Practice Start Test

States of Matter: Gases and Liquids

NEET Chemistry · 61 questions · Page 1 of 7 · Click an option or "Show Solution" to reveal answer

Q1
The correct van der Waals equation for 1 mole of a real gas is :
A (p+aV2)(V−b)=RT\mathrm{\left(p+\dfrac{a}{V^2}\right)(V-b)=R T}
B (p+V2a)(V−b)=RT\mathrm{\left(p+\dfrac{V^2}{a}\right)(V-b)=R T}
C (p+a2V2)(V2−nb)=RT\mathrm{\left(p+\dfrac{a^2}{V^2}\right)\left(V^2-n b\right)=R T}
D (p+an2V)(V−nb)=nRT\mathrm{\left(p+\dfrac{a n^2}{V}\right)(V-n b)=n R T}
Correct Answer
Option A
Solution
(P+aV2)(V−b)=RT\mathrm{\left(P+\frac{a}{V^2}\right)(V-b)=R T}
Q2
The correct option in which the density of argon (Atomic mass = 40) is highest:
A STP
B 0∘^\circC, 2 atm
C 0∘^\circC, 4 atm
D 273∘^\circC, 4 atm
Correct Answer
Option C
Solution
ρ=PMRT\mathrm{\rho=\frac{PM}{RT}}

For maximum density, pressure should be maximum and temperature should be minimum.

Q3
Four gas cylinders containing He, N 2 , CO 2 and NH 3 gases separately are gradually cooled from a temperature of 500 K. Which gas will liquify first? (Given Tc in K −- He : 5.3, N 2 : 126, CO 2 : 304.1 and NH 3 : 405.5)
A NH 3
B He
C N 2
D CO 2
Correct Answer
Option A
Solution

As from the given data NH 3 has highest critical temperature which suggests maximum attractive forces hence NH 3 will get liquified first.

Q4
Which one is not correct mathematical equation for Dalton's Law of partial pressure? Here p = total pressure of gaseous mixture where p i = partial pressure of i th gas χi{\chi _i} = mole fraction of i th gas in gaseous mixture where χi{\chi _i} = mole fraction of i th gas in gaseous mixture piop_i^o = pressure of i th gas in pure state
A p=p1+p2+p3p = {p_1} + {p_2} + {p_3}
B p=n1RTV+n2RTV+n3RTVp = {n_1}{{RT} \over V} + {n_2}{{RT} \over V} + {n_3}{{RT} \over V}
C pi=χip{p_i} = {\chi _i}p
D pi=χipio{p_i} = {\chi _i}p_i^o
Correct Answer
Option D
Solution

∙\bullet Dalton's Law of partial pressure states that the total pressure by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

∙\bullet

pTotal=p1+p2+p3{p_{Total}} = {p_1} + {p_2} + {p_3}

∙\bullet Also,

pi=χip{p_i} = {\chi _i}p

; where p i and

χ\chi

i are the partial pressure and mole fraction of i th gas respectively and p is the total pressure. ∙\bullet

pTotal=p1+p2+p3{p_{Total}} = {p_1} + {p_2} + {p_3}
=n1RTV+n2RTV+n3RTV= {n_1}{{RT} \over V} + {n_2}{{RT} \over V} + {n_3}{{RT} \over V}
=(n1+n2+n3)RTV= ({n_1} + {n_2} + {n_3}){{RT} \over V}
Q5
A 10.0 L flask contains 64 g of oxygen at 27∘^\circC. (Assume O 2 gas is behaving ideally). The pressure inside the flask in bar is (Given R = 0.0831 L bar K −-1 mol −-1 )
A 2.5
B 498.6
C 49.8
D 4.9
Correct Answer
Option D
Solution

We know for ideal gas PV = nRT P = n

RTV{{RT} \over V}

P =

6432×0.0831×30010{{64} \over {32}} \times {{0.0831 \times 300} \over {10}}

P = 4.9 bar Pressure of O 2 gas inside the flask = 4.9 bar

Q6
Choose the correct option for the total pressure (in atm.) in a mixture of 4g O 2 and 2g H 2 confined in a total volume of one litre at 0∘^\circC is : [Given R = 0.082 L atm mol −-1 K −-1 , T = 273 K]
A 26.02
B 2.518
C 2.602
D 25.18
Correct Answer
Option D
Solution
nO2=432=18{n_{{O_2}}} = {4 \over {32}} = {1 \over 8}
nH2=22=1{n_{{H_2}}} = {2 \over 2} = 1
nt=18+1=98{n_t} = {1 \over 8} + 1 = {9 \over 8}
PtV=ntRT{P_t}V = {n_t}RT
Pt=98×0.082×2731=25.18{P_t} = {{{9 \over 8} \times 0.082 \times 273} \over 1} = 25.18

atm

Q7
A mixture of N 2 and Ar gases in a cylinder contains 7g of N 2 and 8g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N 2 is : [Use atomic masses (in g mol − - 1 ) : N = 14, Ar = 40]
A 12 bar
B 15 bar
C 18 bar
D 9 bar
Correct Answer
Option B
Solution
nN2=728=0.25^n{N_2} = {7 \over {28}} = 0.25
nAr=840=0.2^n{A_r} = {8 \over {40}} = 0.2

Now, applying Dalton's law of partial pressure,

PN2=nN2nN2+nAr×PTotal^P{N_2} = {{^n{N_2}} \over {^n{N_2}{ + ^n}{A_r}}} \times {P_{Total}}
=0.250.45×27= {{0.25} \over {0.45}} \times 27

bar = 15 bar

Q8
A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor (Z) is -
A Z < 1 and attractive forces are dominant
B Z < 1 and repulsive forces are dominant
C Z > 1 and attractive forces are dominant
D Z > 1 and repulsive forces are dominant
Correct Answer
Option A
Solution

Compressibility factor (Z ) =

VrealVideal{{{V_{real}}} \over {{V_{ideal}}}}

Here, V real < V ideal . Therefore, Z < 1. In gaseous molecules attractive forces are dominant.

Q9
The correction factor ‘a’ to the ideal gas equation corresponds to
A density of the gas molecules
B volume of the gas molecules
C electric field present between the gas molecules
D forces of attraction between the gas molecules.
Correct Answer
Option D
Solution

In Ideal gases ‘a’ account for the intermolecular attractive forces between gas molecules.

Q10
Given van der Waals’ constant for NH 3 , H 2 , O 2 and CO 2 are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?
A NH 3
B H 2
C O 2
D CO 2
Correct Answer
Option A
Solution

van der Waals’ constant ‘a’ signifies the intermolecular forces of attraction between the particle of gas.

So, higher the value of ‘a’, easier will be the liquefaction of gas.

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