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**Ex.1 When 2 gm of a gaseous substance A is introduced into an initially evacuated flask at 25^{o}C, the pressure is found to be 1 atm. 3 gm another gaseous substance B is then added to it at the same temperature and pressure. The final pressure is found to be 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of the molecular weights of A and B. **

**Sol. **Let M_{A} and M_{B} be the molecular weights of A and B.

Using PV = nRT for A, we get :

...(i)

and using Dalton's Law : P_{Total} = ⇒ 1.5 = ...(ii)

Solving (i) and (ii), we get

**Ex.2 Which of the two gases, ammonia and hydrogen chloride, will diffuse faster and by what factor ? **

**Sol.** By Graham's Law :

⇒

Thus, ammonia will diffuse 1.46 times faster than hydrogen chloride gas.

**Ex.3 The ratio of rate of diffusion of gases A and B is 1 : 4 and their molar mass ratio is 2 : 3. Calculate the composition of the gas mixture initially effusing out. **

**Sol. **By Graham's Law :

⇒ ⇒

⇒ Mole ratio of gas A and B effusing out = [moles ∝ pressure]

**Ex.4 At 30^{o}C and 720 mm of Hg, the density of a gas is 1.5 g/lt. Calculate molecular mass of the gas. Also find the number of molecules in 1 cc of the gas at the same temperature. **

**Sol. **Assuming ideal behaviour and applying ideal gas equation :

PV = nRT

Another form of gas equation is PM_{0} = dRT

⇒ M_{0} = = (T = 30 273 K)

⇒ M_{0} = 39.38

Now number of molecules = n x N_{0}

= 2.29 x 10^{19}

*Ex.5 The pressure exerted by 12 gm of an ideal gas at temperature t ^{o}C in a vessel of volume V litre is one atm. When the temperature is increased by 10^{o} at the same volume, the pressure rises by 10% calculate the temperature t and volume V. (molecular mass of the gas = 120 gm/mole) *

** Sol. **Using Gas equation : PV = nRT

We have, P x V = 0.1 x R x t ...(1)

and 1.1 P x V = 0.1 x R x (t + 10) ...(2)

Using (i) and (ii), we have :

⇒ t = 100 k or t = - 173^{o}C

Putting the value of t in (i), we get :

⇒ 1 x V = 0.1 x 0.0821 x 100 ⇒ V = 0.821 L

**Ex.6 Assuming that the air is essentially a mixture of nitrogen and oxygen in mole ratio of 4 : 1 by volume. Calculate the partial pressures of N_{2} and O_{2} on a day when the atmospheric pressure is 750 mm of Hg. Neglect the pressure of other gases. **

** Sol. **From Dalton's Law of partial pressure, we have

Partial pressure of nitrogen = x P and Partial pressure of oxygen =

Now, , and

⇒ x 750 = 600 mm of Hg and = 150 mm of Hg

**Ex.7 An open vessel at 27^{o}C is heated until three fifth of the air has been expelled, Assuming that the volume of the vessel remains constant, find the temperature to which the vessel has been heated. **

** Sol. **In the given questions, volume is constant. Also, as the vessel is open to atmosphere, the pressure is constant. This means that the gas equation is simply reduced to the following form :

nT = constant (Use PV = nRT)

or n_{1} T_{1} = n_{2} T_{2}

Now let n_{1} = initial moles and n_{2} = final moles

⇒ n_{2} = 2/5 x n_{1} (as 3/5th of the air has been expelled)

⇒ T_{2} = = =

⇒ T_{2} = = 750 K = 477^{o}C** **

**Ex.8 When 3.2 gm of sulphur is vaporized at 450^{o}C and 723 mm pressure, the vapour occupies a volume of 780 m, what is the formula for the sulphur under these conditions ? **

** Sol. **The molecular weight = no. of atoms x atomic mass

So let us find the molecular weight of S from the data given.

⇒ Number of atoms =

Hence, molecular formula of sulphur = S_{8} * *

**Ex.9 A spherical balloon of 21 cm diameter is to be filled with H_{2} at NTP from a cylinder containing the gas at 20 atm at 27^{o}C. If the cylinder can hold 2.80L of water, calculate the number of balloons that can be filled up. **

** Sol. **The capacity of cylinder = 2.80 L

Let n = moles of hydrogen contained in cylinder and n_{0} = moles of hydrogen required to fill one balloon.

n =

(Note : the balloons are being filled at S.T.P.)

= =

⇒ Number of balloons that can be filled = n/n_{0} = 10.50 ≈ 10

**Ex.10 A mixture containing 1.12L of H_{2} and 1.12L of D_{2} (deuterium) at S.T.P. is taken inside a bulb connected to another bulb by a stop-cock with a small opening. The second bulb is fully evacuated; the stop-cock is opened for a certain time and then closed. The first bulb is found to contain 0.05 gm of H_{2}. Determine the % age composition by weight of the gases in the second bulb. **

**Sol. **In the first bulb :

Initial moles of H_{2} = 1.12 / 22.4 = 1/20

Initial moles of D_{2} = 1.12/22.4 = 1/20

Now after opening of stop-cock, mass of H_{2} left in the first bulb = 0.05

⇒ Moles of H_{2} = 0.05/2 = 1/40

⇒ Moles of H_{2} effused into second bulb = 1/20 - 1/40 = 1/40

Let n be number of moles of D_{2} effused.

From Graham's Law :

⇒ = moles of D_{2} in second bulb.

In the second bulb :

The mass of H_{2} gas = 1/40 x 2 = 0.05 gm

The mass of D_{2} gas = √2/80 x 4 = 0.07 gm

⇒ Total mass = 0.05 0.07 = 0.12 gm

⇒ % of H_{2} = 0.05/1.12 x 100 = 41.67 %

⇒ % of D_{2} = 0.07/1.12 x 100 = 58.13 %

**Ex.11** *The pressure in a bulb dropped from 2000 mm to 1500 mm of Hg in 47 min when the contained O _{2} leaked through a small hole. The bulb was then completely evacuated. A mixture of oxygen and another gas (B) of molecular weight 79 in the molar ratio of 1 : 1 at a total pressure of 4000 mm of Hg was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of 74 min.*

**Sol.** Now as P ∝ n (moles), we define the rate of diffusion as the drop in the pressure per second. First we try to find the rate of diffusion of the gas B.

The rate of diffusion of O_{2} = R_{0} = (2000 - 1500)/47 = 10.638 mm/min.

Assuming that gas B was present alone in the bulb. Let the rate of diffusion of B = R_{B}.

From Graham's Law of diffusion, we have :

Now the bulb contains mixture of O_{2} and B in the mole ration of 1 : 1 at total pressure of 4000 mm Hg.

of Hg

As the pressure and temperature conditions are same for both gases in the second case (same bulb), so the rate of diffusion will remain same in the second case also.

Let X_{0} and X_{B} be the final pressure in the bulb after leakage for 74 minutes.

As P x n

⇒ Ratio of moles is given as : X_{O} : X_{B} = 1 : 1235

**Ex.12** *A 672 ml of a mixture of oxygen-ozone at N.T.P. were found to be weigh 1 gm. Calculate the volume of ozone in the mixture.*

**Sol.** Let V ml of ozone are there in the mixture

⇒ (672 - V) m; = vol. of oxygen

Mass of ozone at N.T.P. =

Mass of oxygen at N.T.P.

**Ex.13** *A 20 L flask contains 4.0 gm of O _{2} & 0.6 gm of H_{2} at 100^{o}C. If the contents are allowed to react to form water vapors at 100^{o}C, find the contents of flask and there partial pressures.*

**Sol.** H_{2} reacts with O_{2} to form water [H_{2}O_{(g)}]

⇒ 2 moles of H_{2} = 1 mole of O_{2} = 2 moles of H_{2}O

Here masses of H_{2} and O_{2} are given, so one of them can be in excess. So first check out which of the reactants is in excess.

Now, Moles of O_{2} = 4/32 = 0.125 and Moles of H_{2} = 0.6/2 = 0.3

Since 1 mole of O_{2} ≡ 2 moles of H_{2}

⇒ 0.125 moles of O_{2} ≡ 2 x 0.125 moles of H_{2}

i.e. 0.25 moles of H_{2} are used, so O_{2} reacts completely whereas H_{2} is in excess.

⇒ Moles of H_{2} in excess = 0.3 - 0.25 = 0.05 moles.

Also, 2 moles of H_{2} ≡ 2 moles of H_{2}O

⇒ 0.25 moles of H_{2} ≡ 0.25 moles of H_{2}O are produced.

⇒ Total moles after the reaction = 0.05 (moles of H_{2}) 0.25 (moles of H_{2}O) = 0.3

⇒ The total pressure P_{Total} at the end of reaction is given by :

Now partial pressure of A = mole fraction of A x P_{Total}

_{}

**Ex.14** *The compressibility factor for 1 mole of a van Waals gas at 0 ^{o}C and 100 atm pressure is found to be 0.5. Assuming that the volume of gas molecular is negligible, calculate the van der Waals constant `a'.*

Using van der Waal's equation of state :

Now : V - nb V(given)

⇒ The equation is reduced to :

or

Also,

Substitute the values of V and T :

⇒ a = 1.25 litre^{2} mol^{-2} atm.

**Ex.15** *Calculate the pressure exerted by 5 mole of CO _{2} in one litre vessel at 47^{o}C using van der waals equation. Also report the pressure of gas if it behaves ideally in nature. *

Using van der waals equation of state :

Substituting the given values, we get :

If the gas behaves ideally, then using : PV = nRT

__Solved Examples__

**IIT-JEE MAINS **

**Ex.1 **A gas occupies 300 ml at 27^{o}C and 730 mm pressure. What would be its volume at STP-

(A) 162.2 ml (B) 262.2 ml (C) 362.2 ml (D) 462.2 ml

**Sol. (B) **

Given at T_{1} = 300 K, T_{2} = 273 K (STP)

V_{1} = 300 ml = litre, P_{1} = atm. P_{2} = 1 atm., V_{2} = ?

Q = ,

=

V_{2} = 0.2622 litre = 262.2 ml.

**Ex.2 **A truck carrying oxygen cylinders is filled with oxygen at -23^{o}C and at a pressure of 3 atmosphere in Srinagar, Kashmir. Determine the internal pressure when the truck drives through Madras, Tamil Nadu. Where the temperature is 30^{o}C -

(A) 2.64 atm. (B) 1.64 atm. (C) 1 atm. (D) 3.64 atm.

**Sol. (D)**

P_{1} = 3 atm., P_{2} = ?

T_{1} = - 23 + 273 = 250 K

T_{2} = 273 + 30 = 303 K.

= ,

=

P_{2} = = 3.64 atm.

**Ex.3 The density of a gas at -23 ^{o}C and 780 torr is 1.40 gram per litre. Which one of the following gases is it -**

**(A) CO _{2} (B) SO_{2} (C) Cl_{2} (D) N_{2 }**

**Sol. (D) **

PV = nRT

x 1 = n x .082 x (273 - 23)

n =

n = 0.0501 moles.

Q .0501 moles of the gas weigh = 1.40 gm.

1 mole of gas weigh = = 28 gram.

So gas is N_{2}.

**Ex. 4 Calculate the weight of CH _{4} in a 9 litre cylinder at 16 atm and 27^{o}C (R = 0.08 lit. atm/k) -**

**(A) 96 gm (B) 86 gm (C) 80 gm (D) 90 gm**

**Sol. (A) **

Given P = 16 atm, V = 9 litre.

T = 300 K, m_{CH4} = 16, R = 0.08 litre atm/k.

PV = w/m x R x T

16 x 9 = x 0.08 x 300

w = 96 gm.

**Ex. 5 What is the density of sulphur dioxide (SO _{2}) at STP -**

**(A) 2.86 gm/lit. (B) 1.76 gm/lit (C) 1.86 gm/lit (D) None of these.**

**Sol. (A) **

The gram molecular weight of SO_{2} = 64 gm/mole.

Since 1 mole of SO_{2} occupies a volume of 22.4 litres at S.T.P.

Density of SO_{2} = = 2.86 gm/lit.

**Ex.6 5gm of XeF _{4} gas was introduced into a vessel of 6 litre capacity at 80^{o}C. What is the pressure of the gas in atmosphere -**

**(A) .21 atm (B) .31 atm (C) .11 atm (D) .41 atm.**

**Sol. (C) **

Given V = 6 litre, T = 353 K, R= 0.082, W = 5gm. m= 207.3

PV = x R x T

P x 6 = x 0.082 x (273 80)

P = = .11 atm.

**Ex.7 Calculate the temperature at which 28 gm N _{2} occupies a volume of 10 litre at 2.46 atm-**

**(A) 300 K (B) 320 K (C) 340 K (D) 280 K**

**Sol. (A) **

Given w_{N2} = 28gm ,

P = 2.46 atm, V = 10 litre. m_{N2} = 28,

Q PV = RT

2.46 x 10 = x 0.0821 x T

T = 300 K.

**Ex.8 A mixture of gases at 760 mm pressure contains 65% nitrogen, 15% oxygen and 20% Carbon dioxide by volume. What is the partial pressure of each in mm -**

**(A) 494, 114, 252 (B) 494, 224, 152**

**(C) 494, 114, 152 (D) None of these.**

**Sol. (C) **

P'_{N2} = 760 x = 494 mm

P'_{O2} = 760 x = 114 mm

P'_{CO2} = 760 x = 152 mm.

**Ex.9 0.45 gm of a gas 1 of molecular weight 60 and 0.22 gm of a gas 2 of molecular weight 44 exert a total pressure of 75 cm of mercury. Calculate the partial pressure of the gas 2 -**

**(A) 30 cm of Hg (B) 20 cm of Hg (C) 10 cm of Hg (D) 40 cm of Hg.**

**Sol. (A) **

No. of moles of gas 1 = n_{1} = =

= 0.0075

No. of moles of gas 2 = n_{2} = =

= 0.0050

Total no. of moles = n_{1} n_{2 }

= 0.0075 0.0050 = 0.0125

P_{2 } partial pressure of gas 2 = x 75 = 30 cm of Hg.

**Ex. 10 The total pressure of a sample of methane collected over water is 735 torr at 29^{o}C. The aqueous tension at 29^{o}C is 30 torr. What is the pressure exerted by dry methane -**

**(A) 605 torr (B) 205 torr (C) 405 torr (D) 705 torr**

**Sol. (D)**

P_{total} = P_{dry methane} P_{water}

735 = P_{dry methane} 30

P_{dry methane} = 735 - 30 = 705 torr.

**Ex.11 The odour from a gas A takes six seconds to reach a wall from a given point. If the molecular weight of gas A is 46 grams per mole and the molecular weight of gas B is 64 grams per mole. How long will it take for the odour from gas B to reach the same wall from the same point. Approximately -**

**(A) 6 Sec (B) 7 Sec (C) 8 Sec (D) 9 Sec**

**Sol. (B) **

=

= = 1.18

Time taken for the odour of B to reach the wall = 1.18 x 6 = 7.08 sec 7 sec.

**Ex.12 1 litre of oxygen effuses through a small hole in 60 min. and a litre of helium at the same temperature and pressure effuses through the same hole in 21.2 min. What is the atomic weight of Helium -**

**(A) 2.99 (B) 3.99 (C) 2.08 (D) 1.99**

**Sol. (B)**

= = =

=

Squaring both of sides = = M_{He } = = 3.99

Since Helium is monoatomic so

Atomic weight = Molecular weight = 3.99

**Ex.13 What is the temperature at which oxygen molecules have the same r.m.s. velocity as the hydrogen molecules at 27^{o}C -**

**(A) 3527^{o}C (B) 4227^{o}C (C) 4527^{o}C (D) 4000^{o}C**

**Sol. (C) **

r.m.s. velocity C =

For oxygen Co_{2} =

For CH_{2} at 27**^{o}**C =

When Co_{2} = CH_{2}

=

T = = 4800 K.

Then in **^{o}**C T = 4800 - 273 = 4527

**Ex.14 Calculate the total kinetic energy in joules, of the molecules in 8 gm of methane at 27^{o}C -**

**(A) 1770.5 Joule (B) 1870.5 joule (C) 1970.5 joule (D) 1670.5 joule**

**Sol. (B) **

E_{K} (For 1 mole) = RT = 3741 Joule.

Total energy of 8 gm of methane

= 1/2 mole of methane

= = 1870.5 Joule.

**Ex.15 Calculate the root mean square velocity of SO _{2} at S.T.P. -**

**(A) 3.26 x 10 ^{4} cm/sec (B) 1.26 x 10^{2} cm/sec**

**(C) 1.26 x 10 ^{4} cm/sec (D) 3.26 x 10^{2} cm/sec**

**Sol. (A)**

v_{rms} of SO_{2} =

= = 3.26 x 10^{4} cm/sec.

**Ex.16 Calculate the number of atoms in 1 g of helium**

**(A) 1.506 x 10 ^{23} atoms (B) 1.605 x 10^{32} atoms (C) 1.056 x 10^{25} atoms (D) None of these**

**Sol.** **(A) **

Atomic mass of He = 4

4 g of He contain = 6.023 x 10^{23} atoms

1 g of He contains =

= 1.506 x 10^{23} atoms.

Ex.17 7.00 g of a gas occupies a volume of 4.1 litres at 300 K and 1 atmosphere pressure. Calculate the molecular mass of the gas -

**(A) 40 g mol ^{-1} (B) 42 g mol^{-1} (C) 48 g mol^{-1} (D) 45 g mol^{-1}**

**Sol.** **(B)**

PV = nRT , n = PV/RT

n = = mol

now n = = =

Thus, molecular mass of gas = 7 x 6

= 42 g mol^{-1}

**Ex.18 Calculate density of ammonia at 30^{o}C and 5 atm. pressure -**

**(A) 3.03 g/litres. (B) 3.82 g/litres (C) 3.42 g/litres. (D) 4.42 g/litres.**

**Sol. ** **(C) **

PV = nRT, or PV = RT or P = x or P = d x

d = MP/RT ; d = = 3.42 g/litres.

**Ex.19 3 moles of a gas are present in a vessel at a temperature of 27^{o}C . Calculate the value of gas constant (R) in terms of kinetic energy of the molecules of gas -**

**(A) 7.4 x 10 ^{-4} KE per degree kelvin.**

**(B) 4.5 x 10 ^{-4} KE per degree kelvin.**

**(C) 7.4 x 10 ^{-5} KE per degree kelvin.**

**(D) None of these**

**Sol. [A]** K.E. for 1 mole = 3/2 RT

K.E. for 3 moles = 9/2 RT. or R = KE =

= 7.4 x 10^{-4} KE per degree kelvin.

**Ex.20 In the following diagram, container of NH _{3} gas and container of HCl gas, connected through a long tube, are opened simultaneously at both ends; the white NH_{4}Cl ring first formed will be at Q point. If OP = 40cm, then find OQ -**

**(A) 35 cm (B) 23.74 cm (C) 30 cm (D) 31.25 cm**

**Sol. [B]**

Let OQ = x cm so QP = (40 - x) cm

Diffused volume of NH_{3} gas = Area of T.S. of tube x Distance travelled by NH_{3} gas

= A x OQ = Ax

{Where A is area of T.S. of tube}

Similarly in the same time,

Diffused volume of HCl gas = Area of T.S. of tube x Distance travelled by HCl gas

V_{HCl} = A x QP = A (40 - x)

From Graham's Law of diffusion

⇒ =

⇒ = = 1.46

⇒ = 1.46

⇒ x = 23.74 cm

OQ = 23.74 cm

**Ex.21 The root mean square velocity of hydrogen is times than that of nitrogen. If T is the temperature of the gas, then :**

**(A) = (B) > (C) < (D) = **

**Sol. [C]** V_{rms} = = = ;

=

x =

= =

= 5 = x 14

x 5 = x 14

>

**Ex.22 A jar contains a gas and a few drops of water. The pressure in the jar is 830 mm of Hg. The temperature of the jar is reduced by 1% . The vapour pressure of water at two temperature are 30 and 25 mm of Hg. Calculate the new pressure in jar.**

**(A) 792 mm of Hg (B) 817 mm of Hg (C) 800 mm of Hg (D) 840 mm of Hg**

**Sol. [B]** P_{gas} = P_{dry gas} P_{moisture} at T K

or P_{dry} = 830 - 30 = 800

Now at T_{2} = 0.99 T_{1} ;

at constant volume =

P_{dry} = = 792 mm

P_{gas} = P_{dry} P_{moisture}

= 792 25 = 817 mm

**Ex.23 Calculate relative rate of effusion of O _{2} to CH_{4} through a container containing O_{2} and CH_{4} in 3 : 2 mass ratio.**

**(A) (B) (C) (D) None of these**

**Sol. [B]** = =

= x x =

Ex.24 At what temperature will average speed of the molecule of the second member of the series C_{n}H_{2n} be the same of Cl_{2} at 627* ^{o}*C ?

**(A) 259.4 K (B) 400 K (C) 532.4 K (D) None of these**

**Sol. [C]** Second member of C_{n}H_{2n} series =C_{3}H_{6} = 42

= = = =

T_{2} = 532.4 K

**Ex.25 80 mL of O _{2} takes 2 minute to pass through the hole. What volume of SO_{2} will pass through the hole in 3 minute ?**

**(A) (B) 120 x (C) (D) None of these**

**Sol. [A]** = = = ;

= =; V_{2} =

**Ex.26 The density of a gas filled electric lamp is 0.75 kg/m ^{3}. After the lamp has been switched on, the pressure in it increases from 4 x 10^{4} Pa to 9 x 10^{4} pa. What is increases in U_{RMS} ?**

**(A) 100 (B) 200 (C) 300 (D) None of these**

**Sol. [B]** U_{1} =

DU_{rms} = ( - )

= (300 - 200)

= x 100 = 200

**Ex.27 If one mole each of a monoatomic and diatomic gases are mixed at low temperature then C _{P}/C_{V} ratio for the mixture is :**

**(A) 1.40 (B) 1.428 (C) 1.5 (D) 1.33**

**Sol. [C]** C_{v} = = 2R

C_{p} = = 3R

= 1.5

**Ex.28 If one mole of a monatomic gas (g = 5/3) is mixed with one mole of a diatomic gas (g = 7/5), the value of g for the mixture is :**

**(A) 1.4 (B) 1.5 (C) 1.53 (D) 3.07**

**Sol. [B]**

Thus, for mixture of 1 mole each,

C_{v} = and C_{p} =

Thus, = = 1.5

Ex.29 The average speed at temperature T* ^{o}*C of CH

**(A) 240.55^{o}C (B) -32.45^{o}C (C) 3000^{o}C (D) -24.055^{o}C **

**Sol. [B]** = =

= x 10^{6}

T = = 240.55 K

T**^{o}**C = 240.55 - 273 = -32.45

**Ex.30 The intercept on y-axis and slope of curve plotted between P/T vs T**

**For an ideal gas having 10 moles in a closed rigid container of volume 821. L. (P = pressure in atm and T = Temp. in K , log _{10} 2 = 0.30) are respectively**

**(A) 0.01, 0 (B) 0.1, 1 (C) 0.1,0 (D) 10,1**

**Sol. [B]** Intercept on y-axis

= log_{10}= log_{10}= -1.0

v/s curve =

Intercept = = = 0.1, slope = 0

**Ex.31 Two vessels connected by a valve of negligible volume. One container (I) has 2.8 g of N _{2} at temperature T_{1} (K) The other container (II) is completely evacuated. The container (I) is heated to T_{2} (K) while container (II) is maintained at T_{2}/3(K). volume of vessel (I) is half that of vessel (II). If the valve is opened then what is the weight ratio of N_{2} in both vessel (W_{I}/W_{II}) ?**

**(A) 1 : 2 (B) 1 : 3 (C) 1 : 6 (D) 3 : 1**

**Sol. [C] **

Let x mole of N_{2} present into vessel II and P is final pressure of N_{2}

P(2V) = xR(T_{2}/3) and P (V) = (0.1 - x) RT_{2}

⇒ 2 =

⇒ x = 0.6/7 mole

x 28 ⇒ 2.4 g N_{2}

II has 2.4 g N_{2} and I has 0.4 g of N_{2};

= ⇒ 1 : 6

Ex.32 **7 moles of a tetra-atomic non-linear gas `A' at 10 atm and T K are mixed with 6 moles of another gas B at K and 5 atm in a closed, rigid vessel without energy transfer with surroundings. If final temperature of mixture was K, then gas B is ? (Assuming all modes of energy are active)**

**Sol. **

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