1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

Questions Asked from Gaseous State

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