} \over {2 \times x}}$ = 112 g mol-1
2

### JEE Main 2018 (Online) 16th April Morning Slot

Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is : (Atomic wt. of Cl = 35.5 u)
A
1.46
B
0.46
C
1.64
D
0.64

## Explanation

We know, PV = nRT

n = no. of moles = ${m \over M}$

So,   PV = ${m \over M}RT$

$\Rightarrow $$\,\,\,\, P = {m \over V} \times {{RT} \over M} \Rightarrow$$\,\,\,\,$ P = d $\times$ ${{RT} \over M}$ [ d = density = ${m \over V}$ ]

at constant temperature and pressure d $\propto$ M

Now let d1 and d2 are the density of ammonia and HCl.

$\therefore\,\,\,\,$ ${{{d_1}} \over {{d_2}}} = {{{M_{N{H_3}}}} \over {{M_{HCl}}}}$

$\Rightarrow$ $\,\,\,\,$ ${{{d_1}} \over {{d_2}}}$ = ${{17} \over {36.5}}$ = 0.46
3

### JEE Main 2019 (Online) 9th January Morning Slot

0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m3 at 1000 K. Given R is the gas constant in JK$-$1mol$-$1, x is :
A
${{2R} \over {4 + R}}$
B
${{2R} \over {4 - R}}$
C
${{4 + R} \over {2R}}$
D
${{4 - R} \over {2R}}$

## Explanation

We know,

PV = nRT

Given,

P = 200 Pa

V = 10 m3

T = 1000 K

n = 0.5 + x

$\therefore$   200 $\times$ 10 = (0.5 + x) R $\times$ 1000

$\Rightarrow$    0.5 + x = ${2 \over R}$

$\Rightarrow$    x = ${2 \over R} - {1 \over 2}$

$\Rightarrow$   x = ${{4 - R} \over {2R}}$
4

### JEE Main 2019 (Online) 12th January Morning Slot

The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice than that of gas B at same temperature. The pressures of the gases for equal number of moles are -
A
2PA = 3PB
B
3PA = 2PB
C
PA = 3PB
D
PA = 2PB

## Explanation

Z = PV/nRT

$\Rightarrow$ P = ${{ZnRT} \over V}$

at constant T and mol(n),

P $\propto$ ${Z \over V}$

$\Rightarrow$ ${{{P_A}} \over {{P_B}}} = {{{Z_A}} \over {{Z_A}}} \times {{{V_B}} \over {{V_A}}}$

= $\left( {{3 \over 1}} \right) \times \left( {{1 \over 2}} \right)$ = ${3 \over 2}$

$\therefore$ 2PA = 3PB