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1

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
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
2

JEE Main 2019 (Online) 9th January Morning Slot

MCQ (Single Correct Answer)
English
Hindi
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 = $${{4 - R} \over {2R}}$$

$$0.5$$ mole गैस $$\mathrm{A}$$ तथा $$\mathrm{x}$$ moles गैस $$\mathrm{B}$$ का मिश्रण दिया गया है। $$10 \mathrm{~m}^{3}$$ आयतन के पात्र में $$1000 \mathrm{~K}$$ ताप पर कुल दाब $$200 \mathrm{~Pa}$$ है। तब $$\mathrm{x}$$ ज्ञात कीजिए। $$\mathrm{R},(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$$) सार्वत्रिक गैस नियतांक है।

A
$$\frac{2 R}{4+R}$$
B
$$\frac{2 R}{4-R}$$
C
$$\frac{4+R}{2 R}$$
D
$$\frac{4-R}{2 R}$$
3

JEE Main 2018 (Online) 16th April Morning Slot

MCQ (Single Correct Answer)
English
Hindi
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

आदर्श गैस व्यवहार मानते हुये, एक ही ताप तथा दाब पर अमोनिया तथा हाइड्रोजन क्लोराइड के घनत्वों का अनुपात है : ( $$\mathrm{Cl}$$ का परमाणु भार $$=35.5 \mathrm{~u}$$ )

A
$$1.46$$
B
$$0.46$$
C
$$1.64$$
D
$$0.64$$
4

JEE Main 2017 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)
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

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