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QUESTION: 1

**Direction (Q. Nos. 1-12) This section contains 12 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.**

**Q.
The vapour pressure of liquid water is 23.8 torr at 25°C. By what factor does the molar volume of water increases as it vaporises to form an ideal gas under these conditions? The density of liquid water is 0.997 g cm ^{-3} at 298 K ?**

Solution:

QUESTION: 2

A gas at a pressure of 5.0 bar is heated from 0°C to 546°C and is simultaneously compressed to one-third of its original volume. Hence, final pressure is

Solution:

P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2}

Let the original volume be 3 volume units . The final volume will be 1/3 of 3 which is 1 volume unit

Temperature must be in K - 0 DEG C = 273K and 546 DEG C = 819 K

We want to solve for P1

P_{1}*1/819 = 5*3/273

P_{1} = 819*5*3/273

P_{1} = 45

QUESTION: 3

A 8.3 L cylinder contains 128 . 0 g O_{2} gas at 27°C. What mass of O_{2} gas must be released to reduce the pressure to 6.0 bar ?

Solution:

By applying ideal gas eqn. PV = nRT

Further solving it we get an eq.

w = PVM/RT

Putting values, we get 64g.

Hence, the correct answer is Option A.

QUESTION: 4

N_{2} gas under the given state I is to be changed to state II. Thus, N_{2} escaped from I is

Solution:

Applying pV = nRT in state 1, we have n_{1} = 0.110

Applying pV = nRT in state 1, we have n_{2} = 0.0184

Mole fraction(this fraction is present) = 0.0184/0.110 = 0.16727

So, fraction escaped = (1-0.16727)*100 = 83.33%

QUESTION: 5

1 g H_{2} gas and x g O_{2} gas exert a total pressure of 5 atm. At a given temperature partial pressure of O_{2} gas is 4 atm. This O_{2} in the mixture is

Solution:

QUESTION: 6

The average molar mass of the vapour above solid NH_{4}CI is nearly 26.75 g mol^{-1}.

Thus, vapour consists of (mole fraction)

Solution:

The correct answer is Option C.

The average molar mass of the vapour above solid NH_{4}Cl = 26.5 mol^{-1}

Evaporation product of NH4Cl --->

NH_{4}Cl (s) ----> NH_{3}(g) + HCl(g)

Let Mole fraction of NH_{3} = X_{L}

Mole fraction of HCl = X_{2}

∴ Average molar mass of NH_{3} = 14 + 31 = 17

=> 26.5 = m_{1}x_{1} + m2x2

=> 26.5 = 17x1 + 36.5x_{2} ---(i)

And we know;

x_{1} + x_{2} = 1

From (i) and (ii)

x1 = 0.5 and x_{2} = 0.5

QUESTION: 7

Pressure of 1 g of an ideal gas A at 300 K is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature, pressure becomes 3 bar. Thus,

Solution:

Given pressure(P) of gas A= 2bar Total pressure = 3bar then pressure(P) of gas B = 3-2 =1bar

no. of mole (n) = given mass(m) / molar mass(M)

Gas A

PV = nRT

2×V =( mₐ/Mₐ) RT ------------- (1)

Gas B

PV = nRT

1V = (mb/Mb)RT ---------------- (2)

On dividing eqn 1 by eqn 2

2V/1V = (1/Mₐ) ÷(2/Mb)

2/1 = (Mb/Mₐ) × 1/2

(M_{B}/Mₐ) = (2/1)×(2/1)

M_{B}/Mₐ = 4/1

M_{B} = 4Mₐ

QUESTION: 8

Total pressure of a mixture of H_{2} and O_{2} is 1.00 bar. The mixture is allowed to react to form water, which is completely removed to leave only pure l-l_{2} at a pressure of 0.35 bar. Assuming ideal gas behaviour and that all pressure measurements were made under the same temperature and volume conditions, the composition of the original mixture is

Solution:

Correct answer is option B

Avagadro′s Law says:

Equal volume of two gases at same pressure and temperature conditions have equal number of molecules.

∴1 O_{2} molecule consumes 2 H_{2} molecules.

∴Since after the reaction only H_{2} is left, we know all of O_{2} is consumed.

∴This implies what all consumed 1/3 of it was O_{2}.

Hence, O_{2} in the mix was at a partial pressure of (1 - 0.35)* 1/3 = 0.22

And rest was H_{2} = (1 - 0.22) = 0.78

Hence oxygen- hydrogen composition in the original mix was:0.22 bar O2 and 0.78 bar H_{2}.

QUESTION: 9

The composition of air inhaled by a human is 21 % by volume O_{2} and 0.03% CO_{2} and that of exhaled air is 16% O_{2} and 4.03% CO_{2}. Assuming a typical volume of 6200 L day ^{-1}, moles of O_{2} used and CO_{2} generated by the body at 37°C and 1.00 bar are (assume ideal behaviour)

Solution:

QUESTION: 10

The following arrangement of flasks is set up. Assuming no temperature change, determine the final pressure inside the system after all stopcocks are opened. The connecting tube has zero volume.

Solution:

Applying p1V1 + p2V2 + p3V3 = pnet × Vnet

(0.792)×4 + (1.23)×3 + (2.51)×5 = pnet × 12

On solving, we get p = 1.617 = 1.62 atm

QUESTION: 11

1.00 mole sample of O_{2} and 3.00 moles sample of H_{2} are mixed isothermally in a 125.3 L container at 125°C. Gaseous mixture undergoes reaction to form water vapours. Assume ideal gas behaviour total pressure before and after H_{2}O (g) is formed is are respectivel

Solution:

Correct option is not provided.

2H2 + O2 ------> 2H2O

3 1 0

(-) 2 1 2 mole H_{2}O formed

___________________________

1 0 Total mole = 1+2 = 3

PV = nRT / 100

n = PV / RT

P = nRT/ V

= [4 * 0.083 * (125 + 273)] / 125.3

Pressure = 1.0545 bar after water formed is 2 mole.

P = [3 * 0.083 * 398] /125.3

= 0.7909 bar

= 0.791 bar (appx)

Hence, the answer is 1.0545 bar, 0.7909 bar.

QUESTION: 12

When NO_{2} is cooled to room temperature, some of it reacts to form a dimer N_{2}O_{4} through the reaction,

In a reaction, 15.2 g of NO_{2} is placed in a 10.0 L flask at high temperature and the flask is cooled to 25°C. The total pressure is measured to be 0.500 atm. Thus, partial pressure of NO_{2} is found to be

Solution:

n_{NO2} + 2n_{N2O4}= 0.330 mol

Subtracting n_{NO2}+ n_{N2O4} = 0.204 mol

From n_{NO2}+ 2n_{N2O4} = 0.330 mol

n_{NO2} = 0.126 mol

n_{N2O4} = 0.078 mol

From these results, NO_{2} has a mole fraction of 0.62and a partial pressure of 0.62(0.50 atm) = 0.31 atm;N2O4has a mole fraction of 0.38 and a partial pressure of 0.38(0.50 atm) = 0.19 atm

*Answer can only contain numeric values

QUESTION: 13

**Direction (Q. Nos. 13 and 16) This section contains 4 questions. when worked out will result in an integer from 0 to 9 (both inclusive)**

**Q.
A toy balloon originally held 1.00 g helium gas and had a radius of 10.0 cm.
During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)**

Solution:

We know that

pV = nRT

A/Q , p and T are constant;

So n_{1}/n_{2} = V_{1}/V_{2}

(¼)/(0.25/4) = (10)3/r3

r3 = 750

r ∼ 9

*Answer can only contain numeric values

QUESTION: 14

What will be the temperature difference needed in a hot air balloon to lift 1.00 kg of mass ? Assume that the volume of the balloon is 100 m^{3}, the temperature of the ambient air is 298 K, the pressure is 1.00 bar, and the air is an ideal gas with average molar mass of 29 g mol^{-1}.

Solution:

The correct answer is 2

The solution is as follows:

Δm = m_{cold} − m_{hot}

=[n_{cold}−n_{hot}]M

= PVMR[1T1− 1T2]

Solving we get:

T2−T1 = ΔT = 2

*Answer can only contain numeric values

QUESTION: 15

Iron forms a series of compounds of the type Fe_{x} (CO)_{y}. In air, they are oxidised to Fe_{2}O_{3} and CO_{2} gas. After heating a 0.142 g sample in air, you isolate the CO_{2} in a 1.50 L flask at 25°C. The pressure of the gas is 44.9 mm Fig. What will be the value of y/x in the simplest empirical formula of Fe_{x}(CO)_{y}? [Atomic mass of Fe = 56]

Solution:

The empirical formula is = Fe(CO)_{2}

Carbon dioxide obtained:

Pressure = 44.9 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 44.9 / 760 = 0.0591 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (25 + 273.15) K = 298.15 K

V = 1.50 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value

= 0.0821 L.atm/K.mol

Applying the equation as:

0.0591 atm × 1.50 L

= n × 0.0821 L.atm/K.mol × 298.15 K

⇒n = 0.0036 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.0036 moles

Molar mass of C atom

= 12.0107 g/mol

Mass of C in molecule

= 0.0036 x 12.0107 = 0.0432 g

Given that the compound only contains iron and carbon. So,

Mass of Fe in the sample

= Total mass - Mass of C

Mass of the sample = 0.142 g

Mass of Fe in sample

= 0.142 g - 0.0432 g = 0.0988 g

Molar mass of Fe = 55.845 g/mol

Moles of Fe

= 0.0988 / 55.845 = 0.0018 moles

Taking the simplest ratio for Fe and C as:

0.0018 : 0.0036

= 1 : 2

*Answer can only contain numeric values

QUESTION: 16

To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mole of Fie and 0.1 mole of an unknown compound (vapour pressure 0.68 atm at 0°C) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases 0°C is close to...

**[IITJEE 2011]**

Solution:

Let nx mole of unknown compound be present in vapour form so its partial pressure is

0.68 = (nx/nx+0.1)×1 atm

nx+0.1/nx = 1/0.68 = 1.47

0.1/nx = 0.47

nx = 0.212

Applying ideal gas equation,

1×V = (0.212/0.1)×0.0821×273

V = 7

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