Heat (Magnet) 1

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Section A : Straight Objective Type

1. Answer (1)

312

VP = C

61

PV = C = 6

1C = C

v

1R

= Cv

16

1 R

= Cv

6

5R

C > Cv

2. Answer (3)

As work done by gas is 0 and box is insulated, so

U = 0 T = 0Also, PV = constant

4

PVP V

4

PP

3. Answer (3)

2

3

2

3

rot

trans nRT

nRT

K

K.

4. Answer (1)

Molecules are continuously colliding with one another.

Root mean square speed increases with temperature increase.

5. Answer (2)

PT4 = constant (k)

5

nRTk

V

5

nRTV

k

4 55 5 5V nRT nRT V

T k kT T

...(i)

Heat UNIT 2

210 Heat Success Magnet-Solutions (Part-I)


andV

V V T VT

...(ii)From (i) and (ii),

5

T

6. Answer (1)

PV = nRT

mPV RT

M

mT = constant 8(273 + 7) = m(273 + 47) 8 280 28 8 7 g

320 32m

Gas escaped = 8 7 = 1 g

7. Answer (1)

constant3

1

PV

3

1

.VV

nRT = C

32

TV = C

T 32VT 3 V 2. As volume is increasing, therefore temperature is increasing.

8. Answer (3)

P = m.V

or PV1 = m. By comparison with PV = constant, = 1

Now, C = RRRR

CR

C Vv 222

3

111 .

9. Answer (2)

Wien's displacement law,

m

T = constant

00

33

TT T T

P T4

So, 5

3 243P T

PT

10. Answer (1)

Q = U + Wfor isobaric process

nCpT = nC

vT + nRT

Cp = C

v + R

Cp C

v = R.

211Success Magnet-Solutions (Part-I) Heat


10a. Answer (2, 4) (IIT-JEE 2009)

Diatomic

7 5( ) 6

2 2P V

R RC C R

Monoatomic

5 3( ) 4

2 2P V

R RC C R

2

Diatomic

35( . )

4P V

RC C

2

Monoatomic

15( . )

4P V

RC C p

v

C

C = = 21

f is smaller for diatomic gas

Cp C

v = R = constant

11. Answer (1)

AB is an isothermal process. So, its line parallel toy-axis and BC is an isobaric process. So, V-T graph is a

straight line passing through origin.

12. Answer (3)

In a cyclic process,

U = 0 Q

Cycle = W

Cycle

WCA

= 0 (Isochoric process)

WA B = PV = nRT

= 1 8.31 (450 350)

= 831

Now, 1200 = 831 + WBC

+ 0

WBC

= 2031 J

13. Answer (3)

Q = U + W

23 3

Q QQ U U

Now, U = NCVT

2 53 2

Q RN T

15

4Q NR T NC T

15

4

RC

14. Answer (3)

At a given temperature, energy of A is more than that of B.

So degrees of freedom for A should be more than B.



15. Answer (2)

100

P

P

100x

x

Area of cylinder = = 76 100

A

PV A

P P

x

V x A

= + pressure due to Hg = (76 + )

= (100 )

atm

PV = constant for isothermal process76 100 A = (76 + x) (100 x) A

7600 = 7600 + 100 x 76x x2

x = 24, x = 0

x = 0, 24.

16. Answer (2)

Q = U + W [From 1st law]Here, U = nC

VT

and0 0

6

A

P VT

R

and0 0

30

B

P VT

R

So,0 0

0 0

24336

2

P VU R P V

R

0 0 0 0

1(3 )(3 ) (3 )(3 )

2W P V V P

0 00 0 992

P VP V

0 0272P V

0 0 0 0 0 0

27 9936

2 2Q P V P V P V nC T

0 00 0 24992

P VP V C

R

99 3348 16

R RC

17. Answer (4)

M

Vmp

1

18. Answer (3)

M

RTP



RT

MP. top

= 280

70

R

M

bottom

= 300

76

R

M

99.0

bottom

top

19. Answer (3)

Let the amount of ice be x.

Net heat gained = 336x + (x 50 4.2)

+ (200 x) 4.2 50

+ 100 0.42 50

= 336x + 44100

Net heat lost = (30 4.2 50) + (30 2250)

= 73800 J

Heat gained = Heat lost

336x + 44100 = 73800 x = 88.39

20. Answer (4)

Thermal expansion takes place such that distance between any two points increases.

21. Answer (2)

PV = nRT

VT P

nR

On T-P curve, a process with V = constant is a straight line passing through origin and slope V.

T

P

A

B

B

A

As tan Vtan

A V

A

tan B V

B

VB < V

A and volume decreases continuously

22. Answer (1)

Let the water rises upto a height of H. A

B

( )L H0

L0

HPA = P

B...(i)

P0L

0 = P

A (L

0 H) ...(ii)

PB = P

0 + g (L

0 H) ...(iii)



From equation (ii) )( 0

00

HL

LPPA

)()(

00

0

00 HLgPHL

LP

)(1)(

0

0

0

0 HLgHL

LP

P0H = g (L

0 H)2

105 H = 103 10 (20 102 H)2 H = 0.38 cm

23. Answer (4)

VLiquid

= VVessel

VVessel

= V 3Glass

TV

Liquid = V

Liquid 16

Glass T

So, Liquid

3

16

VV

24. Answer (3)

Q1 = n C

P T

= 1 2

5R (2T T) =

2

5 RT

TTT

V

T

V22

2

2

1

1

Q2 = W = nR (2T) n

2

1

P

P2

2

1 P

P

= 2 RT n 2

QTotal

= (2.5 + 2 n 2) RT

25. Answer (1)

Apply

2

2

1

1

T

P

T

P ( volume is constant) 1

12

2P

TPT

26. Answer (1)

dQ = dU + dW

CT = CvT + W.

12 TTT R

VP

R

VP 1122

R

VP

R

VP

R

VP 000000 5

6

W = Area 00000 32)2(2

1VPVPP



5

33

5500

0000 RCCVP

R

VPC

R

VPC VV

10

21RC

RCV

2

3

27. Answer (2)

dQ = 3dU RCC V2

93

C = 4.5 R28. Answer (1)

Q = U + W.

2

1

22

2

0

2

1

2

2

V

V

RTVVCPdVW ( P

1V

1 = RT

0 to P

2V

2 = 2RT

0)

22

3 00

RTRTQ

Q = 2RT0

29. Answer (3)

PV = RT

2

0( )

RP T V

V

0

20 0

RTdPR

dV V 0TV

2

2 0

0 0 02

TT T V T T

Minimum value of pressure is 0

min 0

0

(2 )2

R TRTP R T

V T

30. Answer (4)

As PV = nRT

U = 2 + 2nRT U = 2nRT C

V = 2R

Since CV = 2R for system,

RCRV

2

5

2

3 So it is mixture of monoatomic and diatomic gas



31. Answer (2)

As 2

2

1

2

V

VP

12

21

P and 5

82 P

P2V

2 P

1V

1 = nRT

KR

T5

11

32. Answer (3)

Net heat gain = 5 4.2 + 0.7

= 0.8 + 0.7

= 1.5 kW

= 1500 W

Now, C = mS = 1500

d

dt

= 1500

0.5 = 3000 J/C

33. Answer (1)

PV mnR R

T M

34. Answer (1)

Work-Energy theorem

Watm

+ Wgas

+ Wext

= K.E. = 0W

ext = W

atm W

gas

0

0

atm 0 0 0( 1)

nV

V

W P dV PV n

and

0

0

gasln( )

nV

V

W PdV RT n

35. Answer (2)

W = Area of rectangle = Area of triangle

W = (2V0 P

0) +

2

1(2V

0 2P

0)

= 4 P0V

0

36. Answer (4)

As initial and final temperatures are same, U = 037. Answer (2)

Pv

1200

H

P ...(1)Let there be x volume of H

2 and y volume of O

2.

mix H O H( ) 16

Hx y x y x y



Hmix

16

( )

x y

x y

H

( )500

( 16 )

P x y

x y

...(2)(1) (2)

2.15xy

38. Answer (4)

PV nRT2

0( )P aV V nRT On differentiation, we get

2

0( 3 )

dTP aV nR

dV

2

20 &

dT d Tve

dV dV

2

00 3

dTaV P

dV or 0

3

Pv

a

0 0

max

2

3 3

P PT

R a

39. Answer (4)

= 6.6 cmAs v = f v = 5 103 6.6 102 m/s

Aslo, M

RTv

and 1

RC

v

40. Answer (1)

300PV nRT Rn

300

PVn

R

1 2,

280 350

P V P Vn n

R R

V V

1 22n n n

2 1 1

300 280 350

PP

5 4

1400P

28'

27P P



41. Answer (2)

For polytropic process, we have

32

1 2 1 ( 1)p v

R R RC C R

x

42. Answer (1)

dT

PdVCC V

PV = nRT

2e 2VV = nRT

)21(2 2 Ve

nR

dT

dV

V

V

nR

dT

PdV

21

(1 2 )V

nRC C

V n

V

RCC V

2143. Answer (1)

T = 300 + 2V

R

PV = 300 + 2V

RV

RP 2

300

Work done

4

2

2

1

2300

V

V

dVRV

RW

W = 300 Rn2 + 4R44. Answer (3)

VP

1200

Calculate P1V

1 and P

2V

2 where V

1 = 2m3

and then nR T = P2V

2 P

1V

1 = 400

So U = 3 400

1

usingTnR

U

= 1200 J

45. Answer (3)

2P T . But from ideal gas equation PV T

2PV P

P

V = constant

Hence, as temperature increases, P also increases.

Density decreases.



46. Answer (2)

U = V 2 TnR

TnCUV

1 3RT = V 23RdT = 2VdV

Now, dT

dVPCC V

V

R

V

RTRC

2

33

2

2

2

33

V

TRRC

RT

TRRC

32

33

2

23

RR

C = 3.5 R

47. Answer (4)

Q = U + W W = 2JnRT = 2J

U = nCVT

= )(2

5TnR

= 5 J

Q = 7 J48. Answer (1)

P

V

2P0

P0

V0 2V0

AC

B

2

3)3(

2

1 0000

VPVPWAB

0 0 0 0

0 0

(4 )3

1

PV PVU PV

00002

93

2

3VPVPQ

AB

002

1VPW

net

9

1AB

net

Q

W



49. Answer (2)

TV

1

PVV

Constant

PV2 = Constant n = 2

n

RCC V

1 C = C

V R C = 2R

Q = CT= 200 R

50. Answer (1)

Q1 : Q

2 : Q

3

900 : T : 400

900

Q1

Q2

T400 K

Q3

W W

Q2 Q

1 = Q

3 Q

2

T 900 = 400 T

2T = 1300

T = 650 K

51. Answer (3)

W = 25 = PV = RT ...(i)

3

4

3

11

6

21

21

f

RR

CV

31

U = C

VT = 3RT ...(ii)

or U = 3(RT) = 3 25 = 75Q = U + W = 75 + 25 = 100 J

52. Answer (2)

The heat do we need to remove from the water is

Q = mC(100C 65C)= 200 g 4.18 J/gC 35C

= 29260 J

The heat in taking a mass m of be cubes to 65C is

Q = mL + mC(65C 0C)= m(L + CT)

m = 29260

(333 4.18 65)

= 29260

604.7

= 48.4 g

which means 48 1 g ice cubes.



53. Answer (1)

Ti = 290 K, V

i = ?, T

f = 703 K, V

f = ?, = 1.40

TiVi

1 = T

fVf

1

fi

V

V=

1

1i

f

T

T

1 1 12.5

1 1.40 1 0.40

=

2.5

2.5

290

703

= 0.109

110

Compression ratio of 10 : 1 is needed.54. Answer (2)

1

21

T

T since 1

2

T

T

is same so will be same.55. Answer (4)

1001

1

2

1

T

T

56. Answer (2)

PT = Constant

P(PV) = K

P 2V = K

KPV 21

2

11

RCCV

37.35 = CV + 2R

2

fR = 37.35 2R

f = 5

57. Answer (1)

W = nRT0 ln

i

f

V

V

W = 3R 300 ln

2

1

W = 5188 J



58. Answer (2)

2 2 41 1

6 3f

W

Q

=

Q U

Q

= 1v

p

nC T

nC T

= 3

14

= 1

4

Q = 4W= 100 J

59. Answer (1)

1T

1 =

2T

2

1 10

2C C

T T

= 20

60. Answer (1)

Isochoric process P T

So,P T

P T

1 1100 T

T = 100 K61. Answer (3)

H

R= k (

R

0)

H 20 = k (20 + 20)

H 10 = k (10 + 40)

or5

4

10

20

H

H

or 5H 100 = 4

H 40

H = 60

62. Answer (1)

i = dx

dTkA

Sokdx

dT 1

or K1 > K

2



63. Answer (1)

Energy received by earth per second is 2

2

2

sun

4

sun

4

R4)T(e

Rd

Energy radiated by earth per second is ]4T[ 24

earth eR . Equating the two values, we get Tearth = 290 K64. Answer (2)

P = T4A (A = dl)

100 = 6.01087

227.01067.5

548 T

or T 4 = 6.08227.067.5

10710013

T = 2018 K

65. Answer (3)

0( )

dk

dt

0

10 0

( )

td

kdt

66. Answer (4)

Since the gas is ideal and there is one mole P0V

0 = RT

When V = V0, P =

0

2

P

Now, 0

02

PV RT

0

0

2RTV

P

0 02

P VT

R

67. Answer (2)

32

1 2

1 1TT

T T

T2

2 = T1T

3

T2 = 327C

68. Answer (1)

E1 =

1T

1

4(4r1

2)

E2 =

2T

2

4(4r2

2)

So

2

2

1

4

2

1

2

1

2

1

r

r

T

T

E

E



69. Answer (4)

P = T 4 (r2)450 = T 4 r2

P =

2

4

2)2(

rT

4

1450 P

P = 1800 W

69(a). Answer (9) IIT-JEE 2010

As per Weins displacement law

3A B

B A

T

T

As per Stefans law

Also,

4 2

4

4 2

(6)(3) 9

(18)

A A A

B B B

P A T

P A T

70. Answer (3)

5 10.052 10 C

25 100A

L

L T

5 10.04

10 C40 100

B

L

L T

1 +

2 = 50 cm ... (i)

1 250.03 cm

A

1T +

B

2T = 0.03

2 105 1 50 + 105

2 50 = 0.03 cm

21 +

2 = 60 cm ... (ii)

From (i) and (ii) 1 = 10 cm

71. Answer (2)

k1x

1 = k

2x

2 and x

1 + x

2 = L T

x1 = 3x

2

x1 =

)(

)(

21

221

kk

kxx

= )3(

3

KK

KTL

= 4

3 TL

72. Answer (4)

1 =

2 [1 + (T

2 T

1)]

v2 = v

1[1 + (T

2 T

1)]

The loss in weight at T1 = v

1

1g w

0 w

1 = v

1

1g

The loss in weight at T2 = v

2

2g w

0 w

2 = v

2

2g

0 1 2 1

0 2 2 1

1 ( )

1 ( )

w w T T

w w T T



73. Answer (1)

50g + 2g = V1g

50g + xg = V2g

where 1 is density of water at 10C and

2 is density of water at 30C

x50

52 =

2

1

= )1(

1

T 52 52 104 20 = 50 + x

(2 x) = 52 104 20x = 0.104 kg

74. Answer (1)

30 = (50 20) [where is constant)

mcdt

d = (40 20)

mcdt

d =

30

3020 2010

1.0 mcHeat capacity = mc = 2000 J/C

75. Answer (4)

2

1

A

S

l

l

21

1 1A

S

l

l

1 21

S A

S

l l

l

11 2

S

A S

l

l l

76. Answer (2)

4Q = nCPT

3Q = nCVT

3

4 f = 6

77. Answer (1)

( 3 ) ( 3)C C

kA T T kA T T

A B

C

T3( ) ( )T 1 32

CT

T

78. Answer (2)

Work done for isobaric process is nRTR (T

0 T

1) + R (T

0 T

2) = W

T0 =

R

TTRW

2

)(21



78(a). Answer (1, 2) IIT-JEE 2010

AB is an isothermal process

Internal energy at A = Internal energy at B.

Applying PV

T= constant between A and B

V0

4V0

A

B

V

T0

TC

0 0 0 0

0 0

4

4 B

B

PV P V PP

T T

As it is not clearly indicated that BC is passing through origin, prediction about point C is not possible.

79. Answer (2)

P = T 4Aso

2

2

4

2

2

1

4

144 rTrT

41

4

2

2

2

2

1

T

T

r

r 2

1

2

2

1

T

T

r

r

79(a). Answer (9) IIT-JEE 2010

As per Weins displacement law

3A B

B A

T

T

As per Stefans law

Also,

4 2

4

4 2

(6)(3) 9

(18)

A A A

B B B

P A T

P A T

80. Answer (2)

012122

kt

1 = 50

2 = 40

0 = 20 t = 5

)2045(5

10 k

202

40

5

40k

225

40

10

10 = 2000 5060 = 2000

C3

100



81. Answer (2)

PV = nRT

T = PV PMnR R

= 2.4 107 K

82. Answer (3)

Rate of cooling is proportional to V

A or

r

1

V1

= 2V2

3

2

3

13

42

3

4rr

312

12

r

r

31

1

2

2

1

of cooling of Rate

of cooling of Rate

r

r

Q

P

31

2

1

83. Answer (4)

Solar constant 21

r

Where r is distance from sun

r = sun of diameter DD

D Sun

Planet

So Solar constant 284. Answer (1)

T = kzjyix 222

)2,1,1()( T = ji 22

85. Answer (3)

Req

= R1 + 2R

2

= AKKAA

30

)001.0(

)01.0(210

4.333

10010010

eq

CuRAK

T

86. Answer (1)

P 2T = constant

3

11

2

5

RRC

P 3V = C C = 4R

3

1x Q = 4P0V

0



87. Answer (2)

88. Answer (1)

constant Tk

(Weidmann-Franz law)

89. Answer (3)

90. Answer (2)

91. Answer (2)

Q2 = mcT + mL

1 4.2 103 27 + 1 3.36 105 = 4.49 105

Q1 =

5 3004.49 10

260

Energy supplied = Q1 Q

2 = 6.9 104 J

92. Answer (4)

93. Answer (2)

and

SNO

QPR

TTT

TTT

A

P S

B

OR

N

Q

So A B

( , , )P Q R ( , , )O N S

R/3 R/3R/6

6

5

363

RRRRReq

94. Answer (1)

95. Answer (2)

For the process A B1 u

3 1

2 /RTM v

or T vSo AB is an isobaric process.

Now 0

32

2

Au RT 04

3

A

uT

R

and

0

AM

V

So 0 0

4

3

AA

A

uRTp

V M and

02

BM

V



Now [ ] AB B A

W P V V

0 00 0

4

3 2

u M M

M

0

2

3 u

96. Answer (3)

x1 + x

2 + x

3 = LT + 3

2 2

LT L T

Kx1

2 Kx2

2 Kx2 3 Kx3

x1 = 2x

2 = 3x

3

x1 +

313

2 3 2

xxL T

1 91

x L T 2 2 2 2 2 2

1

1 81 81

2 121 242E KL T KL T

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Heat (Magnet) 1