(1) 172.6 ml (2) 17.26 ml (3) 192.7 ml (4) 19.27 ml Ans. (1)
Sol. 1 1
2 2 2
V T 200 (273 20) 293
V T
V T V (273 20) 253
2
200 253
V 293
= 172.6 ml
Q.2 Which of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium ?
(1) kinetic energy (2*) momentum (3) density (4) speed
Ans. (2)
Sol. PMVav, As Vav = 0 (in equilibrium )
Pav = 0
Q.3 The average momentum of a molecule in a sample of an ideal gas depends on
(1) temperature (2) number of moles (3) volume (4*) none of these Ans. (4)
Sol. As Pav MVav
, it depend mass and Vav of gas
Q.4 The mean free path of gas molecules depends on (d = molecular diameter)
(1) d (2)d2 (3)d2 (4)d1
Ans. (3)
Sol. Mean free path = 1 2
2 d n d–2 Q.5 A gas behaves more closely as an ideal gas at
(1) low pressure and low temperature (2*) low pressure and high temperature (3) high pressure and low temperature (4) high pressure and high temperature Ans. (2)
Sol. self explainatory
Q.6 Which of the following statements about kinetic theory of gases is wrong (1) The molecules of a gas are in continuous random motion
(2) The molecules continuously undergo inelastic collisions
(3) The molecules do not interact with each other except during collisions (4) The collisions amongst the molecules are of short duration
Ans. (2)
Sol. The collision of molecules of ideal gas is elastic collision
Q.7 Figure shows graphs of pressure vs density for an ideal gas at two temperatures T1and T2.
(1*) T1> T2 (2) T1= T2 (3) T1< T2 (4) any of the three is possible Ans. (1)
RMS and KTG
Q.1 The volume of a gas at 20°C is 200 ml. If the temperature is reduced to – 20°C at constant pressure, its volume will be
EXERCISE-I
Kinetic Theoryof Gasesand Thermodynamics
Sol. Pm = nRTT
slope of T1 > slope of T2
T1> T2
Q.8 If a Vander-Waal’s gas expands freely, then final temperature is (1) Less than the initial temperature
(2) Equal to the initial temperature (3) More than the initial temperature
(4) Less or more than the initial temperature depending on the nature of the gas Ans. (1)
Sol. In free expansion of Vander waal’s gas, its temperature decreases
Q.9 Suppose a container is evacuated to leave just one molecule of a gas in it. Letaandrmsrepresent the average speed and the rms speed of the gas.
(1)a>rms (2)a<rms (3*)a=rms (4)rmsis undefined Ans. (3)
Sol. one molecule has some single value of speed which is equal average speed and rms speed of the gas
Va= V
rms.
Q.10 A gas at the temperature 250 K is contained in a closed vessel. If the gas is heated through 1K, then the percentage increase in its pressure will be
(1) 0.4% (2) 0.2% (3) 0.1% (4) 0.8%
Ans. (1) Sol. (1)
1 1 2 1 2 1
2 2 1 1
P T P P T T
P T
P T P T
P 251 250
% 100 0.4%
P 250
Q.11 The rms speed of oxygen molecules in a gas is. If the temperature is doubled and the O2molecule dissociate into oxygen atoms, the rms speed will become
(1) (2)2 (3*) 2 (4) 4
Ans. (3)
Sol. VRMS= M0
RT 3
=
for oxygen VRMS= 3MR /22T
0
= 2.
Q.12 Consider a mixture of oxygen and hydrogen kept at room tempertaure. As compared to a hydrogen molecule an oxygen molecule hits the wall
(1) With greater average speed (2*) with smaller average speed (3) with greater average kinetic energy (4) with smaller average kinetic energy.
Ans. (2) Sol. V
av M0
1
oxygen molecule hits the wall with smaller average speed
Q.13 The molecular weights of O
2and N
2are 32 and 28 respectively. At 15°C, the pressure of 1 gm O
2will be the same as that of 1 gm N2in the same bottle at the temperature
(1) – 21°C (2) 13°C (3) 15°C (4) 56.4°C
Sol. (1)
For 1 gm gas PV = rT = R M .T
Since P and V are constant T M 2 2
2 2
N N
O O
T M
T M
N2
T 28
252K –21 C (273 15) 32
Q.14 Consider the quantity MkT / pV of an ideal gas where M is the mass of the gas. It depends on the (1) temperature of the gas (2) volume of the gas
(3) pressure of the gas (4*) nature of the gas
Ans. (4) Sol. PV =
M0 M
RT..
PV =M0 M
KN0T
PV
MKT = m1 m1= mass of each molecule it depends nature of gas.
Q.15 The number of molecules in a gas at pressure1.64103atmospheres and temperature 200 K having the volume 1 cc are (1)6.021016 (2)2.631016 (3)3.011019 (4)12.041019
Ans. (1)
Sol. PV = NKT N = PV KT
=
3 5 6
23
(1.64 10 1.01 10 ) (1 10 ) 1.38 10 200
= 60.23 × 1016
Q.16 The ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample containing a mixture of the two gases.
(1*) 14 (2) 7 (3) 28 (4) None of these
Ans. (1) Sol. V
avr= —V = M RT 8
=
M RT 8
— N
— H
2 2
V V
= 1/28 2 /
1 =
1 14
Q.17 Three closed vessels A, B, and C are at the same temperature T and contain gases which obey the Maxwellian distribution of velocities. Vessel A contains only O2, B only N2and C a mixture of equal quantities of O2and N2. If the average speed of O2molecules in vessel A is V1, that of the N2molecules in vessel B is V2 ,the average speed of the O2 molecules in vessel C will be :
(1) (V1+ V2)/2 (2*) V1 (3) (V1V2)1/2 (4) 3kT/M
Ans. (2)
Sol. Vav= M0 RT 8
, VVAV T
For same temp in vessel A, B and C, Average speed of O2molecule is same in vessel A and C and is equal to V1.
Q.18 1 mole of gas occupies a volume of 100 ml at 50 mm pressure. What is the volume occupied by two moles of gas at 100 mm pressure and at same temperature
(1) 50 ml (2) 100 ml (3) 200 ml (4) 500 ml
Ans. (2) Sol. PV = µRT
1 1 1 2
2 2 2 2
P V µ 50 100 1
V 100ml
P V µ 100 V 2
Q.19 A flask is filled with 13 gm of an ideal gas at 27C and its temperature is raised to 52C. The mass of the gas that has to be released to maintain the temperature of the gas in the flask at 52C and the pressure remaining the same is
(1) 2.5 g (2) 2.0 g (3) 1.5 g (4) 1.0 g
Ans. (4)
Sol. PV = mrT since P,V,r remains same
Hence m 1
T 1 2
2 1
m T
m T
2
13 (273 52) m (273 27)
= 325
300
m2= 12 gm
i.e., mass released = 13 gm – 12 gm = 1 gm
Q.20 Which of the following quantities is the same for all ideal gases at the same temperature ? (1) the kinetic energy of 1 mole (2) the kinetic energy of 1 g (3*) the number of molecules in 1 mole (4) the number of molecules in 1 g Ans. (2)
Sol. For an ideal gas, the no of molecules in 1 mole of all gas at is same .
Q.21 Three containers of the same volume contain three different gases. The masses of the molecules are m1, m2 andm3 and the number of molecules in their respective containers are N1, N2 and N3. The gas pressure in the containers areP1, P2 and P3 respectively. All the gases are now mixed and put in one of the containers. The pressure P of mixture will be
(1)P(P1P2 P3) (2)
3
3 2
1 P P
P P (3)PP1 P2 P3 (4)P(P1 P2 P3) Ans. (3)
Sol. According to the Dalton’s law of partial pressures, the total pressure will be P1+ P2+ P3
Q.22 Refer to fig. LetU1andU2be the changes in internal energy of the system in the processes A and B then
(1)U1>U2 (2*)U1=U2 (3)U1<U2 (4)U1 U2 Ans. (2)
Sol. AsU is a state function i.e., it depends initial and final position in process A and B initial and final temp are same.
U, = U2.
Q.23 Air is pumped into an automobile tube upto a pressure of 200 kPa in the morning when the air temperature is 22°C. During the day, temperature rises to 42°C and the tube expands by 2%. The pressure of the air in the tube at this temperature, will be approximately
(1) 212 kPa (2) 209 kPa (3) 206 kPa (4) 200 kPa
Ans. (2)
Sol. 1 1 2 2
1 2
P V P V
PV R(cons tant)
T T T
200 V P2 1.02V (273 22) (273 42)
(V
2= V + 0.02V)
P2= 200 317 275 1.02
= 209 kPa
Q.24 If an ideal gas has volume V at 27°C and it is heated at a constant pressure so that its volume becomes 1.5V. Then the value of final temperature will be
(1) 600°C (2) 177°C (3) 817°C (4) None of these
Ans. (2)
Sol. T2= 2 1 1
V 1.5V
T (273 27) 450 K
V V
177°C
Q.25 To what temperature should the hydrogen at 327°C be cooled at constant pressure, so that the root mean square velocity of its molecules become half of its previous value
(1) – 123°C (2) 123°C (3) – 100°C (4) 0°C
Ans. (1)
Sol.
2
2 2
rms
1 1
T v
3RT 1
v M T v 4
1 2
T (273 327)
T 150K –123 C
4 4
Q.26 The root mean square velocity of a gas molecule of mass m at a given temperature is proportional to
(1) m0 (2) m (3) m (4)
m 1
Ans. (4)
Sol. v
rms= rms
3kT 1
m v m
Q.27 At room temperature, the r.m.s. speed of the molecules of certain diatomic gas is found to be 1930 m/s. The gas is
(1)H2 (2)F2 (3)O2 (4)Cl2
Ans. (1)
Sol. vrms= 2 2
rms
3RT 3RT 3 8.3 300
M M
M v (1920)
= 2 × 10–3kg = 2gm Gas is hydrogen
Q.28 According to the kinetic theory of gases the r.m.s. velocity of gas molecules is directly proportional to
(1) T (2) T (3)T2 (4)1/ T
Ans. (2)
Sol. vrms T
Q.29 r.m.s. velocity of nitrogen molecules at NTP is
(1) 492 m /s (2) 517 m/s (3) 546 m/s (4) 33 m/s
Ans. (2)
Sol. vrms=
3RT 3 8.3 107 300
M 28
= 517 m/sec
Q.30 The root mean square speed of the molecules of a gas is
(1) Independent of its pressure but directly proportional to its Kelvin temperature
(2) Directly proportional to the square roots of both its pressure and its Kelvin temperature
(3) Independent of its pressure but directly proportional to the square root of its Kelvin temperature (4) Directly proportional to both its pressure and its Kelvin temperature
Ans. (1)
Sol. Since Crms<< Ve, hence molecules do not escape out.
Q.31 Let A and B the two gases and given : 4. ;
B B A A
M T M
T whereT is the temperature and M is molecular mass. If CA and CB
are the r.m.s. speed, then the ratio
B A
C C
will be equal to
(1) 2 (2) 4 (3) 1 (4) 0.5
Sol. (1)
A B A B
A B A B
T T T T
4 2
M M M M
A
A B
3RT 3RT
M 2 M CA= 2CB A
B
C 2
C
temperature in degree centigrade to
(1) 54° (2) 270° (3) 327° (4) 600°
Sol. (3)
1 1
2 2 2 2
V T V (273 27) 300
V T
V T 2V T T
T2= 600 K = 327°C
Q.33 A mass of an ideal gas undergoes a reversible isothermal compression. Its molecules will then have compared with initial state, the same
(i) root mean square velocity (ii) mean mometum (iii) mean kinetic energy (1*) (i), (ii), (iii) correct (2) (i), (ii) correct (3) (ii), (iii) correct (4) (i) correct Sol. (1)
As initial and final state are same
TI= TF As Vrms, Pav andKav depends on temperature
all are equal.
Gas Law’s
Q.32 To double the volume of a given mass of an ideal gas at 27°C keeping the pressure constant, one must raise the
Q.34 Volume of gas become four times if
(1) Temperature become four times at constant pressure (2) Temperature become one fourth at constant pressure (3) Temperature becomes two times at constant pressure (4) Temperature becomes half at constant pressure Sol. (1)
V T (as constant pressure)
Q.35 Kinetic theory of gases provide a base for
(1) Charle’s law (2) Boyle’s law
(3) Charle’s law and Boyle’s law (4) None of these Sol. (3)
Boyle’s and Charle’s law follows kinetic theory of gases
(1) Three degrees of freedom (2) Four degrees of freedom (3) Five degrees of freedom (4) Six degrees of freedom Sol. (1)
A monoatomic gas molecule has only three translational degree of freedom.
Q.37 When an ideal diatomic gas is heated at constant pressure , the fraction of the heat energy supplied which increases the internal energy of the gas is .
(1) 5
2 (2)
5
3 (3)
7
3 (4*)
7 5
Sol. dQ = dW + dU dQ = PdV + dU dQ = nRdT + dU dQ = f
dU 2 + dU
dQ dU =
1 f 2 1
dQ dU =
7 5
Q.38 A diatomic molecule has how many degrees of freedom
(1) 3 (2) 4 (3) 5 (4) 6
Sol. (3)
A diatomic molecule has three translational and two rotational degree of freedom.
Hence t otal degree of freedom f = 3 + 2 = 5
Q.39 Boiling water is changing into steam. Under this condition, the specific heat of water is
(1) zero (2) one (3*) Infinite (4) less than one
Sol. (3) s = m T
Q
For changing state T = const orT = 0
s = (infinite )
Internal energy specific heat of gas and degrees of freedom Q.36 A monoatomic gas molecule has
Q.40 If the degree of freedom of a gas are f, then the ratio of two specific heats C /P CV is given by (1) 2 1
f (2)
f
12 (3)
f
11 (4)
f 11
Sol. (1)
P V
C 2
C 1 f
Q.41 Supposing the distance between the atoms of a diatomic gas to be constant, its specific heat at constant volume per mole (gram mole) is
(1*) R 2
5 (2) R
2
3 (3) R (4) R
2 7
Sol. As f = 5 dU = nCvdT =
2 nfRdT
Cv= 2 fR
Cv= 2
R 5
Q.42 A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. TheC /P CV is
(1) 1.67 (2) 1.4 (3) 1.29 (4) 1.33
Sol. (4)
Degree of freedom f = 3 (Translatory) + 2(rotatory) + 1 (vibratory) = 6
P
V
C 2
C 1 f =1 + 2 4 6 3 1
Q.43 For an ideal gas, the heat capacity at constant pressure is larger than than that at constant volume because (1) positive work is done during expansion of the gas by the external pressure
(2*) positive work is done during expasion by the gas against external pressure
(3) positive work is done during expansion by the gas against intermolecular forces of attraction (4) more collisions occur per unit time when volume is kept constant.
Sol. (2)
Self explanatory Q.44 A gas has :
(1) one specific heat only (2) two specific heats only (3*) infinite number of specific heats (4) no specific heat Sol. Gas has different specific heat for different processes
gas has infinite number of specific heats.
Q.45 The relation between two specific heats of a gas is
(1) J
C R
CP V (2)
J C R
CV P (3)CPCV J (4)CV CP J Sol. (1)
When CPand CVare given with calorie and R with Joule then CP– CV= R/J Q.46 For a solid with a small expansion coefficient,
(1) Cp- Cv= R (2) Cp- Cv= R
(3*) Cpis slightly greater than Cv (4) Cpis slightly less than Cv Sol. As compare to gas solid expand very less.
Cpis slightly greater then Cv.
Q.47 The quantity of heat required to raise one mole through one degree Kelvin for a monoatomic gas at constant volume is (1) R
2
3 (2) R
2
5 (3) R
2
7 (4)4R
Sol. (1)
(Q)V= µC
VT (Q)
V= 1 × C
V× 1 = C
V
For monoatomic gas C
V= 3
2R (Q)V= 3 2R
Q.48 One mole of monoatomic gas and three moles of diatomic gas are put together in a container. The molar specific heat (in
1)
1
mol K
J at constant volume is (R8.3JK1mol1)
(1) 18.7 (2) 18.9 (3) 19.2 (4) None of the above
Sol. (1)
ymix=
1 1 2 2
1 2
1 2
1 2
µ µ
1 1
µ µ
1 1
=
5 7
1 1
3 5
5 7
1 1
3 5
1 1
5 7
1 1
3 5
= 3 2 = 1.5
Q.49 The degrees of freedom of a stationary rigid body about its axis will be
(1) One (2) Two (3) Three (4) Four
Sol. (3)
Q.50 A gaseous mixture consists of 16g of helium and 16g of oxygen. The ratio
V P
C C
of the mixture is
(1) 1.4 (2) 1.54 (3) 1.59 (4) 1.62
Sol. (4)
1 1 2 2
1 2
mixture
1 2
1 2
µ µ
1 1
µ µ
1 1
1
µ moles of helium 16 4
4
2
16 1 µ moles of oxygen
32 2
mix
4 5 / 3 1 / 2 7 / 5
5 7
1 1
3 5
4 1 / 2
5 7
1 1
3 5
= 1.62
Q.51 Relationship between P, V, and E for a gas is (1) P EV
2
3 (2)V EP
3
2 (3)PV E
2
3 (4)PV E
3
2
Sol. (4)
P = 2 2 E
(Energy per unit volume)
3 3 V PV=23E
Q.52 The ratio of mean kinetic energy of hydrogen and oxygen at a given temperature is
(1) 1 : 16 (2) 1 : 8 (3) 1 : 4 (4) 1 : 1
Sol. (4)
Kinetic energy is function of temperature
Q.53 The kinetic energy of one gram molecule of a gas at normal temperature and pressure is (R8.31J/moleK) (1) 0.56104 J (2)1.3102J (3)2.7102J (4)3.4103 J
Sol. (4)
Kinetic energy per gm mole E = f 2RT
If nothing is said about gas then we should calculate the translational kinetic energy.
i.e. E
trans= 3 3
RT 8.31 (273 0)
2 2 = 3.4 × 103J
Q.54 The mean kinetic energy of a gas at 300 K is 100 J. The mean energy of the gas at 450 K is equal to
(1) 100 J (2) 3000 J (3) 450 J (4) 150 J
Sol. (4)
Average kinetic energy temperature
1 1
2 2 2
E T 100 300
E T E 450 Þ E2= 150 J
Q.55 Energy of all molecules of a monoatomic gas having a volume V and pressure P is PV 2
3 . The total translational kinetic energy of all molecules of a diatomic gas as the same volume and pressure is
(1) PV 2
1 (2) PV
2
3 (3) PV
2
5 (4)3PV
Sol. (2)
Transalational kinetic energy of all molecules of diatomic g as will also be 3
2 PV, because all the gases have same translational degree of freedom (3)
cycle is 5 J, the work done by the gas in the process C A is
C
A B
10 2
1 V(m3)
P(N/m2)
(1) – 5 J (2) – 10 J (3) – 15 J (4) – 20 J
Sol. (1)
For cyclic process. Total work done
CA BC
AB W W
W
WAB= PV = 10(2 – 1) = 10J and WBC=0 (as V = constant)
From FLOT,Q = U + W
U = 0 (Process ABCA is cyclic)
Q = WAB+WBC+WCA
5 = 10 + 0 + WCA WCA= – 5 J
Q.56 An ideal gas is taken through the cycle A B C A, as shown in the figure. If the net heat supplied to the gas in the
Q.57 In the following figures (1) to (4), variation of volume by change of pressure is shown. A gas is taken along the path ABCDA. The change in internal energy of the gas will be:
(1) (2) (3) (4)
(1) positive in all cases from (1) to (4)
(2) positive in cases (1), (2) and (3) but zero in case (4) (3) negative in cases (1), (2) and (3) but zero in case (4) (4*) zero in all the four cases.
Sol. (4)
As U = nRT For closed path
T = 0
U = 0.
Q.58 In the following indicator diagram, the net amount of work done will be
1 2
V P
(1) Positive (2*) Negative (3) Zero (4) Infinity
Sol. (2)
The cyclic process 1 is clockwise where as process 2 is anticlockwise. Clockwise area represents positive work and anticlockwise area represents negative work. Since negative area (2) > positive area (1), hence net work done is negative.
Q.59 An ideal gas changes from state a to state b as shown in Fig. What is the work done by the gas in the process ?
(1*) zero (2) positive (3) negative (4) infinite
Sol. T P or T
P = constant
As V
nR
P T = constant or V = constant
W = 0.
Q.60 In the cyclic process shown in the figure, the work done by the gas in one cycle is
V1 4 V1
P1
7P1
P
V
(1)28P1V1 (2)14P1V1 (3)18P1V1 (4)9P1V1 Sol. (4)
Work done = Area under curve 2
3 6P1 V1
= 9 P1V1
Q.61 The processU = 0, for an ideal gas can be best represented in the form of a graph :
(1) (2*)
P
V B A
(3) (4)
Sol. (2)
U = 0
T = constant
clearly, option B is correct.
Q.62 An ideal gas is taken around the cycle ABCA as shown in the P-V diagram. The net work done by the gas during the cycle is equal to
3V1
P
V1
P1
3P1
B
A C
(1)12P1V1 (2)6P1V1 (3)3P1V1 (4) 2P1V2
Sol. (4)
Work done 2 1 2 1 2 1 1
2
1 P V PV
Q.63 A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes, after this pressure of the gas in the cylinder
(1) increases (2*) decreases (3) remains constant
(4) increases or decreases depending on the nature of the gas.
Sol. (2)
As Volume decreases
pressure of the gas in the cylinder increases
Q.64 Let Taand Tbbe the final temperature of the samples A and B respectively in the previous question then :
(1*) Ta< Tb (2) Ta= Tb
(3) T
a> T
b (4) The relation between T
aand T
bcannot be deduced.
Sol. (1)
For adiabatic
1 V
1
T V = C ....(i)
For isothermal T = const....(ii) From (i) and (ii)
Ta< Tb
Q.65 In the following V-T diagram what is the relation between P1and P2:
(1) P2= P1 (2) P2> P1 (3*) P2< P1 (4) cannot be predicted Sol. (3)
P nR TV P
1 slope or P slope
1 1
P2< P
1
Q.66 In the isothermal expansion of an ideal gas. Select wrong statement:
(1) there is no change in the temperature of the gas (2) there is no change in the internal energy of the gas
(3) the work done by the gas is equal to the heat supplied to the gas (4*) the work done by the gas is equal to the change in its internal energy Sol. (4)
In isothermal expansion
T = constant U = 0
W =Q
option (4) is correct.
Q.67 In the cyclic process shown on the P – V diagram the magnitude of the work done is :
P1
P2
O V1 V2
V P
(1)
2 1 2
2 P
P
(2)
2 1 2
2 V
V
(3*)
4
(P2– P1) (V2– V1) (4)(P2V2– P1V1)
Sol. (3)
W.D. = × Pressure Radius × volume Radius (area of ellipse)
W =
2
P – P2 1
2
V – V2 1
= 4
(P
2– P
1) (V
2–V
1)
Q.68 A fixed mass of gas undergoes the cycle of changes represented by PQRSP as shown in Figure. In some of the changes, work is done on the gas and in others, work is done by the gas. In which pair of the changes work is done on the gas?
(1) PQ and RS (2) PQ and QR (3) OR and RS (4*) RS and SP.
Sol. (4)
work done on the gas = negative work
W = PdV when V decreases then W = – ve hence option D is correct.
Q.69 Consider two processes on a system as shown in fig. The volumes in the initial states are the same in the two porcesses and the volumes in the final states are also the same. LetW1andW2be the work done by the system in the processes A and B respectively.
(1) W1>W2 (2)W1=W2 (3*)W1<W2
(4) Nothing can be said about the relation betweenW1andW2 Sol. As W = PVV = same is both process
A s P
B> P
A W2> W
1 .
(1)dQdUPdV (2)dQdUPdV (3)dQ(dUdV)P (4)dQPdUdV Sol. (1)
W U Q
and W PV
Q.71 Ideal gas is taken through process shown in figure:
(1) In process AB, work done by system is positive (2*) In process AB, heat is rejected out of the system.
(3) In process AB, internal energy increases
(4) In process AB internal energy decreases and in process BC internal energy increases.
Sol. (2)
In process AB T = constant P = increases P
V 1
or V = decreases Q = W .
W = – ve. or Q = – ve
heat is rejected out of the system
Q.72 If heat is supplied to an ideal gas in an isothermal process,
(1) the internal energy of the gas will increase (2*) the gas will do positive work (3) the gas will do negative work (4) the said process is not possible Sol. (2)
Q = W (T = constant)
if heat is supplied then W = +ve
Q.73 The first law of thermodynamics is concerned with the conservation of
(1) Momentum (2) Energy (3) Mass (4) Temperature
Sol. (2)
Q.74 In a cyclic process shown in the figure an ideal gas is adiabatically taken from B and A, the work done on the gas during the process B A is 30 J, when the gas is taken from A B the heat absorbed by the gas is 20 J. The change in internal energy of the gas in the process A B is :
(1) 20 J (2*) – 30 J (3) 50 J (4) – 10 J
First law of thermodynamics
Q.70 First law of thermnodynamics is given by
Sol. (2) BA
Q = 0
0 = – 30 +UBA
UBA= 30 J
UAB= –UBA= – 30 J
Q.75 A system can be taken from the initial state p1, V1to the final state p2, V2by two different methods. LetQ and W represent the heat given to the system and the work done by the system. Which of the following must be the same in both the methods ?
(1)Q (2)W (3)Q + W (4*)Q - W
Sol. (4)
U = Q – W is same in both methods as it is a state function
Q.76 In thermodynamic process, 200 Joules of heat is given to a gas and 100 Joules of work is also done on it. The change in internal energy of the gas is
(1) 100 J (2) 300 J (3) 419 J (4) 24 J
Sol. (2) QUW; Q200J and W100J
UQW 200 (100)300J
Q.77 The P-V diagram of a system undergoing thermodynamic transformation is shown in figure. The work done on the system in going from A B C is 50 J and 20 cal heat is given to the system. The change in internal energy between A and C is
(1) 34 J (2) 70 J
P
A
C
B V
(3) 84 J (4) 134 J
Sol. (4)
Heat givenQ 20cal204.284J. Work doneW = – 50 J
[As process is anticlockwise]
By first law of thermodynamics
UQW 84(50)134J
Q.78 When a system is taken from state i to a state f along path iaf, Q50J andW 20 J. Along path ibf, Q35 J. If W 13J for the curved return path f i, Q for this path is
P
V a
i b
f
(1) 33 J (2*) 23 J
(3) – 7 J (4) – 43 J
Sol. (2)
Work done = Area enclosed by indicator diagram PV
P P V
V )(4 ) 3 3
2 (
1
(1) Heat content remains constant
(2) Heat content and temperature remain constant (3) Temperature remains constant
(4) None of the above Sol. (3)
In isothermal process temperature remains constant.
Different types of thermodynamics process Q.79 For an ideal gas, in an isothermal process
Q.80 A fixed mass of ideal gas undergoes changes of pressure and volume starting at L, as shown in Figure.
Which of the following is correct :
(1) (2*)
volume M N
Temperature/K L
(3) (4)
Sol. (2)
L M P = constant V T.
MN T = constant
Here, option B is constant
Q.81 For free expansion of a gas in an adiabatic container which of the following is true ? (1*) Q = W = 0 andU = 0 (2) Q = 0, W > 0 andU = Q (3) W = 0, Q > 0 andU = Q (4) W = 0, Q < 0 andU = 0 Sol. (1)
For free expansion Q = 0, W = 0,U = 0 Q.82 The gas law
T
PV constant is true for
(1) Isothermal changes only (2) Adiabatic changes only
(3*) Both isothermal and adiabatic changes (4) Neither isothermal nor adiabatic changes Sol. (3)
Q.83 Starting with the same initial conditions, an ideal gas expands from volume V1to V2in three different ways. The work done by the gas is W
1if the process is isothermal, W
2if isobaric and W
3if adiabatic, then :
(1*) W2> W1> W3 (2) W2> W3> W1 (3) W1> W2> W3 (4) W1> W3> W2 Sol. (1)
As W.D. is isobaric > W.D. in Isothermal > W.D in adiabatic or W2> W
1> W
3
Hence option (1) is correct.
Q.84 The isothermal Bulk modulus of an ideal gas at pressure P is
(1*) P (2)P (3) P / 2 (4) P /
Sol. (1) E P
Q.85 A gas is expanded from volume V0to 2V0under three different processes. Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic. Let
U1,U2 and U3 be the change in internal energy of the gas is these three processes. Then :
(1*)U1>U2>U3 (2)U1<U2<U3 (3)U2<U1<U3 (4)U2<U3<U1 Sol. (1)
Process ...(i) is isobatic
U1=Q – W = positive process (ii) is isothermal
U2= 0
Process (iii) is adiabatic
Q = 0
U = – W = negative
U1>U2>U3
Q.86 The molar heat capacity for the process shown in fig. is
(1) C = Cp (2) C = Cv (3) C > Cv (4*) C = 0
Sol. (4)
For polytropic process C = R1+
x 1
R
As PV= K
Put x = C = 0
Q.87 When an ideal gas in a cylinder was compressed isothermally by a piston, the work done on the gas was found to be joules
104
5 .
1 . During this process about
(1) 3.6103 cal of heat flowed out from the gas (2) 3.6103 cal of heat flowed into the gas (3)1.5104 cal of heat flowed into the gas (4)1.5104 cal of heat flowed out from the gas Sol. (1)
In isothermal compression, there is always an increase of heat. which must flow out the gas.
) 0 (
Q U W Q W U
Q J cal 3cal
4
4 3.6 10
18 . 4
10 5 . 10 1 5 .
1
Q.88 Four curves A, B, C and D are drawn in the Fig. for a given amount of gas. The curves which represent adiabatic and isothermal changes are
(1) C and D respectively (2) D and C respectively (3*) A and B respectively (4) B and A respectively Sol. (3)
Isothermal P V 1
Adiabatic P V
1
Also, slope of adiabatic is more as compare to isothermal
option (3) is correct.
Q.89 If a cylinder containing a gas at high pressure explodes, the gas undergoes (1) Reversible adiabatic change and fall of temperature
(2) Reversible adiabatic change and rise of temperature (3) Irreversible adiabatic change and fall of temperature (4) Irreversible adiabatic change and rise of temperature Sol. (3)
Gas cylinder suddenly explodes is an irreversible adiabatic change and work done against expansion reduces the temperature.
Q.90 A given quantity of a gas is at pressure P and absolute temperature T. The isothermal bulk modulus of the gas is:
(1) 3
2P (2*) P (3)
2
3P (4) 2P
Sol. (2) B = dV
VdP = – dV
) (PdV
(for isothermal process ) B = P
Q.91 The pressure in the tyre of a car is four times the atmospheric pressure at 300 K. If this tyre suddenly bursts, its new temperature will be( 1.4)
(1)300(4)1.4/0.4 (2)
4 . 1 / 4 . 0
4 300 1
(3)300(2)0.4/1.4 (4)300(4)0.4/1.4
Sol. (4)
For adiabatic process 1 P
T constant
1
2 1 1 2
P P T
T 1.4
) 4 . 1 1 ( 2
1 4 300
T 1.4
4 . 0 2300(4) T
Q.92 An ideal gas is allowed to expand freely against a vacuum in a rigid insulated container. The gas undergoes:
(1) an increase in its internal energy (2) a decrease in its internal energy
(3*) neither an increase nor decrease in temperature or internal energy (4) an increase in temperature
Sol. (3)
For free expansion
U = 0 or T = 0
U or T = const
Q.93 The adiabatic bulk modulus of hydrogen gas ( = 1.4) at NTP is :
(1) 1 × 105N/m2 (2) 1 × 10–5N/m2 (3) 1.4 N/m2 (4*) 1.4 × 105N/m2 Sol. B =P (for adiabatic process
B = 1.4 × 1 × 105= 1.4 × 105N/m2
Q.94 An ideal gas at 27oC is compressed adiabatically to 27
8 of its original volume. If 3
5
, then the rise in temperature is
(1) 450 K (2) 375 K (3) 225 K (4) 405 K
Sol. (2)
3 1 2
3 5
2 1
2 1 1 2
8 300 27 8
300 27
V T V T
T
K 2 675 800 3 8
300 27
2 2 3 / 1
K T 675300 375
Q.95 A and B are two adiabatic curves for two different gases. Then A and B corresponds to :
(1) Ar and He respectively (2*) He and H2respectively (3) O2and H2respectively (4) H2and He respectively Sol. (2)
Slope = –
dV dP
As slope of A > slope of B
of A > of B or A Helium
B Hydrogen
Q.96 Which is the correct statement
(1) For an isothermal change PV = constant
(2) In an isothermal process the change in internal energy must be equal to the work done
(3) For an adiabatic change
1 2 1 2
V V P
P , where is the ratio of specific heats
(4) In an adiabatic process work done must be equal to the heat entering the system Sol. (1)
In thermodynamic processes.
Work done = Area covered by PV diagram with V-axis
V P
isothermal Adaibatic
From graph it is clear that(Area)iso (Area)adi
WisoWadi
Q.97 When an ideal gas undergoes an adiabatic change causing a temparture changeT (i) there is no heat gained or lost by the gas
(ii) the work done by the gas is equal to change in internal energy (iii) the change in internal energy per mole of the gase is C
vT, where Cvis the molar heat capacity at constnat volume.
(1) (i), (ii), (iii) correct (2) (i), (ii) correct (3*) (i), (iii) correct (4) (i) correct Sol. Adiabatic process
Q = 0
For any process
V= nC
VT
Hence, option (3) is correct.
Q.98 The amount of work done in an adiabatic expansion from temperature T to T1 is
(1)R(TT1) (2) ( )
1 T T1
R
(3) RT (4)R(TT1)(1)
Sol. (2) WadiR1(TiTf)R1(TT1)
Q.99 In an adiabatic expansion of a gas initial and final temperatures areT1 and T2 respectively, then the change in internal energy of the gas is
(1) 1(T2 T1)
R
(2) 1(T1 T2)
R
(3)R(T1T2) (4) Zero
Sol. (1)
) 1 (
) ( 1 2
T T W R
U 1
) ( 2 1
T T R
Q.100 A gas expands under constant pressure P from volume V1toV2. The work done by the gas is (1)P(V2V1) (2) P(V1V2) (3)P(V1V2) (4)
1 2
2 1
V V
V P V
Sol. (1)
Work done
Q.101 Entropy of a thermodynamic system does not change when this system is used for (1) Conduction of heat from a hot reservoir to a cold reservoir
(2) Conversion of heat into work isobarically
(3) Conversion of heat into internal energy isochorically (4) Conversion of work into heat isochorically
Sol. (4)
Entropy of a reversible process does not change.
Q.102 Which relation is correct for isometric process
(1) QU (2)WU (3)QW (4) None of these
Sol. (1)
For isochoric process V0 W = 0 From FLOTQUWQU
Q.103 A cyclic process ABCD is shown in the figure P-V diagram. Which of the following curves represent the same process
(1) C
B A
D V
P
(2)
A B
C P
T
D (3)
A B
C P
T
D (4)
A B
C P
T D
Sol. (1) AB is isobaric process, BC is isothermal process, CD is isometric process and DA is isothermal process These process are correctly represented by graph (1).
supplied to the engine from source per cycle
(1) 1800 J/cycle (2) 1000 J/cycle (3) 2000 J/cycle (4) 1600 J/cycle Sol. (4)
Q W T
T T
1 2
1 Q TTT W
2 1
1
) 800 300 600 (
600
=1600 J
Q.105 A Carnot engine works between 600 K and 300 K. In each cycle of operations, the engine draws 1000joule of energy from the source at 600 K. The efficiency of the engine is -
(1) 20% (2*) 50% (3) 70% (4) 90%
Sol. (2)
Efficiency () =
1 2
T –T
1 × 100 =
600 –300
1 × 100 = 50%
Heat Engines
Q.104 A Carnot engine working between300 K and 600K has work output of 800 J per cycle. What is amount of heat energy
Q.106 In the above problem, the output work of second engine is
(1*) 2.93 × 103calorie (2) 12.3 × 103calorie (3) 12.3 × 103joule (4) 2.93 × 103calorie Sol. (1)
Work = output of second engine W2= Heat given ×1
= 273
200 10 4 3
cal = 2.93 × 103cal.
Q.107 In the above problem, the ratio of outputs of two engines is -
(1*) 0.577 (2) 0.377 (3) 0.777 (4) 0.177
Sol. (1)
cal 10 93 . 2
cal 10 69 . 1 W W
3 3
2 1
= 0.577
Q.108 In the above problem, the useful work done by the engine is -
(1) 100 joule (2*) 500 joule (3) 1000 joule (4) 150 joule
Sol. (2)
given Heat Total
done Work
× 100 =
W = 1000
10050 = 500 J
Q.109 The coefficient of performance of a carnot refrigertor working between 30° C and 0° C is
(1) 10 (2) 1 (3*) 9 (4) 0
Sol. (3)
30 9 273 273 303
273 T
T K T
2 1
2
Q.110 If the door of a refrigerator is kept open then which of the following is ture
(1) Room is cooled (2*) Room is heated
(3) Room is either cooled or heated (4) Room is neither cooled nor heated Sol. (2)
In a refrigerator, the heat dissipated in the atmophere is more then that taken from the cooling chamber, therefore the room is heated if the door of a refrigerator is kept open.
Q.111 An Ideal gas heat engine operated in a carnot's cycle between 227° C and 127° C . It absorbs 6 × 104J at high temperature.
The amount of heat converted into work is
(1) 4.8 × 104J (2) 3.5 × 104J (3) 1.6 × 104J (4*) 1.2 × 104J Sol. (4)
Q W 5 1 Q W 5
1 500 1 400 T 1 T
1
2
10 1.2 10 J
5 6 5
W Q 4 4
Q.112 An ideal gas heat engine exhausting heat at 77° C is not have a 30% efficiency. It must take heat at
(1) 127° C (2*) 227° (3) 327° C (4) 673°C
Sol. (2)
1 1
2
T 1350 100
30 T
1T
T 500K 227 C
10 7 100
70 100 1 50 T 350
1 1
Q.113 An ideal gas is taken around ABCA as shown in the above P-V diagram. The work done during a cycle is
B
C (P, 3V) P
E D V
A (P,V)
(3P, 3V)
(1) 2PV (2)PV (3) 1/2PV (4) Zero
Sol. (1)
Work done = Area enclosed by triangle ABC AC BC (3V V) (3P P) 2PV 2
1 2
1
Q.114 In the above problem, the energy rejected to the sink is
(1) 100 joule (2*) 500 joule (3) 1000 joule (4) 300 joule
Sol. (2)
Heat rejected = total heat given – W
= 1000 – 500 = 500 J
Q.115 An ideal gas of mass m in a state A goes to another state B via three different processes as shown in figure. IfQ1, Q2 and Q3 denote the heat absorbed by the gas along the three paths, then
V 1 2 3
B P A
(1)Q1 Q2 Q3 (2)Q1Q2 Q3 (3)Q1 Q2 Q3 (4)Q1 Q2 Q3 Sol. (1)
Initial and final states are same in all the process.
Hence U = 0; in each case.
By FLOT;Q = W = Area enclosed by curve with volume axis.
(Area)1< (Area)2< (Area)1 Q1< Q2< Q3- .
Q.116 A Carnot engine works between ice point and steam point. Its efficiency will be -
(1*) 26.81 % (2) 53.36 % (3) 71.23 % (4) 85.42 %
Sol.
=
1 2
T – T
1 × 100 T
2= 0ºC = 273 K
=
373 – 273
1 × 100 T1= 100ºC = 373 K
= 26.81%
Q.117 The effciency of Carnot's engine operating between reservoirs, maintained at temperatures 27° C and 123° C, is
(1*) 50% (2) 24% (3) 0.75% (4) 0.4%
Sol. (1)
% 2 50 1 300 1 150 ) 27 273 (
) 123 273 1 ( T T
1
2
Q.118 A Carnot engine operates between 227° C and 27°C. Efficiency of the engine will be (1) 3
1 (2*)
5
2 (3)
4
3 (4)
5 3
Sol. (2)
5 2 500 1 300 T 1 T
1
2
Q.119 A carnot engine has the same efficiency between 800 K to 500 K and x K ot 600 K. The value of x is
(1) 1000 K (2*) 960 K (3) 846K (4) 754 K
Sol. (2)
In first case, (1) = 1 – 800 500=
8 3
and in Second case, (2) = 1 – x 600
Since n
1= n
2 therefore 8 3= 1
x 600
or x 600 = 1
8 2=
8 5 or x =
5 8 600
= 960 K
Q.120 A scientist says that the efficiency of his heat engine which operates at source temperature 127°C and sink temperature 27°
C is 26% then
(1*) It is impossible (2) It is possible but less probable (3) It is quite probable (4) Data are incomplete ]
Sol. (1)
% 4 25 1 400 1 300 T 1 T
1 2
max
So 26 % efficiency is impossibel
Q.121 A Carnot's engine is made to work between 200° C and 0°C first and then between 0°C and –200°C. The ratio of efficiencies of the engine in the two cases is
(1*) 1.73 :1 (2) 1:1.73 (3) 1:1 (4) 1 : 2
Sol. (1)
In first case
473 200 ) 200 273 (
) 0 273 1 (
T 1 T
1 2
1
In second case
2= 1–
473 200 ) 0 273 (
) 0 273
(
1:1.73
273 473 1
2
1
Q.122 A thermodynamic process is shown in the figure. The pressures and volumes corresponding to some points in the figure are : PA 3104Pa,PB 8104Pa and VA 2103m3,VD 5103m3
In process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The change in internal energy of the system in process AC would be
(1) 560 J (2) 800 J V
P
D A
B C
O
(3) 600 J (4) 640 J
Sol. (1)
By adjoining graphWAB 0 and J WBC 8104[52]103 240
WAC WAB WBC 0240 240J Now,QAC QAB QBC 600200800J From FLOTQAC UAC WAC
800 UAC 240 UAC 560 J.
Q.123 In the above problem, in order to increase the efficiency of the engine by 20%, its sink temperature will have to be changed by -
(1) increase by 20 K (2) decrease by 293 K (3) increase by 20ºC (4*) decrease by 20ºC Sol.
= 26.81 + 20
= 46.81 =
1 2
T –T
1 × 100
100 81 .
46 =
1 2
T –T 1
1 2
T T
= 100
81 . –46 1
T2= 373
100 81 . – 46 1
T2=
100 19 . 53 373
= 198.39 K
Q.124 A Cannot engine works between 200ºC and 0ºC. Another Carnot engine works between 0ºC and –200ºC. In both cases the working substance absorbs 4 kilocalories of heat from the source. The efficiency of first engine will be -
(1)100
473 (2*) 200
473 (3) 200
273 (4)
373 273
Sol.
=
1 2
T – T
1 × 100
=
473 – 273
1 × 100
= 473
200 × 100 in % or = 473
200 in fraction
Q.125 In the above problem, the ratio of efficiencies of two engines will be -
(1) 0.18 (2) 0.38 (3*) 0.58 (4) 0.78
Sol. (3)
273 /200 473 200
2 1
= 473 273 = 0.58
