Solved Examples - States of Matter Class 11 Notes | EduRev

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Ex.1 When 2 gm of a gaseous substance A is introduced into an initially evacuated flask at 25oC, 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 MA and MB be the molecular weights of A and B.

Using PV = nRT for A, we get :

Solved Examples - States of Matter Class 11 Notes | EduRev ...(i)

and using Dalton's Law : PTotal = Solved Examples - States of Matter Class 11 Notes | EduRev ⇒ 1.5 = Solved Examples - States of Matter Class 11 Notes | EduRev ...(ii)

Solving (i) and (ii), we get Solved Examples - States of Matter Class 11 Notes | EduRev


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

Sol. By Graham's Law :

Solved Examples - States of Matter Class 11 Notes | EduRev ⇒ Solved Examples - States of Matter Class 11 Notes | EduRev

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 :

Solved Examples - States of Matter Class 11 Notes | EduRev ⇒ Solved Examples - States of Matter Class 11 Notes | EduRev ⇒ Solved Examples - States of Matter Class 11 Notes | EduRev

⇒ Mole ratio of gas A and B effusing out = Solved Examples - States of Matter Class 11 Notes | EduRev [moles ∝ pressure]


Ex.4 At 30oC 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 PM0 = dRT

⇒ M0 = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev (T = 30 273 K)

⇒ M0 = 39.38

Now number of molecules = n x N0

Solved Examples - States of Matter Class 11 Notes | EduRev = 2.29 x 1019


Ex.5 The pressure exerted by 12 gm of an ideal gas at temperature toC in a vessel of volume V litre is one atm. When the temperature is increased by 10o 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 : Solved Examples - States of Matter Class 11 Notes | EduRev

⇒ t = 100 k or t = - 173oC

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 N2 and O2 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 = Solved Examples - States of Matter Class 11 Notes | EduRev x P and Partial pressure of oxygen = Solved Examples - States of Matter Class 11 Notes | EduRev

Now, , and

Solved Examples - States of Matter Class 11 Notes | EduRev x 750 = 600 mm of Hg and Solved Examples - States of Matter Class 11 Notes | EduRev = 150 mm of Hg


Ex.7 An open vessel at 27oC 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 n1 T1 = n2 T2

Now let n1 = initial moles and n2 = final moles

⇒ n2 = 2/5 x n1 (as 3/5th of the air has been expelled)

⇒ T2 = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

⇒ T2 = Solved Examples - States of Matter Class 11 Notes | EduRev = 750 K = 477oC 


Ex.8 When 3.2 gm of sulphur is vaporized at 450oC 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.

Solved Examples - States of Matter Class 11 Notes | EduRev

⇒ Number of atoms = Solved Examples - States of Matter Class 11 Notes | EduRev

Hence, molecular formula of sulphur = S8  


Ex.9 A spherical balloon of 21 cm diameter is to be filled with H2 at NTP from a cylinder containing the gas at 20 atm at 27oC. 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 n0 = moles of hydrogen required to fill one balloon.

n = Solved Examples - States of Matter Class 11 Notes | EduRev

Solved Examples - States of Matter Class 11 Notes | EduRev (Note : the balloons are being filled at S.T.P.)

= Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

⇒ Number of balloons that can be filled = n/n0 = 10.50 ≈ 10


Ex.10 A mixture containing 1.12L of H2 and 1.12L of D2 (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 H2. Determine the % age composition by weight of the gases in the second bulb. 

Sol. In the first bulb :

Initial moles of H2 = 1.12 / 22.4 = 1/20

Initial moles of D2 = 1.12/22.4 = 1/20

Now after opening of stop-cock, mass of H2 left in the first bulb = 0.05

⇒ Moles of H2 = 0.05/2 = 1/40

⇒ Moles of H2 effused into second bulb = 1/20 - 1/40 = 1/40

Let n be number of moles of D2 effused.

From Graham's Law :

 ⇒ Solved Examples - States of Matter Class 11 Notes | EduRev = moles of D2 in second bulb.

In the second bulb :

The mass of H2 gas = 1/40 x 2 = 0.05 gm

The mass of D2 gas = √2/80 x 4 = 0.07 gm

⇒ Total mass = 0.05 0.07 = 0.12 gm

⇒ % of H2 = 0.05/1.12 x 100 = 41.67 %

⇒ % of D2 = 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 O2 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 O2 = R0 = (2000 - 1500)/47 = 10.638 mm/min.

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

From Graham's Law of diffusion, we have :

Solved Examples - States of Matter Class 11 Notes | EduRev

Solved Examples - States of Matter Class 11 Notes | EduRev

Now the bulb contains mixture of O2 and B in the mole ration of 1 : 1 at total pressure of 4000 mm Hg.

Solved Examples - States of Matter Class 11 Notes | EduRev 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 X0 and XB be the final pressure in the bulb after leakage for 74 minutes.

Solved Examples - States of Matter Class 11 Notes | EduRev

Solved Examples - States of Matter Class 11 Notes | EduRev

Solved Examples - States of Matter Class 11 Notes | EduRev

As P x n

⇒ Ratio of moles is given as : XO : XB = 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. = Solved Examples - States of Matter Class 11 Notes | EduRev

Mass of oxygen at N.T.P. Solved Examples - States of Matter Class 11 Notes | EduRev

Solved Examples - States of Matter Class 11 Notes | EduRev


Ex.13 A 20 L flask contains 4.0 gm of O2 & 0.6 gm of H2 at 100oC. If the contents are allowed to react to form water vapors at 100oC, find the contents of flask and there partial pressures. 

Sol. H2 reacts with O2 to form water [H2O(g)]

Solved Examples - States of Matter Class 11 Notes | EduRev

⇒ 2 moles of H2 = 1 mole of O2 = 2 moles of H2O

Here masses of H2 and O2 are given, so one of them can be in excess. So first check out which of the reactants is in excess.

Now, Moles of O2 = 4/32 = 0.125 and Moles of H2 = 0.6/2 = 0.3

Since 1 mole of O2 ≡ 2 moles of H2

⇒ 0.125 moles of O2 ≡ 2 x 0.125 moles of H2

i.e. 0.25 moles of H2 are used, so O2 reacts completely whereas H2 is in excess.

⇒ Moles of H2 in excess = 0.3 - 0.25 = 0.05 moles.

Also, 2 moles of H2 ≡ 2 moles of H2O

⇒ 0.25 moles of H2 ≡ 0.25 moles of H2O are produced.

⇒ Total moles after the reaction = 0.05 (moles of H2) 0.25 (moles of H2O) = 0.3

⇒ The total pressure PTotal at the end of reaction is given by :

Solved Examples - States of Matter Class 11 Notes | EduRev

Now partial pressure of A = mole fraction of A x PTotal


Ex.14 The compressibility factor for 1 mole of a van Waals gas at 0oC 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 :

Solved Examples - States of Matter Class 11 Notes | EduRev

Now : V - nb Solved Examples - States of Matter Class 11 Notes | EduRev V(given)

⇒ The equation is reduced to : Solved Examples - States of Matter Class 11 Notes | EduRev

or Solved Examples - States of Matter Class 11 Notes | EduRev

Solved Examples - States of Matter Class 11 Notes | EduRevSolved Examples - States of Matter Class 11 Notes | EduRev

Also, Solved Examples - States of Matter Class 11 Notes | EduRev

Substitute the values of V and T :

⇒ a = 1.25 litre2 mol-2 atm.


Ex.15 Calculate the pressure exerted by 5 mole of CO2 in one litre vessel at 47oC using van der waals equation. Also report the pressure of gas if it behaves ideally in nature. (a = 3.592 atm litre2 mol-2, b = 0.0427 litre mol-1)

Using van der waals equation of state :

Solved Examples - States of Matter Class 11 Notes | EduRev

Substituting the given values, we get :

Solved Examples - States of Matter Class 11 Notes | EduRev

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

Solved Examples - States of Matter Class 11 Notes | EduRev


Solved Examples 


IIT-JEE MAINS 

Ex.1 A gas occupies 300 ml at 27oC 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 T1 = 300 K, T2 = 273 K (STP)

V1 = 300 ml = Solved Examples - States of Matter Class 11 Notes | EduRev litre, P1 = Solved Examples - States of Matter Class 11 Notes | EduRevatm. P2 = 1 atm., V2 = ?

Q Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev,

Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

V2 = 0.2622 litre = 262.2 ml.


Ex.2 A truck carrying oxygen cylinders is filled with oxygen at -23oC 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 30oC -

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

Sol. (D)

P1 = 3 atm., P2 = ?

T1 = - 23 + 273 = 250 K

T2 = 273 + 30 = 303 K.

Solved Examples - States of Matter Class 11 Notes | EduRev= Solved Examples - States of Matter Class 11 Notes | EduRev,

Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

P2 = Solved Examples - States of Matter Class 11 Notes | EduRev = 3.64 atm.


Ex.3 The density of a gas at -23oC and 780 torr is 1.40 gram per litre. Which one of the following gases is it -

(A) CO2 (B) SO2 (C) Cl2 (D) N

Sol. (D) 

PV = nRT

Solved Examples - States of Matter Class 11 Notes | EduRev x 1 = n x .082 x (273 - 23)

n = Solved Examples - States of Matter Class 11 Notes | EduRev

n = 0.0501 moles.

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

1 mole of gas weigh = Solved Examples - States of Matter Class 11 Notes | EduRev = 28 gram.

So gas is N2.


Ex. 4 Calculate the weight of CH4 in a 9 litre cylinder at 16 atm and 27oC (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, mCH4 = 16, R = 0.08 litre atm/k.

PV = w/m x R x T

16 x 9 = Solved Examples - States of Matter Class 11 Notes | EduRev x 0.08 x 300

w = 96 gm.


Ex. 5  What is the density of sulphur dioxide (SO2) 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 SO2 = 64 gm/mole.

Since 1 mole of SO2 occupies a volume of 22.4 litres at S.T.P.

Density of SO2 =Solved Examples - States of Matter Class 11 Notes | EduRev = 2.86 gm/lit.


Ex.6 5gm of XeF4 gas was introduced into a vessel of 6 litre capacity at 80oC. 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 = Solved Examples - States of Matter Class 11 Notes | EduRev x R x T

P x 6 = Solved Examples - States of Matter Class 11 Notes | EduRev x 0.082 x (273 80)

P = Solved Examples - States of Matter Class 11 Notes | EduRev = .11 atm.


Ex.7 Calculate the temperature at which 28 gm N2 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 wN2 = 28gm ,

P = 2.46 atm, V = 10 litre. mN2 = 28,

Q PV = Solved Examples - States of Matter Class 11 Notes | EduRevRT

2.46 x 10 = Solved Examples - States of Matter Class 11 Notes | EduRev 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 Solved Examples - States of Matter Class 11 Notes | EduRev = 494 mm

P'O2 = 760 x Solved Examples - States of Matter Class 11 Notes | EduRev = 114 mm

P'CO2 = 760 x Solved Examples - States of Matter Class 11 Notes | EduRev= 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 = n1 = = Solved Examples - States of Matter Class 11 Notes | EduRev

= 0.0075

No. of moles of gas 2 = n2 = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

= 0.0050

Total no. of moles = n1 n

= 0.0075 0.0050 = 0.0125

P partial pressure of gas 2 = Solved Examples - States of Matter Class 11 Notes | EduRev x 75 = 30 cm of Hg.


Ex. 10 The total pressure of a sample of methane collected over water is 735 torr at 29oC. The aqueous tension at 29oC 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)

Ptotal = Pdry methane Pwater

735 = Pdry methane 30

Pdry 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) 

Solved Examples - States of Matter Class 11 Notes | EduRev= Solved Examples - States of Matter Class 11 Notes | EduRev

Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = 1.18

Time taken for the odour of B to reach the wall = 1.18 x 6 = 7.08 sec Solved Examples - States of Matter Class 11 Notes | EduRev 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)

Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

= Solved Examples - States of Matter Class 11 Notes | EduRev

Squaring both of sides = Solved Examples - States of Matter Class 11 Notes | EduRev= Solved Examples - States of Matter Class 11 Notes | EduRev MHe  = Solved Examples - States of Matter Class 11 Notes | EduRev = 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 27oC -

(A) 3527oC  (B) 4227oC  (C) 4527oC  (D) 4000oC

Sol. (C) 

r.m.s. velocity C = Solved Examples - States of Matter Class 11 Notes | EduRev

For oxygen Co2 = Solved Examples - States of Matter Class 11 Notes | EduRev

For CH2 at 27oC = Solved Examples - States of Matter Class 11 Notes | EduRev

When Co2 = CH2

Solved Examples - States of Matter Class 11 Notes | EduRev= Solved Examples - States of Matter Class 11 Notes | EduRev

Solved Examples - States of Matter Class 11 Notes | EduRev

T = Solved Examples - States of Matter Class 11 Notes | EduRev = 4800 K.

Then in oC  T = 4800 - 273 = 4527oC 


Ex.14 Calculate the total kinetic energy in joules, of the molecules in 8 gm of methane at 27oC -

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

Sol. (B) 

EK (For 1 mole) = RT = 3741 Joule.

Total energy of 8 gm of methane

= 1/2 mole of methane

= Solved Examples - States of Matter Class 11 Notes | EduRev = 1870.5 Joule.


Ex.15 Calculate the root mean square velocity of SO2 at S.T.P. -

(A) 3.26 x 104 cm/sec (B) 1.26 x 102 cm/sec

(C) 1.26 x 104 cm/sec (D) 3.26 x 102 cm/sec

Sol. (A)

vrms of SO2 =

= Solved Examples - States of Matter Class 11 Notes | EduRev = 3.26 x 104 cm/sec.


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

(A) 1.506 x 1023 atoms (B) 1.605 x 1032 atoms (C) 1.056 x 1025 atoms (D) None of these

 Sol. (A) 

Atomic mass of He = 4

4 g of He contain = 6.023 x 1023 atoms

1 g of He contains =

Solved Examples - States of Matter Class 11 Notes | EduRev = 1.506 x 1023 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 = Solved Examples - States of Matter Class 11 Notes | EduRev= mol

now n = Solved Examples - States of Matter Class 11 Notes | EduRev= = Solved Examples - States of Matter Class 11 Notes | EduRev

Thus, molecular mass of gas = 7 x 6

= 42 g mol-1


Ex.18 Calculate density of ammonia at 30oC  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 = Solved Examples - States of Matter Class 11 Notes | EduRevRT or P = Solved Examples - States of Matter Class 11 Notes | EduRevx Solved Examples - States of Matter Class 11 Notes | EduRev or P = d x Solved Examples - States of Matter Class 11 Notes | EduRev

d = MP/RT ; d = Solved Examples - States of Matter Class 11 Notes | EduRev= 3.42 g/litres.


Ex.19 3 moles of a gas are present in a vessel at a temperature of 27oC  . 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 = Solved Examples - States of Matter Class 11 Notes | EduRevKE = Solved Examples - States of Matter Class 11 Notes | EduRev

= 7.4 x 10-4 KE per degree kelvin.


Ex.20 In the following diagram, container of NH3 gas and container of HCl gas, connected through a long tube, are opened simultaneously at both ends; the white NH4Cl ring first formed will be at Q point. If OP = 40cm, then find OQ -

Solved Examples - States of Matter Class 11 Notes | EduRev

(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 NH3 gas = Area of T.S. of tube x Distance travelled by NH3 gas

Solved Examples - States of Matter Class 11 Notes | EduRev = 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

VHCl = A x QP = A (40 - x)

From Graham's Law of diffusion

Solved Examples - States of Matter Class 11 Notes | EduRev= Solved Examples - States of Matter Class 11 Notes | EduRev

Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev= 1.46

Solved Examples - States of Matter Class 11 Notes | EduRev= 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) Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev (B) Solved Examples - States of Matter Class 11 Notes | EduRev> Solved Examples - States of Matter Class 11 Notes | EduRev (C) Solved Examples - States of Matter Class 11 Notes | EduRev < Solved Examples - States of Matter Class 11 Notes | EduRev (D) Solved Examples - States of Matter Class 11 Notes | EduRev= Solved Examples - States of Matter Class 11 Notes | EduRev

 

Sol. [C] Vrms = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev;

=

Solved Examples - States of Matter Class 11 Notes | EduRev x Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

= Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

= 5 = Solved Examples - States of Matter Class 11 Notes | EduRev x 14

Solved Examples - States of Matter Class 11 Notes | EduRev x 5 = Solved Examples - States of Matter Class 11 Notes | EduRevx 14

Solved Examples - States of Matter Class 11 Notes | EduRev > Solved Examples - States of Matter Class 11 Notes | EduRev


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] Pgas = Pdry gas Pmoisture at T K

or Pdry = 830 - 30 = 800

Now at T2 = 0.99 T1 ;

at constant volume = Solved Examples - States of Matter Class 11 Notes | EduRev

Pdry = Solved Examples - States of Matter Class 11 Notes | EduRev = 792 mm

Pgas = Pdry Pmoisture

= 792 25 = 817 mm

 

Ex.23 Calculate relative rate of effusion of O2 to CH4 through a container containing O2 and CH4 in 3 : 2 mass ratio.

(A) Solved Examples - States of Matter Class 11 Notes | EduRev (B) Solved Examples - States of Matter Class 11 Notes | EduRev (C) Solved Examples - States of Matter Class 11 Notes | EduRev (D) None of these

Sol. [B]Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

= Solved Examples - States of Matter Class 11 Notes | EduRev x Solved Examples - States of Matter Class 11 Notes | EduRev x Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev


Ex.24 At what temperature will average speed of the molecule of the second member of the series CnH2n be the same of Cl2 at 627oC  ?

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

 Sol. [C] Second member of CnH2n series =C3H6 = 42

= Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

T2 = 532.4 K

 

Ex.25 80 mL of O2 takes 2 minute to pass through the hole. What volume of SO2 will pass through the hole in 3 minute ?

(A) Solved Examples - States of Matter Class 11 Notes | EduRev (B) 120 x Solved Examples - States of Matter Class 11 Notes | EduRev (C) Solved Examples - States of Matter Class 11 Notes | EduRev (D) None of these

Sol. [A]Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev;

Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev =Solved Examples - States of Matter Class 11 Notes | EduRev; V2 = Solved Examples - States of Matter Class 11 Notes | EduRev


Ex.26 The density of a gas filled electric lamp is 0.75 kg/m3. After the lamp has been switched on, the pressure in it increases from 4 x 104 Pa to 9 x 104 pa. What is increases in URMS ?

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

Sol. [B] U1 = Solved Examples - States of Matter Class 11 Notes | EduRev

DUrms = Solved Examples - States of Matter Class 11 Notes | EduRev(Solved Examples - States of Matter Class 11 Notes | EduRev - Solved Examples - States of Matter Class 11 Notes | EduRev)

= Solved Examples - States of Matter Class 11 Notes | EduRev (300 - 200)

= Solved Examples - States of Matter Class 11 Notes | EduRev x 100 = 200


Ex.27 If one mole each of a monoatomic and diatomic gases are mixed at low temperature then CP/CV ratio for the mixture is :

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

Sol. [C] Cv = Solved Examples - States of Matter Class 11 Notes | EduRev = 2R

Cp = Solved Examples - States of Matter Class 11 Notes | EduRev = 3R

Solved Examples - States of Matter Class 11 Notes | EduRev = 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]Solved Examples - States of Matter Class 11 Notes | EduRev

Solved Examples - States of Matter Class 11 Notes | EduRev

Thus, for mixture of 1 mole each,

Cv = Solved Examples - States of Matter Class 11 Notes | EduRev and Cp = Solved Examples - States of Matter Class 11 Notes | EduRev

Thus, Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = 1.5


Ex.29 The average speed at temperature ToC  of CH4(g) is Solved Examples - States of Matter Class 11 Notes | EduRev x 103 ms-1. What is the value of T ?

(A) 240.55oC  (B) -32.45oC  (C) 3000oC  (D) -24.055oC 

Sol. [B]Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev x 106

T = Solved Examples - States of Matter Class 11 Notes | EduRev = 240.55 K

ToC  = 240.55 - 273 = -32.45oC 


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 , log10 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

= log10Solved Examples - States of Matter Class 11 Notes | EduRev= log10Solved Examples - States of Matter Class 11 Notes | EduRev= -1.0

Solved Examples - States of Matter Class 11 Notes | EduRev v/s curve Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev

Intercept = Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev = 0.1, slope = 0



Ex.31 Two vessels connected by a valve of negligible volume. One container (I) has 2.8 g of N2 at temperature T1 (K) The other container (II) is completely evacuated. The container (I) is heated to T2 (K) while container (II) is maintained at T2/3(K). volume of vessel (I) is half that of vessel (II). If the valve is opened then what is the weight ratio of N2 in both vessel (WI/WII) ?

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

Sol.   [C]       Solved Examples - States of Matter Class 11 Notes | EduRev

Let x mole of N2 present into vessel II and P is final pressure of N2

P(2V) = xR(T2/3) and P (V) = (0.1 - x) RT2

⇒ 2 = Solved Examples - States of Matter Class 11 Notes | EduRev

⇒ x = 0.6/7 mole

Solved Examples - States of Matter Class 11 Notes | EduRev x 28 ⇒ 2.4 g N2

II has 2.4 g N2 and I has 0.4 g of N2;

Solved Examples - States of Matter Class 11 Notes | EduRev = Solved Examples - States of Matter Class 11 Notes | EduRev ⇒ 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 Solved Examples - States of Matter Class 11 Notes | EduRev K, then gas B is ? (Assuming all modes of energy are active)

Sol.  
Solved Examples - States of Matter Class 11 Notes | EduRev

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