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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 1023 298 ~ 6.17 1021 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)

32 103

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 127C 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

28 103

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

32 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

32 103

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

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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 1023 298 ~ 6.17 1021 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)

32 103

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 127C 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

28 103

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

32 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

32 103

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

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