Heat and Thermodynamics
811 Questions
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Evening Shift
For an adiabatic expansion of an ideal gas, the fractional change in its pressure is equal to (where $\gamma$ is the ratio of specific heats) :
A.
$ - {1 \over \gamma }{{dV} \over V}$
B.
$ - \gamma {V \over {dV}}$
C.
$ - \gamma {{dV} \over V}$
D.
${{dV} \over V}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Evening Shift
An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is S1 and that of the other part is S2. Given that S1 > S2. If the piston is removed then the total entropy of the system will be :
A.
S1 $-$ S2
B.
${{{S_1}} \over {{S_2}}}$
C.
S1 $\times$ S2
D.
S1 + S2
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
The P-V diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process CD is (use $\gamma$ = 1.4) :
A.
$-$500 J
B.
$-$400 J
C.
400 J
D.
200 J
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T? (kB is Boltzmann constant)
A.
${1 \over 2}{k_B}T$
B.
${2 \over 3}{k_B}T$
C.
${3 \over 2}{k_B}T$
D.
${k_B}T$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
If one mole of the polyatomic gas is having two vibrational modes and $\beta$ is the ratio of molar specific heats for polyatomic gas $\left( {\beta = {{{C_P}} \over {{C_V}}}} \right)$ then the value of $\beta$ is :
A.
1.02
B.
1.35
C.
1.2
D.
1.25
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
Which one is the correct option for the two different thermodynamic processes?
A.
(a) only
B.
(c) and (d)
C.
(b) and (c)
D.
(c) and (a)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
A polyatomic ideal gas has 24 vibrational modes. What is the value of $\gamma$?
A.
1.37
B.
1.30
C.
1.03
D.
10.3
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
Two ideal polyatomic gases at temperatures T1 and T2 are mixed so that there is no loss of energy. If F1 and F2, m1 and m2, n1 and n2 be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is :
A.
${{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{F_1} + {F_2}}}$
B.
${{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{n_1}{F_1} + {n_2}{F_2}}}$
C.
${{{n_1}{T_1} + {n_2}{T_2}} \over {{n_1} + {n_2}}}$
D.
${{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{n_1} + {n_2}}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
Two identical metal wires of thermal conductivities K1 and K2 respectively are connected in series. The effective thermal conductivity of the combination is :
A.
${{2{K_1}{K_2}} \over {{K_1} + {K_2}}}$
B.
${{{K_1} + {K_2}} \over {{K_1}{K_2}}}$
C.
${{{K_1} + {K_2}} \over {2{K_1}{K_2}}}$
D.
${{{K_1}{K_2}} \over {{K_1} + {K_2}}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
A Carnot's engine working between 400 K and 800 K has a work output of 1200 J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is :
A.
2400 J
B.
1600 J
C.
1800 J
D.
3200 J
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Evening Shift
Calculate the value of mean free path ($\lambda$) for oxygen molecules at temperature 27$^\circ$C and pressure 1.01 $\times$ 105 Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (k = 1.38 $\times$ 10$-$23 JK$-$1)
A.
32 nm
B.
58 nm
C.
86 nm
D.
102 nm
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Evening Shift
A bimetallic strip consists of metals A and B. It is mounted rigidly as shown. The metal A has higher coefficient of expansion compared to that of metal B. When the bimetallic strip is placed in a cold bath, it will :
A.
Neither bend nor shrink
B.
Bend towards the left
C.
Not bend but shrink
D.
Bend towards the right
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
In thermodynamics, heat and work are :
A.
Path functions
B.
Point functions
C.
Extensive thermodynamics state variables
D.
Intensive thermodynamic state variables
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is :
A.
${{3RT} \over V}$
B.
${{4RT} \over V}$
C.
${{88RT} \over V}$
D.
${5 \over 2}{{RT} \over V}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Evening Shift
The internal energy (U), pressure (P) and volume (V) of an ideal gas are related as U $=$ 3PV + 4. The gas is :
A.
either monoatomic or diatomic.
B.
monoatomic only.
C.
polyatomic only.
D.
diatomic only.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
The temperature $\theta$ at the junction of two insulating sheets, having thermal resistances R1 and R2 as well as top and bottom temperatures $\theta$1 and $\theta$2 (as shown in figure) is given by :
A.
${{{\theta _1}{R_2} + {\theta _2}{R_1}} \over {{R_1} + {R_2}}}$
B.
${{{\theta _1}{R_1} + {\theta _2}{R_2}} \over {{R_1} + {R_2}}}$
C.
${{{\theta _1}{R_2} - {\theta _2}{R_1}} \over {{R_2} - {R_1}}}$
D.
${{{\theta _2}{R_2} - {\theta _1}{R_1}} \over {{R_2} - {R_1}}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
Given below are two statements :
Statement I : In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution.
Statement II : In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule.
In the light of the above statements, choose the correct answer from the options given below :
Statement I : In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution.
Statement II : In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule.
In the light of the above statements, choose the correct answer from the options given below :
A.
Both Statement I and Statement II are true
B.
Both Statement I and Statement II are false
C.
Statement I is true but Statement II is false.
D.
Statement I is false but Statement II is true.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If V2 = 2V1 then the ratio of temperature T2/T1 is :
A.
$\sqrt 2 $
B.
${1 \over {\sqrt 2 }}$
C.
${1 \over 2}$
D.
2
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : When a rod lying freely is heated, no thermal stress is developed in it.
Reason R : On heating, the length of the rod increases.
In the light of the above statements, choose the correct answer from the options given below :
Assertion A : When a rod lying freely is heated, no thermal stress is developed in it.
Reason R : On heating, the length of the rod increases.
In the light of the above statements, choose the correct answer from the options given below :
A.
A is true but R is false
B.
A is false but R is true
C.
Both A and B are true but R is NOT the correct explanation of A
D.
Both A and R are true and R is the correct explanation of A
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
A diatomic gas, having ${C_p} = {7 \over 2}R$ and ${C_v} = {5 \over 2}R$, is heated at constant pressure. The ratio dU : dQ : dW :
A.
5 : 7 : 3
B.
3 : 7 : 2
C.
5 : 7 : 2
D.
3 : 5 : 2
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
If one mole of an ideal gas at (P1, V1) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B $ \to $ C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net workdone by the gas is equal to :
A.
$ - {{RT} \over {2(\gamma - 1)}}$
B.
$RT\left( {\ln 2 - {1 \over {2(\gamma - 1)}}} \right)$
C.
$RT\ln 2$
D.
$0$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
On the basis of kinetic theory of gases, the gas exerts pressure because its molecules :
A.
continuously lose their energy till it reaches wall.
B.
are attracted by the walls of container.
C.
suffer change in momentum when impinge on the walls of container.
D.
continuously stick to the walls of container.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
Match List I with List II.
Choose the correct answer from the options given below :
| List I | List II | ||
|---|---|---|---|
| (a) | Isothermal | (i) | Pressure constant |
| (b) | Isochoric | (ii) | Temperature constant |
| (c) | Adiabatic | (iii) | Volume constant |
| (d) | Isobaric | (iv) | Heat content is constant |
Choose the correct answer from the options given below :
1.
(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
2.
(a) - (ii), (b) - (iv), (c) - (iii), (d) - (i)
3.
(a) - (iii), (b) - (ii), (c) - (i), (d) - (iv)
4.
(a) - (i), (b) - (iii), (c) - (ii), (d) - (iv)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes.
A $ \to $ B : Isothermal expansion at temperature T so that the volume is doubled from V1 to V2 = 2V1 and pressure charges from P1 to P2
B $ \to $ C : Isobaric compression at pressure P2 to initial volume V1.
C $ \to $ A : Isochoric change leading to change of pressure from P2 to P1.
Total workdone in the complete cycle ABCA is :
A $ \to $ B : Isothermal expansion at temperature T so that the volume is doubled from V1 to V2 = 2V1 and pressure charges from P1 to P2
B $ \to $ C : Isobaric compression at pressure P2 to initial volume V1.
C $ \to $ A : Isochoric change leading to change of pressure from P2 to P1.
Total workdone in the complete cycle ABCA is :
A.
nRTln 2
B.
0
C.
$nRT\left( {\ln 2 - {1 \over 2}} \right)$
D.
$nRT\left( {\ln 2 + {1 \over 2}} \right)$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
Each side of a box made of metal sheet in cubic shape is 'a' at room temperature 'T', the coefficient of linear expansion of the metal sheet is '$\alpha$'. The metal sheet is heated uniformly, by a small temperature $\Delta$T, so that its new temperature is T + $\Delta$T. Calculate the increase in the volume of the metal box.
A.
3a3$\alpha$$\Delta$T
B.
4$\pi$a3$\alpha$$\Delta$T
C.
${{4 \over 3}}$$\pi$a3$\alpha$$\Delta$T
D.
4a3$\alpha$$\Delta$T
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
The temperature of 3.00 mol of an ideal diatomic gas is increased by 40.0$^\circ$C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is ${x \over {10}}$. Then the value of x (round off to the nearest integer) is ___________. (Given R = 8.31 J mol$-$1 K$-$1)
Correct Answer: 25
Explanation:
Given, the number of diatomic moles, n = 3 mol
The increase in temperature of the diatomic mole,
$\Delta$T = 40$^\circ$C
Now, the degree of freedom
f = linear + rotational + no oscillation
f = 3 + 2 + 0 $\Rightarrow$ f = 5
Change in internal energy,
$\Delta$U = nCv$\Delta$T .... (i)
where, ${C_v} = {f \over 2}R = {5 \over 2}R$
Substituting the value in Eq. (i), we get
$\Delta U = {{5R} \over 2}nR\Delta T$
Now, work done by the gas for isobaric process,
$W = p\Delta V = nR\Delta T$
The ratio of the change in internal energy to the work done by the gas,
${{\Delta U} \over W} = {{{5 \over 2}nR\Delta T} \over {nR\Delta T}}$
$ = {{\Delta U} \over W} = {5 \over 2}$
Multiply and divide the above equation with 5, we get
${{\Delta U} \over W} = {{5 \times 5} \over {2 \times 5}} = {{25} \over {10}}$
Comparing with given equation, ${{\Delta U} \over W} = {x \over {10}}$
The value of the x = 25.
The increase in temperature of the diatomic mole,
$\Delta$T = 40$^\circ$C
Now, the degree of freedom
f = linear + rotational + no oscillation
f = 3 + 2 + 0 $\Rightarrow$ f = 5
Change in internal energy,
$\Delta$U = nCv$\Delta$T .... (i)
where, ${C_v} = {f \over 2}R = {5 \over 2}R$
Substituting the value in Eq. (i), we get
$\Delta U = {{5R} \over 2}nR\Delta T$
Now, work done by the gas for isobaric process,
$W = p\Delta V = nR\Delta T$
The ratio of the change in internal energy to the work done by the gas,
${{\Delta U} \over W} = {{{5 \over 2}nR\Delta T} \over {nR\Delta T}}$
$ = {{\Delta U} \over W} = {5 \over 2}$
Multiply and divide the above equation with 5, we get
${{\Delta U} \over W} = {{5 \times 5} \over {2 \times 5}} = {{25} \over {10}}$
Comparing with given equation, ${{\Delta U} \over W} = {x \over {10}}$
The value of the x = 25.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
The average translational kinetic energy of N2 gas molecules at .............$^\circ$C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given kB = 1.38 $\times$ 10$-$23 J/K) (Fill the nearest integer).
Correct Answer: 500
Explanation:
Given, the average translational kinetic energy of dinitrogen (N2) = Kinetic energy of an electron .... (i)
Translational kinetic energy of dinitrogen (N2)
$KE = {3 \over 2}{K_B}T$
Here, T = temperature of the gas,
and KB = Boltzmann constant.
Kinetic energy of an electron = eV
Given, the potential differential of an electron, V = 0.1 V
Substituting the values in the Eq. (i), we get
${3 \over 2}{K_B}T = eV$
$ \Rightarrow {3 \over 2} \times 1.38 \times {10^{ - 23}} \times T = 1.6 \times {10^{ - 19}} \times (0.1)$
$T = 773K = 773 - 273^\circ C = 500^\circ C$
Translational kinetic energy of dinitrogen (N2)
$KE = {3 \over 2}{K_B}T$
Here, T = temperature of the gas,
and KB = Boltzmann constant.
Kinetic energy of an electron = eV
Given, the potential differential of an electron, V = 0.1 V
Substituting the values in the Eq. (i), we get
${3 \over 2}{K_B}T = eV$
$ \Rightarrow {3 \over 2} \times 1.38 \times {10^{ - 23}} \times T = 1.6 \times {10^{ - 19}} \times (0.1)$
$T = 773K = 773 - 273^\circ C = 500^\circ C$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
A sample of gas with $\gamma$ = 1.5 is taken through an adiabatic process in which the volume is compressed from 1200 cm3 to 300 cm3. If the initial pressure is 200 kPa. The absolute value of the workdone by the gas in the process = _____________ J.
Correct Answer: 480
Explanation:
v = 1.5
p1v1v = p2v2v
(200) (1200)1.5 = P2 (300)1.5
P2 = 200 [4]3/2 = 1600 kPa
$\left| {W.D.} \right| = {{{p_2}{v_2} - {p_1}{v_1}} \over {v - 1}} = \left( {{{480 - 240} \over {0.5}}} \right) = 480$ J
p1v1v = p2v2v
(200) (1200)1.5 = P2 (300)1.5
P2 = 200 [4]3/2 = 1600 kPa
$\left| {W.D.} \right| = {{{p_2}{v_2} - {p_1}{v_1}} \over {v - 1}} = \left( {{{480 - 240} \over {0.5}}} \right) = 480$ J
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
A heat engine operates between a cold reservoir at temperature T2 = 400 K and a hot reservoir at temperature T1. It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be ______________ K.
Correct Answer: 500
Explanation:
Qin = 300 J ; Qout = 240 J
Work done = Qin $-$ Qout = 300 $-$ 240 = 60 J
Efficiency = ${W \over {{Q_{in}}}} = {{60} \over {300}} = {1 \over 5}$
efficiency = $1 - {{{T_2}} \over {{T_1}}}$
${1 \over 5} = 1 - {{400} \over {{T_1}}} \Rightarrow {{400} \over {{T_1}}} = {4 \over 5}$
T1 = 500 k
Work done = Qin $-$ Qout = 300 $-$ 240 = 60 J
Efficiency = ${W \over {{Q_{in}}}} = {{60} \over {300}} = {1 \over 5}$
efficiency = $1 - {{{T_2}} \over {{T_1}}}$
${1 \over 5} = 1 - {{400} \over {{T_1}}} \Rightarrow {{400} \over {{T_1}}} = {4 \over 5}$
T1 = 500 k
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
A rod CD of thermal resistance 10.0 KW$-$1 is joined at the middle of an identical rod AB as shown in figure. The end A, B and D are maintained at 200$^\circ$C, 100$^\circ$C and 125$^\circ$C respectively. The heat current in CD is P watt. The value of P is ................. .
Correct Answer: 2
Explanation:
Rods are identical so
RAB = RCD = 10 Kw$-$1
C is mid-point of AB, so
RAC = RCB = 5 Kw$-$1
at point C
${{200 - T} \over 5} = {{T - 125} \over {10}} + {{T - 100} \over 5}$
2(200 $-$ T) = T $-$ 125 + 2(T $-$ 100)
400 $-$ 2T = T $-$ 125 + 2T $-$ 200
$T = {{725} \over 5}$ = 145$^\circ$C
${I_h} = {{145 - 125} \over {10}}w = {{20} \over {10}}w$
Ih = 2w
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
A system consists of two types of gas molecules A and B having same number density 2 $\times$ 1025/m3. The diameter of A and B are 10 $\mathop A\limits^o $ and 5 $\mathop A\limits^o $ respectively. They suffer collision at room temperature. The ratio of average distance covered by the molecule A to that of B between two successive collision is ____________ $\times$ 10$-$2
Correct Answer: 25
Explanation:
$\because$ mean free path
$\lambda = {1 \over {\sqrt 2 \pi {d^2}n}}$
${{{\lambda _1}} \over {{\lambda _2}}} = {{d_2^2{n_2}} \over {d_1^2{n_1}}}$
$ = {\left( {{5 \over {10}}} \right)^2} = 0.25 = 25 \times {10^{ - 2}}$
$\lambda = {1 \over {\sqrt 2 \pi {d^2}n}}$
${{{\lambda _1}} \over {{\lambda _2}}} = {{d_2^2{n_2}} \over {d_1^2{n_1}}}$
$ = {\left( {{5 \over {10}}} \right)^2} = 0.25 = 25 \times {10^{ - 2}}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
The area of cross-section of a railway track is 0.01 m2. The temperature variation is 10$^\circ$C. Coefficient of liner expansion of material of track is 10$-$5/$^\circ$C. The energy stored per meter in the track is ____________ J/m.
(Young's modulus of material of track is 1011 Nm$-$2)
(Young's modulus of material of track is 1011 Nm$-$2)
Correct Answer: 05
Explanation:
As the tracks won't be allowed to expand linearly, the rise in temperature would lead to developing thermal stress in track.
${{(Stress)} \over y} = \alpha \Delta T$ or $\sigma = Y\alpha \Delta T$
Energy stored per unit volume = ${1 \over 2}{\sigma \over Y}$
$\Rightarrow$ Energy stored per unit length = ${{A{\sigma ^2}} \over {2Y}}$
$ = {A \over 2} \times Y{\alpha ^2}\Delta {T^2}$
$ = {{{{10}^{ - 2}} \times {{10}^{11}} \times {{10}^{ - 10}} \times 100} \over 2} = 5$ J/m
${{(Stress)} \over y} = \alpha \Delta T$ or $\sigma = Y\alpha \Delta T$
Energy stored per unit volume = ${1 \over 2}{\sigma \over Y}$
$\Rightarrow$ Energy stored per unit length = ${{A{\sigma ^2}} \over {2Y}}$
$ = {A \over 2} \times Y{\alpha ^2}\Delta {T^2}$
$ = {{{{10}^{ - 2}} \times {{10}^{11}} \times {{10}^{ - 10}} \times 100} \over 2} = 5$ J/m
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
One mole of an ideal gas at 27$^\circ$ is taken from A to B as shown in the given PV indicator diagram. The work done by the system will be _________ $\times$ 10$-$1 J. [Given : R = 8.3 J/mole K, ln2 = 0.6931] (Round off to the nearest integer)
Correct Answer: 17258
Explanation:
Process of isothermal
$W = nRT\ln \left( {{{{V_2}} \over {{V_1}}}} \right)$
= 1 $\times$ 8.3 $\times$ 300 $\times$ ln2
= 17258 $\times$ 10$-$1 J
$W = nRT\ln \left( {{{{V_2}} \over {{V_1}}}} \right)$
= 1 $\times$ 8.3 $\times$ 300 $\times$ ln2
= 17258 $\times$ 10$-$1 J
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
In the reported figure, heat energy absorbed by a system in going through a cyclic process is ___________ $\pi$J.
Correct Answer: 100
Explanation:
Consider the given diagram,
Here, r1 = 10 $\times$ 103 m and r2 = 10 $\times$ 10-3 m
We know that for a complete cyclic process, change in internal energy, ($\Delta$U) = 0 .... (i)
According to 1st law of thermodynamics,
$\Delta$Q = $\Delta$U + W ..... (ii)
From Eqs. (i) and (ii), we get
$\Rightarrow$ $\Delta$Q = 0 + W
$\Rightarrow$ $\Delta$Q = W .... (iii)
$\because$ W = Area = $\pi$r1r2
= $\pi$ $\times$ (10 $\times$ 103) $\times$ (10 $\times$ 10$-$3)
W = 100 $\pi$ J .... (iv)
From Eqs. (iii) and (iv), we get
$\Delta$Q = 100 $\pi$ J
Here, r1 = 10 $\times$ 103 m and r2 = 10 $\times$ 10-3 m
We know that for a complete cyclic process, change in internal energy, ($\Delta$U) = 0 .... (i)
According to 1st law of thermodynamics,
$\Delta$Q = $\Delta$U + W ..... (ii)
From Eqs. (i) and (ii), we get
$\Rightarrow$ $\Delta$Q = 0 + W
$\Rightarrow$ $\Delta$Q = W .... (iii)
$\because$ W = Area = $\pi$r1r2
= $\pi$ $\times$ (10 $\times$ 103) $\times$ (10 $\times$ 10$-$3)
W = 100 $\pi$ J .... (iv)
From Eqs. (iii) and (iv), we get
$\Delta$Q = 100 $\pi$ J
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
For an ideal heat engine, the temperature of the source is 127$^\circ$C. In order to have 60% efficiency the temperature of the sink should be ___________$^\circ$C. (Round off to the Nearest Integer)
Correct Answer: -113
Explanation:
Given, temperature of source, TH = 127$^\circ$C
= 273 + 127 = 400 K
Efficiencies, $\eta $ = 60% = 0.6
The efficiency of Carnot (ideal) heat engine is given by
$\eta = \left( {1 - {{{T_L}} \over {{T_H}}}} \right)$
where, TL = temperature of sink,
$0.6 = \left( {1 - {{{T_L}} \over {{T_H}}}} \right) \Rightarrow {{{T_L}} \over {{T_H}}} = 1 - 0.6$
${{{T_L}} \over {{T_H}}} = 0.4 \Rightarrow {T_L} = 0.4 \times {T_H}$
$ = 0.4 \times 400 = 160K$
$ = 160 - 273 = - 113^\circ C$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
1 mole of rigid diatomic gas performs a work of ${Q \over 5}$ when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is ${xR \over 8}$. The value of x is _________. [R = universal gas constant]
Correct Answer: 25
Explanation:
From thermodynamics law:
$Q = \Delta U + W$
$Q = \Delta U + {Q \over 5}$
$\Delta U = {{4Q} \over 5}$
$n{C_v}\Delta T = {4 \over 5}nC\Delta T$
${5 \over 4}{C_v} = C$
$C = {5 \over 4}\left( {{f \over 2}} \right)R = {5 \over 4}\left( {{5 \over 2}} \right)R$
$C = {{25} \over 8}R$
$X = 25$
$Q = \Delta U + W$
$Q = \Delta U + {Q \over 5}$
$\Delta U = {{4Q} \over 5}$
$n{C_v}\Delta T = {4 \over 5}nC\Delta T$
${5 \over 4}{C_v} = C$
$C = {5 \over 4}\left( {{f \over 2}} \right)R = {5 \over 4}\left( {{5 \over 2}} \right)R$
$C = {{25} \over 8}R$
$X = 25$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
The volume V of a given mass of monoatomic gas changes with temperature T according to the relation $V = K{T^{{2 \over 3}}}$. The workdone when temperature changes by 90K will be xR. The value of x is _________. [R = universal gas constant]
Correct Answer: 60
Explanation:
We know that work done is
$W = \int {PdV} $ .... (1)
$ \Rightarrow P = {{nRT} \over V}$ .... (2)
$ \Rightarrow W = \int {{{nRT} \over V}dv} $ .... (3)
and given $V = K{T^{2/3}}$ .... (4)
$ \Rightarrow W = \int {{{nRT} \over {K{T^{2/3}}}}.dv} $ .... (5)
$ \Rightarrow $ from (4) : $dv = {2 \over 3}K{T^{ - 1/3}}dT$
$ \Rightarrow W = \int\limits_{{T_1}}^{{T_2}} {{{nRT} \over {K{T^{2/3}}}}{2 \over 3}K{1 \over {{T^{1/3}}}}} dT$
$ \Rightarrow W = {2 \over 3}nR \times \left( {{T_2} - {T_1}} \right)$ .... (6)
$ \Rightarrow {T_2} - {T_1} = 90K$ .... (7)
$ \Rightarrow W = {2 \over 3}nR \times 90$
$ \Rightarrow W = 60nR$
Assuming 1 mole of gas
n = 1
So, W = 60R
$W = \int {PdV} $ .... (1)
$ \Rightarrow P = {{nRT} \over V}$ .... (2)
$ \Rightarrow W = \int {{{nRT} \over V}dv} $ .... (3)
and given $V = K{T^{2/3}}$ .... (4)
$ \Rightarrow W = \int {{{nRT} \over {K{T^{2/3}}}}.dv} $ .... (5)
$ \Rightarrow $ from (4) : $dv = {2 \over 3}K{T^{ - 1/3}}dT$
$ \Rightarrow W = \int\limits_{{T_1}}^{{T_2}} {{{nRT} \over {K{T^{2/3}}}}{2 \over 3}K{1 \over {{T^{1/3}}}}} dT$
$ \Rightarrow W = {2 \over 3}nR \times \left( {{T_2} - {T_1}} \right)$ .... (6)
$ \Rightarrow {T_2} - {T_1} = 90K$ .... (7)
$ \Rightarrow W = {2 \over 3}nR \times 90$
$ \Rightarrow W = 60nR$
Assuming 1 mole of gas
n = 1
So, W = 60R
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
A container is divided into two chambers by a partition. The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contain 4.0 moles of gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x $\times$ 10$-$1 atm. Value of x is ________.
Correct Answer: 25
Explanation:
By energy conservation
${3 \over 2}{n_1}R{T_1} + {3 \over 2}{n_2}R{T_2} = {3 \over 2}({n_1} + {n_2})RT$
Using PV = nRT
P1V1 + P2V2 = P(V1 + V2)
$P = {{{P_1}{V_1} + {P_2}{V_2}} \over {{V_1} + {V_2}}} = {{2 \times 4.5 + 3 \times 5.5} \over {4.5 + 5.5}}$
$P = {{9 + 16.5} \over {10}} = {{25.5} \over {10}}$
$ \approx 25 \times {10^{ - 1}}$ atm
${3 \over 2}{n_1}R{T_1} + {3 \over 2}{n_2}R{T_2} = {3 \over 2}({n_1} + {n_2})RT$
Using PV = nRT
P1V1 + P2V2 = P(V1 + V2)
$P = {{{P_1}{V_1} + {P_2}{V_2}} \over {{V_1} + {V_2}}} = {{2 \times 4.5 + 3 \times 5.5} \over {4.5 + 5.5}}$
$P = {{9 + 16.5} \over {10}} = {{25.5} \over {10}}$
$ \approx 25 \times {10^{ - 1}}$ atm
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52K, its efficiency is doubled. The temperature in Kelvin of the source will be __________.
Correct Answer: 208
Explanation:
$\because$ $n = {w \over {{Q_{in}}}} = {1 \over 4}$
${1 \over 4} = 1 - {{{T_1}} \over {{T_2}}}$
${{{T_1}} \over {{T_2}}} = {3 \over 4}$
When the temperature of the sink is reduced by 52K then its efficiency is doubled.
${1 \over 2} = 1 - {{\left( {{T_1} - 52} \right)} \over {{T_2}}}$
${{{T_1} - 52} \over {{T_2}}} = {1 \over 2}$
${{{T_1}} \over {{T_2}}}{{ - 52} \over {{T_2}}} = {1 \over 2}$
${3 \over 4} - {{52} \over {{T_2}}} = {1 \over 2}$
${{52} \over {{T_2}}} = {1 \over 4}$
${T_2} = 208$ K
${1 \over 4} = 1 - {{{T_1}} \over {{T_2}}}$
${{{T_1}} \over {{T_2}}} = {3 \over 4}$
When the temperature of the sink is reduced by 52K then its efficiency is doubled.
${1 \over 2} = 1 - {{\left( {{T_1} - 52} \right)} \over {{T_2}}}$
${{{T_1} - 52} \over {{T_2}}} = {1 \over 2}$
${{{T_1}} \over {{T_2}}}{{ - 52} \over {{T_2}}} = {1 \over 2}$
${3 \over 4} - {{52} \over {{T_2}}} = {1 \over 2}$
${{52} \over {{T_2}}} = {1 \over 4}$
${T_2} = 208$ K
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
A monoatomic gas of mass 4.0 u is kept in an insulated container. Container is moving with velocity 30 m/s. If container is suddenly stopped then change in temperature of the gas (R = gas constant) is ${x \over {3R}}$. Value of x is ___________.
Correct Answer: 3600
Explanation:
To find the change in temperature of a monoatomic gas when the container it is in is suddenly stopped, we can follow these steps:
Relationship Between Kinetic Energy and Internal Energy Change:
The kinetic energy lost by the container when it stops is converted into the internal energy of the gas:
$ \Delta KE = \Delta U $
Expression for Internal Energy Change:
The change in internal energy can be expressed in terms of the change in temperature:
$ \Delta U = n C_v \Delta T $
where $ n $ is the number of moles, $ C_v $ is the molar heat capacity at constant volume, and $ \Delta T $ is the change in temperature.
Using the Kinetic Energy Formula for the Change:
For a monoatomic gas, the molar heat capacity at constant volume $ C_v $ is $\frac{3R}{2}$. Thus, we equate the kinetic energy to the change in internal energy:
$ \frac{1}{2} m v^2 = \frac{3}{2} n R \Delta T $
Solving for $\Delta T$:
Rearrange the equation to solve for the change in temperature:
$ \Delta T = \frac{m v^2}{3 n R} $
Substituting Given Values:
Here, mass $ m = 4 $ u, velocity $ v = 30 $ m/s, and the number of moles $ n = 1 $. Substitute these into the equation:
$ \Delta T = \frac{4 \times (30)^2}{3 \times 1 \times R} $
Simplifying:
$ \Delta T = \frac{3600}{R} $
Relating to Given Expression:
According to the problem, $\Delta T = \frac{x}{3R}$. Therefore, equating:
$ \frac{x}{3R} = \frac{3600}{R} $
Solve for $x$:
Simplifying gives:
$ x = 3600 $
Thus, the value of $x$ is 3600.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
In a certain thermodynamical process, the pressure of a gas depends on its volume as kV3. The work done when the temperature changes from 100$^\circ$C to 300$^\circ$C will be ___________ nR, where n denotes number of moles of a gas.
Correct Answer: 50
Explanation:
In this thermodynamical process, the pressure $ P $ of a gas is related to its volume $ V $ by the equation $ P = kV^3 $. Thus, we have:
$ PV^{-3} = k $
This implies that the value of $ x $ is $-3$ in the relationship. To find the work done ($ W $) when the temperature changes from 100°C to 300°C, we can use the formula:
$ W = \frac{nR(T_1 - T_2)}{x - 1} $
Substitute the given temperatures and the value of $ x $:
$ W = \frac{nR(100 - 300)}{-3 - 1} $
Calculate the expression:
$ W = \frac{nR(-200)}{-4} $
Which simplifies to:
$ W = 50 \, nR $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
The root mean square speed of molecules of a given mass of a gas at 27$^\circ$C and 1 atmosphere pressure is 200 ms$-$1. The root mean square speed of molecules of the gas at 127$^\circ$C and 2 atmosphere pressure is ${{x \over {\sqrt 3 }}}$ ms$-$1. The value of x will be _________.
Correct Answer: 400
Explanation:
Given, T1 = 27$^\circ$C = 27 + 273 = 300K, p1 = 1 atm, v1 = 200 ms$-$1, T2 = 127$^\circ$C = 400 K, p2 = 12 atm, v2 = ?
As we know that,
Root mean square speed, ${v_{rms}} = \sqrt {{{3RT} \over m}} $
$\therefore$ ${{{v_1}} \over {{v_2}}} = \sqrt {{{{T_1}} \over {{T_2}}}} = \sqrt {{{300} \over {400}}} = \sqrt {{3 \over 4}} $
$ \Rightarrow {v_2} = \sqrt {{4 \over 3}} {v_1} = {2 \over {\sqrt 3 }} \times 200 = {{400} \over {\sqrt 3 }}$ ms$-$1
$ \Rightarrow {x \over {\sqrt 3 }} = {{400} \over {\sqrt 3 }} \Rightarrow x = 400$
As we know that,
Root mean square speed, ${v_{rms}} = \sqrt {{{3RT} \over m}} $
$\therefore$ ${{{v_1}} \over {{v_2}}} = \sqrt {{{{T_1}} \over {{T_2}}}} = \sqrt {{{300} \over {400}}} = \sqrt {{3 \over 4}} $
$ \Rightarrow {v_2} = \sqrt {{4 \over 3}} {v_1} = {2 \over {\sqrt 3 }} \times 200 = {{400} \over {\sqrt 3 }}$ ms$-$1
$ \Rightarrow {x \over {\sqrt 3 }} = {{400} \over {\sqrt 3 }} \Rightarrow x = 400$
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 2 Online
A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially the bottle is sealed at atmosphere pressure p0 = 105 Pa so that the volume of the trapped air is v0 = 3.3 cc. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure p0 + $\Delta$p without changing its orientation. At this pressure, the volume of the trapped air is v0 $-$ $\Delta$v. Let $\Delta$v = X cc and $\Delta$p = Y $\times$ 103 Pa.
The value of X is _______________.
The value of X is _______________.
Correct Answer: 0.3
Explanation:
When the tube just sinks/floats, then average density = density of water
${{Mass} \over {total\,volume}}$ = 1 gm/cc
$ \Rightarrow {{5gm} \over {total\,volume}}$ = 1 gm/cc
$\Rightarrow$ Total volume = 5 cc
$\Rightarrow$ Volume of tube + final volume of air in the tube = 5 cc
$ \Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$
$\Rightarrow$ Vf = 5 $-$ 2 = 3 cc
$\Rightarrow$ $\Delta$V = 0.3 cc
${{Mass} \over {total\,volume}}$ = 1 gm/cc
$ \Rightarrow {{5gm} \over {total\,volume}}$ = 1 gm/cc
$\Rightarrow$ Total volume = 5 cc
$\Rightarrow$ Volume of tube + final volume of air in the tube = 5 cc
$ \Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$
$\Rightarrow$ Vf = 5 $-$ 2 = 3 cc
$\Rightarrow$ $\Delta$V = 0.3 cc
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 2 Online
A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially the bottle is sealed at atmosphere pressure p0 = 105 Pa so that the volume of the trapped air is v0 = 3.3 cc. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure p0 + $\Delta$p without changing its orientation. At this pressure, the volume of the trapped air is v0 $-$ $\Delta$v. Let $\Delta$v = X cc and $\Delta$p = Y $\times$ 103 Pa.
The value of Y is _______________.
The value of Y is _______________.
Correct Answer: 10
Explanation:
When the tube just sinks/floats, then average density = density of water
${{Mass} \over {total\,volume}}$ = 1 gm/cc
$ \Rightarrow {{5gm} \over {total\,volume}}$ = 1 gm/cc
$\Rightarrow$ Total volume = 5 cc
$\Rightarrow$ Volume of tube + final volume of air in the tube = 5 cc
$ \Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$
$\Rightarrow$ Vf = 5 $-$ 2 = 3 cc
$\Rightarrow$ $\Delta$V = 0.3 cc
For isothermal process,
${p_i}{V_i} = {p_f}{V_f} \Rightarrow {p_f} = {10^5} \times {{3.3} \over 3}$
pf = 1.1 $\times$ 105
pf $-$ pi = 1.1 $\times$ 105 $-$ 105
= 0.1 $\times$ 105 = 10 $\times$ 103 Pa $\Rightarrow$ Y = 10
${{Mass} \over {total\,volume}}$ = 1 gm/cc
$ \Rightarrow {{5gm} \over {total\,volume}}$ = 1 gm/cc
$\Rightarrow$ Total volume = 5 cc
$\Rightarrow$ Volume of tube + final volume of air in the tube = 5 cc
$ \Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$
$\Rightarrow$ Vf = 5 $-$ 2 = 3 cc
$\Rightarrow$ $\Delta$V = 0.3 cc
For isothermal process,
${p_i}{V_i} = {p_f}{V_f} \Rightarrow {p_f} = {10^5} \times {{3.3} \over 3}$
pf = 1.1 $\times$ 105
pf $-$ pi = 1.1 $\times$ 105 $-$ 105
= 0.1 $\times$ 105 = 10 $\times$ 103 Pa $\Rightarrow$ Y = 10
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 1 Online
A small object is placed at the center of a large evacuated hollow spherical container. Assume that the container is maintained at 0 K. At time t = 0, the temperature of the object is 200 K. The temperature of the object becomes 100 K at t = t1 and 50 K at t = t2. Assume the object and the container to be ideal black bodies. The heat capacity of the object does not depend on temperature. The ratio (t2/t1) is ____________.
Correct Answer: 9
Explanation:
Ts = 0 K
Ti = 200 K, e = 1
$ - Ms{{dT} \over {dt}} = {{dQ} \over {dt}} = \sigma eA{T^4}$
$ - {{dT} \over {dt}} = {{\sigma A{T^4}} \over {Ms}}$
${{\sigma A} \over {Ms}}\int\limits_{{t_i}}^{{t_f}} {dt = - \int\limits_{{T_i}}^{{T_f}} {{{dT} \over {{T^4}}}} } $
${{\sigma A} \over {Ms}}({t_f} - {t_i}) = {1 \over 3}\left( {{1 \over {T_f^3}} - {1 \over {T_i^3}}} \right)$
${{{{\sigma A} \over {Ms}}({t_1} - 0) = {1 \over 3}\left( {{1 \over {{{(100)}^3}}} - {1 \over {{{(200)}^3}}}} \right)} \over {{{\sigma A} \over {Ms}}({t_2} - 0) = {1 \over 3}\left( {{1 \over {{{(50)}^3}}} - {1 \over {{{(200)}^3}}}} \right)}}$
${{{t_1}} \over {{t_2}}} = {{{{{{(200)}^3} - {{(100)}^3}} \over {{{(100)}^3}{{(200)}^3}}}} \over {{{{{(200)}^3} - {{(50)}^3}} \over {{{(50)}^3}{{(200)}^3}}}}}$
$\therefore$ ${{{t_2}} \over {{t_1}}} = {9 \over 1}$
Ti = 200 K, e = 1
$ - Ms{{dT} \over {dt}} = {{dQ} \over {dt}} = \sigma eA{T^4}$
$ - {{dT} \over {dt}} = {{\sigma A{T^4}} \over {Ms}}$
${{\sigma A} \over {Ms}}\int\limits_{{t_i}}^{{t_f}} {dt = - \int\limits_{{T_i}}^{{T_f}} {{{dT} \over {{T^4}}}} } $
${{\sigma A} \over {Ms}}({t_f} - {t_i}) = {1 \over 3}\left( {{1 \over {T_f^3}} - {1 \over {T_i^3}}} \right)$
${{{{\sigma A} \over {Ms}}({t_1} - 0) = {1 \over 3}\left( {{1 \over {{{(100)}^3}}} - {1 \over {{{(200)}^3}}}} \right)} \over {{{\sigma A} \over {Ms}}({t_2} - 0) = {1 \over 3}\left( {{1 \over {{{(50)}^3}}} - {1 \over {{{(200)}^3}}}} \right)}}$
${{{t_1}} \over {{t_2}}} = {{{{{{(200)}^3} - {{(100)}^3}} \over {{{(100)}^3}{{(200)}^3}}}} \over {{{{{(200)}^3} - {{(50)}^3}} \over {{{(50)}^3}{{(200)}^3}}}}}$
$\therefore$ ${{{t_2}} \over {{t_1}}} = {9 \over 1}$
2021
JEE Advanced
MCQ
JEE Advanced 2021 Paper 2 Online
The value of ${{{T_R}} \over {{T_0}}}$ is
A.
$\sqrt 2 $
B.
$\sqrt 3 $
C.
2
D.
3
2021
JEE Advanced
MCQ
JEE Advanced 2021 Paper 2 Online
The value of ${Q \over {R{T_0}}}$ is
A.
$4(2\sqrt 2 + 1)$
B.
$4(2\sqrt 2 - 1)$
C.
$(5\sqrt 2 + 1)$
D.
$(5\sqrt 2 - 1)$
2021
JEE Advanced
MCQ
JEE Advanced 2021 Paper 1 Online
An ideal gas undergoes a four step cycle as shown in the P-V diagram below. During this cycle, heat is absorbed by the gas in
A.
steps 1 and 2
B.
steps 1 and 3
C.
steps 1 and 4
D.
steps 2 and 4
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 20th August Evening Shift
A glass vessel of volume $V_o$ is completely filled with a liquid and its temperature is raised by $\Delta T$. What volume of the liquid will flow over, if the coefficient of linear expansion of glass is $\alpha_g$ and coefficient of volume expansion of the liquid is $\gamma_l$ ?
A.
$V_0 \Delta T\left(\gamma_1-3 \alpha_g\right)$
B.
$V_0 \Delta T\left(3 \alpha_g-\gamma_1\right)$
C.
$\left(\lambda_l-3 \alpha_g\right) \Delta T$
D.
$\left(3 \alpha_g-\gamma_1\right) \Delta T$
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 20th August Evening Shift
A Carnot engine whose heat sink is at 27$^\circ$C has an efficiency of 40%. By how much should its source temperature be changed, so as to increase its efficiency to 60%?
A.
250 K
B.
100 K
C.
500 K
D.
350 K