6. HEAT & THERMODYNAMICS-SOLUTIONS-PART TEST-1
SOLUTIONS TO PART TEST-01 (PT-01) TOPIC : HEAT & THERMODYNAMICS (PHYSICS)
1. In an ideas gas internal energy =
f nRT
2
U = 5 × 2RT + 4 ×
2
3 RT = 11 RT.
2
2. Higher is the temperature greater is the most probable velocity.
3. According to law of equipartition of energy, energy equally distributes among its degree of freedom, Let translational and rotational degree of freedom be f1 and f2.
KT 3
and K + K = U
KR 2 T R
Hence the ratio of translational to rotational degrees of freedom is 3:2. Since translational degrees of freedom is 3, the rotational degrees of freedom must be 2.
Internal energy (U) = 1 × (f
+ f2) ×
1 RT
2
1 5 8.3 100
U = 2
4. PV = n RT
= U = 2075 J
12 4 8.31 300
nRT
P = V
= 4 2
20 103
= 6.25 × 105 Pa
5. Total KE = U = 2 nRT
In case of H2 degree of freedom is greatest and number of moles n is highest So this is the case of maximum kinetic energy.
6. P = P
T = T
V1 V2
2
=
1 2 1 2
n1 n2
n1 n2
M1(2 – ) = M2
or =
2M1
M1 M2 =
16
15
7. When speed of 5 molecules which are selected randomly are recorded, then the average is most likely to be equal to the most probable speed.
The average of these values is most likely equal to .
8. TF 32
180
= TK 273
100
32
= 180
273
= 100
5 – 32 × 5 = 9 – 273 × 9
= = 574.25
9. For minimum value of m, the final temperature of the mixture must be 0°C.
20 ×
1 × 10 + 20 × 80 = m 540 + m.1. 100
2
m =
1700
640 =
85
32 gm
10. Applying energy conservation : mgh = mLf
Lf
h = =
80cal / gm
=
80 4.2 1000
J/kg
336 103 N s2
=
= 33.6 km.
g 10m/ s2
10 m/s2
10
kg
11. Work done by atmosphere = P
atm
V = P
V (i)
atm 2
Initially gas in container is in thermodynamic equilibrium with its surroundings.
Pressure inside cylinder = Patm
& PV = nRT PatmV = nRT or V =
nRT Patm
Putting in (1), W =
12. As p V2 T V3
nRT 2
i.e. if temperature increases, volume also increases hence work done will be positive.
13. For process at constant pressure
5 2
Q = nCp T = 2 nR T and W= PV = nRT = 5 Q
14. Net power given to N2 gas = 100 – 30 = 70 cal/s The nitrogen gas expands isobarically.
Q = n Cp
dT
dt or 70 = 5 x
7 R dT
2 dt
dT
dt = 2k/sec
15. As PV = nRT
P V
For n = constant :
T
= constant
for all changes. Hence (C)
16. By symmetry
AB BC AD DC
No current in BO and OD
T = T = T
dQ
17. dt
= – KA (1 – 2) [Newton’s law of cooling]
Rate of loss of heat is same
from
dQ = ms d
d = KA ( – )
dt dt
dt ms s
Since, ‘m’ is larger for sphere , rate of cooling will be less for sphere.
18. The colour of an object indicates the rate at which energy is emitted and hence indicates the temperature.
19. Heat radiated (at temp same temp) A
Q 4R2 and Q' (4R2 + 2 × R2)
Q' 6R2
Q 4R2
= 1.5
Here R2 is extra surface area of plane surface of one of the hemisphere.
20. assuming rate of heat transfer to be constant.
s (m + w) T = Q where s = specific heat of water = 1 g/cc s (m + w)T = Q w = water equivalent of calorimeter
Q1
Q2
(50 w)s
= (100 w) s
50 w
= 100 w
Q1 t1 10 50 w
but Q2
= t2 = 18 = 100 w
50 = 4w w = 12.5 g
21. Thermal resistance R = 𝑙
KA
for same temperature difference, thermal current 1
R
i1 = R2 =
i2 R1
𝑙 2
K 2 A 2 /
𝑙1
K1A1
𝑙 2
= 𝑙1
A1 K1
A 2 K 2
1 12 1
= 2 x 22 x 1 = 1/8
22. Since elasticity of balloon is negligible, pressure inside ballon –~ pressure outside baloon = Patm.
W = P V Vin = 10 litre.
Vin
Vfinal
VinTfinal
Tfinal
58
= V =
litre. W = P V
1
105 × 10–2
= 200 J
Tin
Tfinal
final
Tin
atm
in
Tin
290
23. U in a – b process = 0 (isothermal process) in b – c process Q = 0 (adiabatic process)
U = – W = – 4 J
U in cyclic process = 0
U in c – a process = 4J
24. Let initial temperature and volume be T0 and V0. Since the process is adiabatic, the final temperature and
volume is TV –1 = T V –1 ( =
5
for monoatomic gas)
0 0 3
2
V0 3
T = T0 V / 8
= 4T0
0
percentage increase in temperature of gas is 4 × 100 – 100 = 300%
25. Let initial pressure, volume, temperature be P0, V0, T0 indicated by state A in P-V diagram. The gas is then isochorically taken to state B (2P0, V0, 2T0) and then taken from state B to state C (2P0, 2V0, 4T0)
isobarically.
Total heat absorbed by 1 mole of gas
Q = C (2T – T ) + C (4T – 2T )
= 5 R T +
2 0
7 R × 2T =
2 0
19 RT
2 0
Total change in temperature from state A to C is
T = 3T
Molar heat capacity =
Q
T =
19 RT0
2
3T0
19
= 6 R.
1/
V 1/ 2
1
T2 P2
2rms T2
V2rms
P2 2
26.
T P
and V
T
V P
1 1
1rms
1
1rms 1
1.41
= 221.4
= 21/7
27. Vr.m.s. =
28. (B)
= 3.5 .
PV = nRT P(aT – b) = nRT P =
nRT aT – b
nR
= b
a – T
a – b
now as T increases then
increases hence P decreases
29. At equilibrium :
dQ1 =
dt
dQ2 dt
KA(100 )
t =
4nk An( 0) 2n2t
= 100/3.
30. Moles in ‘A’ initially & finally will be same
PA (L / 3) P Ax
.... (1)
RT RT
Moles of ‘B’ remain same
2PA (L / 3) PA (L 2x)
.... (2)
RT
Divide (1) by (2)
R T
1 x
x = L/4
2 L 2x
SOLUTIONS TO PART TEST-02 (PT-02) TOPIC : HEAT & THERMODYNAMICS (PHYSICS)
1. External Work
= work done on both gases
= – (work done by the gases)
4 / 3 V0 2 / 3 V0 8 9
= – P0 V0𝑙n
V0
P0 V0𝑙n
0
= – P0V0𝑙n 9
= P0V0𝑙n 8
2. B = –
p
dv
v
dv = – pv vT = pv T = p
3. K1
0°C
70 0
B
70°C
100 70
B B
K2
100°C
R2 K1
R1 = R2
R1
= 3/7
K 2 = 3/7
0°C K2
T K1
100°C
T 0
R2
100 T
= R1
100 T = R1
T R2
K 2
= K1
= 7/3
300 – 3 T = 7T T = 30°C
4. n = 7v – v2 – 10
for most probable velocity dn 0
dv
n is maximum at this velocity
7–2v = 0 v =
5. As dQ = msdT
7 = 3.5 m/sec
2
dQ ms dT
dt dt
From question : S T or S = K1T. (K1 being proportionality constant)
d
Also, dt
dT
= constant = K2 (say) ms dt
K1 T2
dT
= K2 m(K1T) dt
= K2
m
K 2 2
= t T
Hence, the graph will be parabolic
6.
AC = 3V0
AB = 3V0 cos 37º in triangle ABN AN = 3V cos2 37º
P0
P = V0 V
P0
PB = V0
(V0
+ 3V0
cos2 37º)
P0 V
3V
16
PB =
0
V0
0
25
1 48
=
25
= P0(73/25)
7. Wein's displacement law is : .T = b
1
i.e. T m
Here, becomes half.
Temperature doubles.
Also E = T4
E T 4
1 1
E2 T2
T 4
E =
2
T1
.E1
= (2)4 . 16
= 256 J m–2 s–1
8. dQ KAT
T 10
= = J /sec.
dt 2𝑙
2𝑙 120
KA
dQ
New rate dt
T
= 𝑙
2KA
20
= t J /sec. ;
20
So time taken is t = 40
× 120 sec. = 60 sec.
9. PV =
m RT M0
RT
P = M0
P R T
M0
Slope of the curve Temperature
Hence cd and ab are isothermal processes.
1
V
i.e. bc and da are constant volume process
(A) and (B) are true.
Temp. in cd process is greater than ab.
Net work done by the gas in the cycle is negative, as is clear by the PV-diagram.
10. Since , e = a = 0.2 (Since, a = (1 – r – t) = 0.2 for the body B) E = (100) (0.2) = 20 W/m2
Power emitted = E.A = 20 × 10 = 200 Watt
11. Heat obviously flows from higher temperature to lower temperature in steady state. A is true.
Temperature gradient
1 in steady state. B is false.
cross sec tion area
Thermal current through each cross section area is same. C is true.
Temperature decreases along the length of the rod from higher temperature end to lower temperature end. D is false.
12. no. of collisions per unit area
1
time between two collision area
Vrms
n• /A
Vr m s
distance b/w walls A
n• T1/ 2
n• P
distance b/w walls × A = volume So ,
A V A
•
if both T and V are halved, increases
A
T1/ 2
13. An isothermal process is an example to show the possibility "for both the pressure and volume of a monoatomic ideal gas to change simultaneously without causing the internal energy of the gas to change". Hence Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
14. For monoatomic gas the degrees of freedom are only translational and hence do not depend on temperature. Therefore molar heat capacity of an ideal monoatomic gas at constant volume is a constant at all temperatures.
15. Heat energy stored in 1 g of steam at 100°C is more in comparison to 1 g of water at 100°C. Hence burns caused by steam are more severe.
16. v
rms =
, if the absolute temperature T doubles, v
rms
must increase by
(m and k are constants).
17. In 40 min. temperture of water has come down by 40°C.
mS T
Therefore rate P = t
0.60 4200 40
= 40 60
= 42.0 W
18. Sample of ice has been receiving heat at constant rate P from water. Its temperature has increased by 30°C in time 60 min.
Therefore mi si Ti
P
= 60 min.
mi =
Alternate :
(60 60s) (42 W) (2100J / kg).(30C)
= 2.4 kg
mwSwTw + m'wL = miSi 30° (m'w = mass of water which is converted to ice)
19. Thermal equilibrium reaches after 60 min. Ice conversion takes place for 20 min. During this time water at 0°C continues to give heat at rate P.
m'w × Lf = P × (20 × 60s) m'w =
42 20 60
3.3 105
kg = 0.15 kg
3 V2 2 V1
20. W = Area under the curve = 2
P1V1
P2 2P1
and P1V1 = nRT1
3 . P V
W
Therefore nRT1 =
2 1 1
P V
1 1
21. Q = dU + W dU = nCv dT
For final state P2V2 = 2P1 2V1 = 4P1V1 = nR(4T1) Hence final temp. is 4T1
dU = n .
3 R . 3T =
2 1
9 nRT
2 1
Q = 3 . nRT +
2 1
9 nRT
2 1
= 6nRT1
Q
nRT1 = 6
22. nC T = Q nCT = 6n RT1 dT = 4T1 – T1 = 3T1
n . C . 3T1 = 6nRT1
C = 2
R
23. In (A), V is on vertical axis.
As V is icreasing, W is positive.
As negative work in part-II is greater than positive work in part-I, net work during the process is negative.
Using PV = nRT and as Vremains same for initial and final points of the process, it is obvious that final temp. is greater than initial temperature as pressure has increased. Therefore dU is positive. Hence option (S) is connected with (A).
Similar arguments can be applied to other graphs.
24. (A) Initially more heat will enter through section A but the metal will absorb some heat and less heat will leave from C.
(B) At steady state heat accumulatron = )
dQ
dt is same for all sections
dQ 1 dQ
(C) At steady state
= kA or
=
dt kA dt
is inversely proportional to area of cross-section. Hence is maximum at B and minimum at C
(D) At steady state heat accumulation = 0
dT
So dt = 0 for any section.
SOLUTIONS TO PART TEST-03 (PT-03) TOPIC : HEAT & THERMODYNAMICS (PHYSICS)
1. Heat released by steam inconversion to water at 100°C is Q1 = mL = 4 × 540 = 2160 cal.
Heat required to raise temperature of water from 46°C t 100°C is Q = mS = 20 × 1 × 54 = 1080 J
Q1 > Q2
Q1
and Q2 = 2
Hence all steam is not converted to water only half steam shall be converted to water
Final mass of water = 20 + 2 = 22 gm
2. Q = U + W WAB = 0
AB =
AB =
nR T
2
5 ( PV)
2
( PV)
2
QAB = 2.5 P0 V0
Process BC
BC BC BC
QBC = 0 + 2P0 V0 𝑙n 2
= 1.4 P0 V0
Qnet = QAB + QBC = 3.9 P0 V0
3. Y =
Stress Strain =
T / A
𝑙 / 𝑙
T = Y.𝑙 A
𝑙
= Y.A T
In both the rods tension will be same so T1 = T2 Hence Y A T = Y A T
A1 A 2
Y22
= Y11
4. V2
rms
= < V2 > =
V2dN
V 2 V 2 V 2 ...........
N
dN
= dN
here dV
= N(V)
2
rms
1 N(V) V2 dV
0
1 V0 3N 3
= 0 .V2 V2 dV = V 2 V =
.V .
N0 V 3 5 0
rms 0
0 0
2𝑙
d𝑙
5. As ; d𝑙 = 𝑙 dT 𝑙
𝑙
T
= aT dT
0
T2 𝑙n4 1/ 2
𝑙n2 = a
2
T =
a
6. For a block body, wavelength for maximum intensity :
1
T
& P T4
P 1
4
P = 16 P. ∴ P T = 32PT P’ = 16 P.
7. PV = nRT PdV = nRdT
= 1 dV
and
dV nR
= 1
V dT dT P T
1
For given temperature T , = T0
8. The rate of heat loss by a thin hollow sphere of thickness ‘x’ , mean radius ‘r’ and made of density ‘’ is given by
dT
mS dt
= – A(T4 – T 4 )
( 4r2x) S
dT = 4r2 (T4 – T 4 )
dt
dT
dt
(T 4 T 4 )
= – 0
Sx
is independent of radius
Hence rate of cooling is same for both spheres.
P2
9. Equation of process = constant = C (1)
Equation of State P R T (2)
M
From 1 and 2 PT = constant C is false, D is true.
As -changes to 2
Hence T changes to
P changes to
2T . B is true.
from equation (1) A is false.
A
10. A
× 100 =
𝑙
2 A × 100 % increase in Area = 2 × 0.2 = 0.4
V
V × 100 = 3 × 0.2 = 0.6 %
Since l = l T
𝑙
× 100 = T × 100 = 0.2 = 0.25 × 10–4 / ºC
𝑙
11. Area under the curve is equal to number of molecules of the gas sample. Hence
N = 1 . a . V
2 0
aV0
= 2N
V0 a 2 V
V = 1 v N(V)dV
= 1 C .
.V dV
= V
avg 2
=
avg N 0
N 0 v0
3 0 V0 3
v0 a
2 V 1
2
r ms
= 1 V 2 N(V)dV
N 0
= 1 V 2
N 0 V0
.V
dV =
V0
2
rms
V0 =
Area under the curve from 0.5 V0
to V0
is 3 of total area.
4
12 V
r.m.s. =
Since PV = nRT therefore P and V both can change simultaneously keeping temperature constant.
13. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
14. From Wein's law T = constant i.e., peak emission wavelength 1 . Hence as T increase
decreases.
m m T m
15. to 17. TP(1 – /) = K. (given)
dT
Lapse Rate is dy , therefore differentiating w.r.t. y
1 1
dT
1
1 dP
dT T 1 dP
. P T.
dy
.P
. 0
dy
dy P
.
dy
Using PV =
M RT M
P =
RT M
T M
P R
and
dP
dy = – g
dT M 1
1 Mg
.
dy R
( g) 1 R
dT
using given value dy = – 9.8 K/km.
18. (A) PV = nRT
P = (nRT)
1 = (constant) 1 V V
T = constant i.e. isothermal process V increases, W is positive and Q = U + U = W > 0
(B) Q = 0 pdV = W = positive
(C) PV = nRT
As volume increases, T also increases
i.e., U > 0 PdV = W > 0 So Q > 0
(D) For cyclic process U = 0
W < 0 (anticlockwise)
Q < 0.
𝑙
1 dx
19. dR = k (r x tan )2
O 1
𝑙
Solving R = k r1 r2
G = 1 =
R
k r1 r2
𝑙
20. Due to the heating pressure inside is not changed. Let inside pressure be . Then for equilibrium of the system :
P(A1 – A2) = P0(A1 – A2) + (m1 + m2)g
PV = (p0A + mg)𝑙
𝑙 is displacement of the piston.
P . V = nRT. 1g
T = PV
nR
= (p0A mg)𝑙
nR
PA1 T
PA2
(105Pa x 103 m2 5 x 10)(50 102 )
= 1 x R
m2g
P0A2
T =
75 K.
R
21. The maximum temperature of the gas will be during process BC. Process BC can be represented by straight line, y = m x + C
So P = m V + C Putting point B & C gives
3 P = 2 m V + C (1) P = 6 m V + C (2)
So subtracting 2 P = 4 m V So m = P
P
From (2) P = 2 V . 6 V + C C = 4 P
P
Hence we get equation as, y = 2 V
x + 4 P (4)
where y is pressure and x is volume of gas. Putting y from above,
P
now we have x y = n R T 2 V x2 + 4 P x = n R T (3)
d T
For maximum temperature
d x
Hence x = 4 V
= 0 2 V
x + 4 P = 0
8 PV
Putting in (3) we get n R Tmax = 2 P (4 V) = 8 P V So, Tmax = n R
22. Before heating let the pressure of gas be P1 from the equilibrium piston, PA = kx1
PA nRT A
1 8.3 100 102
x = K
= =
K
0.83 200
= 0.05 m
Since during heating process,
The spring is compressed further by 0.1 m
x = 0.15 m
1 1
work done by gas = 2 .200((0.15)2 – (0.05)2) = 2 .200 (.02) = 2J
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