Jee 2014 Booklet2 Hwt Solutions Gaseous State

Vidyamandir Classes

Solutions to Home Work Test/Chemistry Gaseous State 3.(A)

HWT - 1

1 atm = 1.01  105 N / m 2  760 mm of Hg  1.01  105 Pa Note : Pound is a unit of mass and not force. Pressure is force per unit area and not mass per unit area.)

4.(B)

Nitrogen is third most electronegative and hydrogen is highly electropositive and thus strong dipoles develop in

NH 3 molecule causing strong dipole-dipole interactions. Note that carbon is less electronegative as compared to nitrogen and hence the dipoles (positive – negative charge centers developed in CH 4 molecule are not as strong as they are in NH 3 and in CO2 the dipoles developed will be even weaker as both oxygen and carbon are electronegative atoms and hence the electronegativity differences will be less compared to NH 3 and CH 4 .

5.(C)

3.36 lt O2 (at S.T.P.) 

 6.(B)

7.(D)

3.36 mol O2  0.15 O2 22.4

0.15 mol H 2 required  0.15  2  0.3g H2 required.

(M 0 )gas  2  V.D  2  11.2  22.4 11.2g gas 

11.2  0.5 mol gas  11.2lt (at STP.) 22.4

V1 T  1 V2 T2

T  313  V2   2  V1   128  136.7 ml T 293  1

(At P, n  const.) 8.(C)

Use

9.(D)

r

~ 137 ml

P1 T  1 (at n, v  const.) P2 T2

 T2  40  2  4 atm  P1  T 20  1

 P2  

Pgas M gas

Note : Pgas is the pressure of the gas in that container from which gas is effusing out. In this question since nothing is mentioned about the pressure we will assume pressure of both gases to be the same.

rgas



10.(B)

rH 2 rO2

rO2 



M O2 M H2

M O2 M gas



32 1 1   4 2 128



nH2 / t



g H 2 /(M 0 ) H 2 32  4 gO2 /(M 0 )O2 2

n O2 / t



32 2 

gH2 / 2 4 / 32

Gaseous State



g H 2  1g

HWT - 2

5  V1 PV T 819 3 P2  1 1  2    45 atm T1 V2 273 1 ( V1 / 3)

1.(C)

PV PV Use 1 1  2 2  T1 T2

4.(C)

(n gases ) total  n H 2  n He  n NO  

4

1 2 3    0.5  0.5  0.1  1.1 2 4 30 nRT 1.1  0.08  300 P  ~ 24atm V 1.1

VMC/Gaseous State

12

HWT-Solutions/Chemistry

Vidyamandir Classes 6.(A) 7.(D)

Total K.E =

C

3 nRT 2

T M





u1 



T and u 2  m1

u1 m1  u 2 m2

rH 2 

50ml  20 min

rO2 

40ml  t min

T m2

M H2 PO2



M O2



9.(D)

Cavg 

10.(B)

K.Eavg per molecule 



Cavg 

50 / 20 32  40 / t 2 50t 4 40  20 t = 64 min T M



Check

T M

3 3 KT   1.38  1023  298 ~ 6.17  1021 J 2 2

Gaseous State 3  8.314  300

3.(C)

 Crms O2



4.(D)

Crms 

12  22  22  3

HWT - 3

(Note : M0 has to be in kg/mol and not g/mol)

32  103

9  3  1.73 m / s 3

Gaseous State 2.(A)

Open container



n1T1  n 2 T2





1600  533.33 K 3

(PH 2  PO2  similar conditions)

PH 2



RT M

THe 

m1u12  m2 u 22



8.(B)



0.3  THe  0.4  400

HWT - 4

P and V are const. (Think why ?)

nT n  300 3 n2  1 1  1  n1 T2 400 4

3 th of the gas is present inside the container at 127C 4

3.(C)

Pdry air  dry air  Ptotal  0.98  1.2  1.176 atm

4.(A)

T  Molecular attractions decrease and mean free path increases on increasing temp     . Though Rate of collisions among particles P 

increases on increasing temp but Pressure in a container is not due to collisions of particles amongst themselves but it is due to collision of particles with the walls of the container (so option B is also incorrect). 9.(D)

K.E is only function of temp. Translational K.E. (per molecule) 

VMC/Gaseous State

3 KT 2

13

HWT-Solutions/Chemistry

Vidyamandir Classes

Gaseous State 1.(C)

HWT - 5

(Crms ) N 2  7 (Crms ) N 2



3R TH 2



2 10

3



TH 2



TH 2  TN 2

1

 7

T  7  H2 284

3R TN 2

28  103

 TH 2 

TN 2 2

2  0.08  540  1.97 ~ 2atm 44.8

2.(B)

P

8.(C)

1  M 0  d  R  300 , Now 1  M 0  0.75d  R  T





d  300  0.75d  T

T

300  400K 0.75

Gaseous State 1.(C)

2.(A)

PM 0  dRT  d  rgas rHe

HWT - 6

PM 0 2  16  ~ 1.3 g / lt RT 0.08  300

M He (assu min g Pgas  PHe ) M gas



Now rgas 

1 1 rHe   5 5

3.(D)

P1V1 P2 V2  T1 T2

4.(B)

n CH 4 

4 M gas

PV   V2   1 1    T1 



M gas  25  4  100

T2 PV 2T1  1 1  = 4V1 P2 T1 P1 / 2

PV 16  9  6 RT 0.08  300 gCH 4  n CH 4  (M 0 )CH 4 = 6  16  96g



Gaseous State

2.(B)

PCH 4 PH 2 



n CH 4 nH2



gCH 4 /(M 0 )CH 4

PH 2

g H 2 /(M 0 ) H 2

PH 2  PCH 4

VMC/Gaseous State





HWT - 7

2 1  16 8

8 9

14

HWT-Solutions/Chemistry

Vidyamandir Classes

Gaseous State 2.(A)

T for Cavg to be same M

Cavg 

TCH 4



M CH 4

TO2

32 1 1   n O2   6  1023 = 1.5  1023 0.08  300 4 4

n O2 

7.(D)

324g A 

rgas rN 2O

324 12  3  12mol A  mol Cl 2  18 mol Cl 2  18  22.4 lt of Cl 2 at S.T.P = 403.2 lt 27 2

M N 2O



M gas

(assu min g PN 2O  Pgas )



Ax / t 44 1    A(   x) / t 176 2



2x    x 13  4.33 3

x

9.(B)

300  16  150 K 32

 TCH 4 

MO2

5.(B)

8.(A)

HWT - 8

x 1  x 2



3x    13



  x  13  4.33  8.66



9th row from front.

M eff  2  V.D  2  14.4  28 .8   N 2 M N 2  O2  M 0   N 2  28  (1   N 2 )  32



28.8  32  4  N 2 

N2 

3.2  0.8  %N 2   N 2  100  80% 4

Gaseous State 3.(B)

PM 0  dRT  M 0 

7.(C)

 Crms O2

8.(C)

1 V  P 



1.15  0.08  373  44 0.8

3  8.3  300

3RT  M

HWT - 9

32  103

 483.5 ~ 484 m / s

V  P  4 atm 4

Gaseous State 3.(C)

 Crms O3  Crms O2



M O2 M O3



32  48

HWT - 10

2 3

7.(B)

As pressure increases, the tendency of liquification of a gas increases (or in other words the tendency of vapourisation of the gas decreases) Boiling point of the liquid (corresponding to the gas) increases. This is what happens in pressure cooker. 

10.(D)

d  0.001293 g / ml = 1.293g / lt .

Now (M 0 )air 

d  R T P

 1.293  224  28.96

V.D. 

VMC/Gaseous State

(M 0 )air 28.96   14.48 2 2

15

HWT-Solutions/Chemistry