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Gaseous State Hints and solutions

Gaseous State

 

Hints to Subjective Problems

 

LEVEL – I

  1. Apply Charles law
  2. PV = nRT
  3. Rate of effusion
  4. Apply
  5. Calculate Meff. for air

 

LEVEL – II

  1. Difference of in energy  = R(T2 – T1) per mole 

 

  1. Pressure inside the balloon is pressure due to gas + atmospheric pressure 

 

  1. Vc =

 

  1. Calculate Mef. for air

 

  1. Calculate empirical formula first 

 

  1. Use b = 4NA × V

 

  1. Apply PV = nRT

 

  1. Apply Graham’s law

LEVEL – III

  1. Calculate total no. of moles at equilibrium 

 

  1. Total pressure  = Pressure due to mercury + atmospheric pressure  

 

  1. Calculate water ratio first

 

  1. First calculate total no. of moles of N2 and I2 in mixture 

 

  1. Taking 34.45 as average weight calculate mole fraction of two isotopes 

 

  1. Use Dalton’s law of partial pressure

 

  1. Make equation for oxidation of H2 and CO then solve6. Find molar ratio of two isotopes

 

  1. NH3 will decompose to N2 and H2 then apply V n

 

  1. Use equation for line then apply (pV)max = nRTmax

 

  1. Use Graham’s law of diffusion

 

  1. Find the number of moles then apply pV = nRT

 

  1. Use Dalton’s law of partial pressure

 

Solution to Subjective Problems

 

LEVEL – I

 

  1. In order to triple the volume, temperature or Kelvin scale will have to be tripled (Charles’ law) so the required temperature will be 3(273 + 27) i.e., 900°K or (900 – 273)°C = 627°C

 

2.

 

  1. PV =

⇒ V = 25.565 litre

 

  1. PV =

16 × 9 =

W=96 g

 

5. K.E. of one mole of a gas = = = 900 calories  

 

  1. PM = dRT

⇒ d = 1.5124 g/lt

 

  1. H2 > He > N2

Rate of diffusion of the gas and

 

  1. Vav =

=

= 1.502 × 103 m/s

  1. At 273 + 27 = 300K (T1), let the number of moles of air be 1 and at T2 the number of moles would be 1 – 3/5 = 2/5 = 0.4 moles. Pressure is constant, because the vessel is open and the volume of the vessel is also constant (given).

Using the gas equation: PV = n1RT1 = n2RT2

Whence, = T1/T2 = n2/n1 or T2 = T1 × n1/n2

= = 750K

 

  1. Since the root mean square velocity u (T)1/2

Hence =

or or T1 = 9T2

 

  1. ρ = = = 0.780 g/lf

 

  1. Mol. wt. (M)  = ×22.4; V = Vol. In lit. apply pV = RT)

w = Wt. of the gas in g at 473 K and 16 atm]

 

13 According to gas equation,

P2 = = 1051 mm

 

14.

ratio of mole fraction =

 

= =

1: 24

 

  1. We have PV =

and also, under identical condition of temperature and pressure volume % of a gas  = mole percent of the gas 

Av molecular weight of air  M= = 28.84

weight of air in the tyre = = 5.815

 

LEVEL – II

  1. Change in translation  kinetic energy per mole

= = = 300 cals.

Change in translation  kinetic energy for 3.45 g Ne

= = 51.29 cal.

 

  1. Let the volume of cylinder = V lt

PV = nRT

30 × V =

⇒ v = 9.43 lt

let the final moles be n2

P2V2 = n2RT2 (V2 = V0 = 9.43)

1 × 9.43 = n2 × 0.0821 × 358

n2 = 0.32

wt = 0.32×32 = 10.26

mass of O2 escaped = 370-10.26 = 359.7 g

 

  1. (since no. of moles of gas are constant)

⇒ P2 = 0.031atm

 

  1. Vc = 3b = and also

Vc =

So

⇒ r3 = 4.25 × 10-30

⇒ r = 1.62 Ao

 

  1. Vescape = = O2

use R = 8:314

T = = = 1.6 ×105K

6. The aqueous tension remains same in both the flask. Also flask are at same temperature

∴ P1V1 = P2V2

where P1 = 200 – 93 = 107

V1 = 1 litre

V2 = 2 litre

∴107 × 1 = P × 2

P = 53.5 mm

Since aqueous tension is also present in flask equivalent to 93 mm

∴Pressure of gaseous mixture = 93 + 53.5 = 146.5 mmHg

  1. Molecular mass of HCl = 36.5

Molecular mass of NH3 = 17.0

According to Graham’s law of diffusion,

Let the distance between P and X be d cm.

= 0.6824

or d = (0.6824) (200 – d)

or d + 0.6824 d = 0.6824 × 200

or d = = 81.12 cm

 

  1. At STP vol. of 1 mole He = 22400 C.C.

But b = 24 cm3

b = 4 ×NA× actual volume of single He atom = 24 cm3

i.e, effective volume occupied by 1 mole He gas = 24 cm3

fraction of total volume He

actually occupied = = 0.00107 or 0.107%

Which is very less 

So b can be neglected as compared to Vreal.

Reduced vander Waal equation 

V = nRT

 

9.

Dividing by (V-nb) and solving for P,

Substituting n = 1, R = 0.0821 L atm mol-1 K-1, V = 0.25 L, T = 300 K and the values of a and b, we have

 

From the ideal gas equation,

P = 

 

  1. Let weight of Cx H8 = y gm weight of CxH12 = (41.4–y) g 

1.56 ×10 = ×0.082 ×317 ——— (1)

% of hydrogen  = = 14 ——— (2)

On solving equation (I) and (2) we get 

x =  5 and  y = 23.8

(1) Gases are C5H8 and C5H12

(2) C5H8 = 0.35 mole C5H12 = 0.244 mole 

 

  1. a) for ideal gas nature Z = 1

PV = nRT

V =

=

V = 0.0448 litre

  1. b) for real gas

PV = ZnRT

V =

=

V = 8.987 × 10-3 litre.

 

  1. PV =

P × 0.75 =

P = 14.827 atm

Pressure inside the bottle = P + atm pressure = 14.827 + 1

= 15.827 atm

  1. =

1033 × 16 = 0.623  m

(a) 0.623 m

(b) no where

 

  1. a)

*

V= 0.37 lt

  1. b) PV =

m = 181.54

  1. c) PV =

P × 0.418 =

P = 2.27 atm

  1. d) the volume of vessel at 300 K = volume of gas at 300 K

= 0.418 lt

 

  1. We know that b = 4 v N

=

4.2 × 10-2 × 10-3 =

∴ r3 = 4.16 × 10-30 

∴ r= 1.6 × 10-10 m

2 r = closest distance between two nuclei = 3.2 Ao

 

LEVEL – III

  1. We know that PV = nRT

Volume occupied by 1 mol gaseous water at 100 oC and 1 atm pressure

= = 30.62 litre

volume of 1 mol liquid water = = = 1.88 × 10-2 litre

% of volume occupied by liquid water

= = 0.0613 %

% of free volume = 100 – 0.0613 = 99.93 %

 

  1. N2   2N

0  (at initial)

Total moles =

PV = nRT

P ×

P= 1.92 atm

 

3. Initially at lower end

P = 76 cm of Hg + 76 cm of air

= 152  cm, T = 300K, 

V = (where V1 is the volume of cylinder)

Finally at lower end

P = 76 cm of air + 38 cm of Hg

= 114 cm, T = ?, V =

⇒ T = 337.5 K

 

4. For flask 1 P1V1 = n1RT1

For flask 2 P2V2 = n2RT2

Since both the flask are of equal volume and have been joined by a narrow tube, hence pressure is also same in both the flask

P1V1 = P2V2

n1RT1 = n2RT2

n1 + n2 = 0.6

∴ n1 = 0.4 mol at 300 K

n2 = 0.2 mol at 600 K

volume of flask =

( initial conditions)

= 1.48 lt

∴final pressure =

  = 0.66 atm

  1. Let the moles of propane (C3H8) = n1

Moles of methane CH4 = n2

We know that PV = nRT

∴320 × V = (n1+n2) RT

after combustion

total moles of CO2 formed = (3n1+n2)

Once again we have PV = nRT

448 ×V = ….(2)

Dividing eqn (2) by (1) we have

= 0.25

mol fraction of C3H8 =

= = 0.2

  1. 10 gms of I2 (solid) and Nitrogen at 10 atm pressure and at 25°C is taken in one litre flask. The pressure is due to only Nitrogen as I2 is solid at 25° C

Applying gas equation

PV = nRT, we determine the moles of Nitrogen taken

P = 10 atm V = 1 litre R = 0.0821 T = 25 + 273 = 298 K

10 × 1 = × 0.0821 × 298

= 0.4087 moles

Moles of I2(s) taken = = 0.0394

Total moles of I2(s) and Nitrogen in I flask of volume 1 litre is,

= 0.4087 +   0.0394 =  0.4481 moles

On heating the I flask to 523 K I2(s) vaporises and we get a mixture of I2 (vapours) and Nitrogen. Out of 0.4481 moles, let n moles migrate to second flask kept at 473 K. Let Pnew be the pressure in the flask developed.

Applying gas equation

Pnew × 1 = (0.4481 –n) × R × 523 …(1)

Pnew × 40 = n × R × 473 …(2)

Dividing (1) by (2) we get,

On solving we get,

n = 0.4382 moles

Substituting the value of ‘n’ in equation (2) we get,

Pnew × 40 = 0.4382 × 0.0821 × 473

Pnew = 0.4254 atm

 

  1. Given pressure of N2 = 0.965 atm

Volume of N2 = 10 litre. Temperature of N2 = 298 K

For N2 when bag is fully expanded,

Volume of N2 (alone) = 30 litre at 298 K

P1V1 = P2V2

0.965 × 10 = P2 × 30

(alone) in 30 litre bag at 298 K = 0.322 atm

Now PM = +

0.990 = + 0.332

= 0.668 atm

 

  1. Combustion equations :

(i) 2H2 + O2 2H2O

2 vols. + 1 vol Zero (liquid)

(ii) 2CO + O2 2CO2

2 vols. 1 vol. 2 vols.

 

From the equations, it is clear that 12 ml. of H2 present in 24 ml. of water gas will react with 6 ml. of oxygen to produce negligible volume of liquid water at room temperature and 12 ml. of CO present in the water gas will react with 6 ml. of O2 to produce 12 ml. of CO2.

Thus, the total volume of oxygen used = 6 + 6 = 12 ml.

Volume of O2 present in 80 ml. of air containing 20% oxygen

=

Hence, the volume of gases at the end of the reaction will be:

Oxygen unused = 16 −12 = 4ml.

Nitrogen unused = 80 − 16 = 64 ml.

and carbon dioxide produced = 12 ml.

[NOTE: Volume of steam produced in equation(i) above will be the same as the volume of H2 used, but when it is condensed at room temperature to water, the volume occupied by it is negligible.]

 

  1. Since 34.45 is the average atomic weight of 35Cl (at. wt. = 35) and 37Cl (at. wt. = 37) we have,

= 35.45

where n1 and n2 are the number of moles of 35Cl and 37Cl respectively.

= 3.44; n1 : n2  = 3.44:1

∴ Ratio of lengths is 3.44:1

Since the pressures on both the sides of the partition is equal, the pressure before and after partition will be same (no.of moles per unit volume being same).

 

  1. Since Ammonia on sparking will decompose to form Nitrogen and Hydrogen only, the volume of 20 ml. after sparking consists of a mixture of Nitrogen and Hydrogen. Let the volume of Hydrogen be x ml., therefore, the volume of Nitrogen will be (20 − x) ml. When mixed with 30 ml. of oxygen and exploded, the hydrogen of the mixture reacted with oxygen and a contraction of (30 + 20−27.5) = 22.5 ml. took place. According to the equation.

2H2 + O2 2H2O

2 vols. 1 vol. zero vol. (after cooling)

or x ml. x/2 ml.

Thus, contraction in volume   =  

    ∴ x = 15

Thus, 10 ml. of ammonia, on decomposition, produced 15 ml. of hydrogen and 20−15=5 ml of Nitrogen. or 2 vols. of Ammonia on decomposition, produced 3 vols. of Hydrogen and 1 vol. of Nitrogen.

 

By Avogadro’s hypothesis:

 

2 molecules of ammonia on decomposition produce 3 molecules or 6 atoms of hydrogen and 1 molecule or 2 atoms or nitrogen.

Thus the formula of ammonia should be NH3.

V.D. of ammonia = 8.5

∴ Mol. wt. of ammonia = 8.5 × 2 = 17

Mol. wt. is also in accordance with the formula of ammonia – NH3 

 

  1. Eqn. For the line is ; 11p + 8V = 142

For (pV)max, 11p =

or then apply

(pV)max = nRTmax

 

  1. Volume of H2 left in the first bulb after diffusion = litres at N.T.P.

Volume of H2 diffused = 1.12 -0.56 = 0.56 litres.

Let the volume of D2 diffused be v2.

According to Graham’s law of diffusion, or (at. mass of D2 being 2)

hence, v2 = 0.396 litres at N.T.P.

Wt. of H2 in second bulb = 0.05 g, Wt. of D2 in the second bulb

= g.

Percentage of H2 = 41.67. Percentage of D2 = 100-41.67 = 58.33

  1. At 273°C, the wt. of A= = 48 g = 1  mole (being 50%)

Also at 273°C, the wt of A = 48 g = moles = 0.5 mole 

(since Mol. wt. of  A2 = 48 × 2 = 96)

Total No. of moles at 273° C = 1 +0.5 = 1.5 moles.

By gas equation PV = nRT, where n = No. of moles or

= 2atm 

 

  1. Volume of container = 0.731 ml

Temperature = 23 + 273 = 296 K

Initial pressure = = 1.74 mm …(1)

At  –750C, H2O is frozen out and the mixture on retaining 23°C has pressure P

P = = 1.32 mm …(2)

At –95°C, CO2 is also frozen out and the mixture on retaining 23°C has pressure

= 0.53 mm …(3)

By (2) and (3) = 1.32  0.53 = 0.79 mm

By (1) and (2) = 1.74  1.32 = 0.42 mm

Now using PV = nRT for each gas separately

For N2: = 2.1 × 10-8

For CO2: n = = 3.1× 10-8

For H2O v n =

= = 1.7 × 10–8 

 

  1. For the equilibrium Cl2 2Cl

Initial moles 1 0

At equilibrium   1-x   2x (x is the degree of dissociation) Molecular weight of the mixture of Cl2 and Cl at equilibrium , i.e., Mmix is calculated as 

Now we have, = 1.16 x = 0.14         

 

  Objective Problems

 

LEVEL – I

 

  1. Differentiate Charle’s Equation 

 

2.

 

5.

 

  1. Easily liquified gases have large Vander Waal’s attraction and have high value of ‘a’. NH3 is easily liquified and hence posses maximum Vander Waals constant ‘a’.

 

  1. Ethyne (C2H2) has molecular mass of 26 which H2 has molecular mass of 2. When dipped in H2 gas, H2 will diffuse faster into the balloon and hence ballloon will get enlarged. 

 

  1. Factual question

 

  1. = =

Mcycloalkene = 27 × 2 = 54

(B)

 

  1. Hint: Most probable velocity of gas increases with temperature but the fraction of total gas molecules decreases.

 

  1. Volume of 1 mole ideal gas at 27°C & 1 atm  =

= 24.6 Lt.

Volume occupied by simple molecule

= = 4.08 ×10–20 c.c.

r = = 2.35 ×10–7 cm

average distance = 2 × 2.35 ×10–7  = 4.7 ×10–7 cm

(D)

[Note: For solving objective problem, just see the order of distance. Don’t go for entire calculation or do it by reverse path follow with order of magnitude only.]

 

  1. Boyle’s constant = PV = nRT = × 0.082 ×300

= 12.3 Litre atm 

∴ (A)

 

  1. Let V1 be lts initial volume at 300K

Let V2 be the volume of container after heating 

Initial Temperature = 27°C = 300 K

Final temperature = 127 = 400 K

Since the vessel is open P = 1 atm 

∴ V1 = (V2)

Gases remaining

 

  1. Since volume is constant under same temperature and pressure, V ∝ n (the no. of moles).

Moles of methane ∝ volume

Moles of methane = Moles of oxygen

= 2 times that of methane

 

  1. Intensive property is a property which is independent of quantity of matter. Temperature is one such property. Hence ‘D’ is correct choice.

 

LEVEL – II

 

  1. Apply

 

  1. K.E. ∝ T

 

  1. Calculate Meff.
  2. Vrms
  3. where Ax & Ay are the area of cross section of orifice.

r = radius of a radius orifice

= length of square orifice

  1. For H2, r2 =

Where V is volume of H2

∴ V =

  1. a) For He V = = 5
  2. b) For O2 = V = 20
  3. c) For CO V =
  4. d) For O2 = V =

Since V is identical for X and O2, B is carried choice.

 

  1. Vo = Vrms =

=

= Vo 

(D)

  1. Partial pressure = Total pressure × Mole fraction

Mole fraction of oxygen = =

∴ MF of nitrogen = 1 – 0.7 = 0.3

 

  1. PV =

PV = P = 100 kPa

100 × V = V =

At constant T, P1 = 0.1 kPa

 

100 × V = 0.1 RT

0.01 × V = n1RT

n1 = 10–5

1 mole will contain 6.023 × 1023 molecules

∴ 10–5 mole will contain

6.023 × 1018 moles 

 

  1. Volume = 33.6 L

Temperature = 273 + 273 = 546 K

Molar mass of monomer = 48

Dimer = 96

Mole of monomer = 1

Mole of dimer = 1/2

PV = Nmix × R × T

P × 33.6 = 1 ½ × 0.0821 × 546

P = = 2.00 atm

 

17.

m = 79.9

(A)

 

  1. 22.4 L of H2 at STP contains 1 mole.

∴ 0.224 L of H2 at STP contain

= = 0.01 moles

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