Heat and Thermodynamics
486 Questions
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$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
In a dilute gas at pressure P and temperature T, the mean time between successive collisions of a
molecule varies with T as :
A.
$\sqrt T $
B.
T
C.
${1 \over T}$
D.
${1 \over {\sqrt T }}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
Three rods of identical cross-section and lengths are made of three different materials of thermal
conductivity K1
, K2
and K3
, respecrtively. They are joined together at their ends to make a long rod
(see figure). One end of the long rod is maintained at 100oC and the other at 0oC (see figure). If
the joints of the rod are at 70oC and 20oC in steady state and there is no loss of energy from the
surface of the rod, the correct relationship between K1
, K2
and K3
is :
A.
K1 : K3 = 2 : 3,
K2 : K3 = 2 : 5
K2 : K3 = 2 : 5
B.
K1 < K2 < K3
C.
K1 : K2 = 5 : 2,
K1 : K3 = 3 : 5
K1 : K3 = 3 : 5
D.
K1 > K2 > K3
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
Molecules of an ideal gas are known to have three translational degrees of freedom and two
rotational degrees of freedom.The gas is maintained at a temperature of T. The total internal
energy, U of a mole of this gas, and the value of
$\gamma \left( { = {{{C_p}} \over {{C_v}}}} \right)$ are given, respectively by:
$\gamma \left( { = {{{C_p}} \over {{C_v}}}} \right)$ are given, respectively by:
A.
U = ${5 \over 2}RT$ and $\gamma = {7 \over 5}$
B.
U = 5RT and $\gamma = {6 \over 5}$
C.
U = 5RT and $\gamma = {7 \over 5}$
D.
U = ${5 \over 2}RT$ and $\gamma = {6 \over 5}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
In an adiabatic process, the density of a
diatomic gas becomes 32 times its initial value.
The final pressure of the gas is found to be n
times the initial pressure. The value of n is :
A.
128
B.
32
C.
326
D.
${1 \over {32}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
Two different wires having lengths L1 and L2,
and respective temperature coefficient of linear
expansion $\alpha $1 and $\alpha $2, are joined end-to-end.
Then the effective temperature coefficient of
linear expansion is :
A.
$2\sqrt {{\alpha _1}{\alpha _2}} $
B.
$4{{{\alpha _1}{\alpha _2}} \over {{\alpha _1} + {\alpha _2}}}{{{L_2}{L_1}} \over {{{\left( {{L_2} + {L_1}} \right)}^2}}}$
C.
${{{\alpha _1} + {\alpha _2}} \over 2}$
D.
${{{\alpha _1}{L_1} + {\alpha _2}{L_2}} \over {{L_1} + {L_2}}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
Three different processes that can occur in an
ideal monoatomic gas are shown in the P vs V
diagram. The paths are labelled as A $ \to $ B,
A $ \to $ C and A $ \to $ D. The change in internal
energies during these process are taken as
EAB, EAC and EAD and the work done as WAB,
WAC and WAD.
The correct relation between these parameters are :
The correct relation between these parameters are :
A.
EAB < EAC < EAD, WAB > 0, WAC > WAD
B.
EAB = EAC = EAD, WAB > 0, WAC = 0, WAD < 0
C.
EAB > EAC > EAD, WAB < WAC < WAD
D.
EAB = EAC < EAD, WAB > 0, WAC = 0, WAD < 0
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
Number of molecules in a volume of 4 cm3 of
a perfect monoatomic gas at some temperature
T and at a pressure of 2 cm of mercury is close
to?
(Given, mean kinetic energy of a molecule
(at T) is 4 $ \times $ 10–14 erg, g = 980 cm/s2, density of
mercury = 13.6 g/cm3)
(Given, mean kinetic energy of a molecule
(at T) is 4 $ \times $ 10–14 erg, g = 980 cm/s2, density of
mercury = 13.6 g/cm3)
A.
5.8 $ \times $ 1018
B.
4.0 $ \times $ 1016
C.
5.8 $ \times $ 1016
D.
4.0 $ \times $ 1018
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
A bullet of mass 5 g, travelling with a speed of
210 m/s, strikes a fixed wooden target. One half
of its kinetic energy is converted into heat in
the bullet while the other half is converted into
heat in the wood. The rise of temperature of
the bullet if the specific heat of its material is
0.030 cal/(g – oC) (1 cal = 4.2 × 107 ergs) close
to :
A.
87.5 oC
B.
83.3 oC
C.
38.4 oC
D.
119.2 oC
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
Match the thermodynamic processes taking place in a system with the correct conditions. In the
table : $\Delta $Q is the heat supplied, $\Delta $W is the work done and $\Delta $U is change in internal energy of the
system.
| Process | Condition |
|---|---|
| (I) Adiabatic | (1) $\Delta $W = 0 |
| (II) Isothermal | (2) $\Delta $Q = 0 |
| (III) Isochoric | (3) $\Delta $U $ \ne $ 0, $\Delta $W $ \ne $ 0, $\Delta $Q $ \ne $ 0 |
| (IV) Isobaric | (4) $\Delta $U = 0 |
A.
(I) - (1), (II) - (1), (III) - (2), (IV) - (3)
B.
(I) - (2), (II) - (4), (III) - (1), (IV) - (3)
C.
(I) - (1), (II) - (2), (III) - (4), (IV) - (4)
D.
(I) - (2), (II) - (1), (III) - (4), (IV) - (3)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
The specific heat of water
= 4200 J kg-1K-1 and the latent heat of
ice = 3.4 $ \times $ 105 J kg–1. 100 grams of ice at
0oC is placed in 200 g of water at 25oC. The
amount of ice that will melt as the temperature
of water reaches 0oC is close to (in grams) :
= 4200 J kg-1K-1 and the latent heat of
ice = 3.4 $ \times $ 105 J kg–1. 100 grams of ice at
0oC is placed in 200 g of water at 25oC. The
amount of ice that will melt as the temperature
of water reaches 0oC is close to (in grams) :
A.
63.8
B.
61.7
C.
69.3
D.
64.6
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
Match the ${{{C_P}} \over {{C_V}}}$ ratio for ideal gases with different type of molecules :
| Molecule Type | CP/CV |
|---|---|
| (A) Monatomic | (I) 7/5 |
| (B) Diatomic rigid molecules | (II) 9/7 |
| (C) Diatomic non-rigid molecules | (III) 4/3 |
| (D) Triatomic rigid molecules | (IV) 5/3 |
A.
(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
B.
(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
C.
(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
D.
(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
To raise the temperature of a certain mass of gas by 50oC at a constant pressure, 160 calories of
heat is required. When the same mass of gas is cooled by 100oC at constant volume, 240 calories
of heat is released. How many degrees of freedom does each molecule of this gas have (assume
gas to be ideal)?
A.
6
B.
7
C.
5
D.
3
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
A calorimeter of water equivalent 20 g contains 180 g of water at 25oC. ‘m’ grams of steam at
100oC is mixed in it till the temperature of the mixure is 31oC. The value of ‘m’ is close to :
(Latent heat of water = 540 cal g–1, specific heat of water = 1 cal g–1 oC–1)
(Latent heat of water = 540 cal g–1, specific heat of water = 1 cal g–1 oC–1)
A.
2.6
B.
2
C.
4
D.
3.2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
A balloon filled with helium (32oC and 1.7 atm.)
bursts. Immediately afterwards the expansion
of helium can be considered as
A.
Irreversible adiabatic
B.
Reversible adiabatic
C.
Irreversible isothermal
D.
Reversible isothermal
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
Consider a gas of triatomic molecules. The
molecules are assumed to be triangular and
made of massless rigid rods whose vertices
are occupied by atoms. The internal energy of
a mole of the gas at temperature T is :
A.
${3 \over 2}RT$
B.
${9 \over 2}RT$
C.
${5 \over 2}RT$
D.
3RT
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
A heat engine is involved with exchange of heat
of 1915 J, – 40J, + 125 J and –Q J, during one
cycle achieving an efficiency of 50.0%. The
value of Q is
A.
980 J
B.
40 J
C.
400 J
D.
640 J
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
When the temperature of a metal wire is
increased from 0oC to 10oC, its length
increases by 0.02%. The percentage change in
its mass density will be closest to :
A.
0.008
B.
0.06
C.
0.8
D.
2.3
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
An ideal gas in a closed container is slowly
heated. As its temperature increases, which of
the following statements are true?
(A) the mean free path of the molecules decreases.
(B) the mean collision time between the molecules decreases.
(C) the mean free path remains unchanged.
(D) the mean collision time remains unchanged.
(A) the mean free path of the molecules decreases.
(B) the mean collision time between the molecules decreases.
(C) the mean free path remains unchanged.
(D) the mean collision time remains unchanged.
A.
(C) and (D)
B.
(A) and (D)
C.
(B) and (C)
D.
(A) and (B)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
A gas mixture consists of 3 moles of oxygen
and 5 moles of argon at temperature T.
Assuming the gases to be ideal and the oxygen
bond to be rigid, the total internal energy (in
units of RT) of the mixture is :
A.
11
B.
20
C.
15
D.
13
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
Two gases-argon (atomic radius 0.07 nm,
atomic weight 40) and xenon (atomic radius
0.1 nm, atomic weight 140) have the same
number density and are at the same
temperature. The raito of their respective mean
free times is closest to :
A.
2.3
B.
1.83
C.
4.67
D.
3.67
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
Consider two ideal diatomic gases A and B at
some temperature T. Molecules of the gas A are
rigid, and have a mass m. Molecules of the gas
B have an additional vibrational mode, and
have a mass ${m \over 4}$
. The ratio of the specific heats ($C_V^A$ and $C_V^B$ ) of gas A and B, respectively is :
A.
7 : 9
B.
5 : 7
C.
3 : 5
D.
5 : 9
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
Which of the following is an equivalent cyclic
process corresponding to the thermodynamic cyclic
given in the figure ? where, 1 $ \to $ 2 is adiabatic.
(Graphs are schematic and are not to scale)
(Graphs are schematic and are not to scale)
A.
B.
C.
D.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
Consider a mixture of n moles of helium gas
and 2n moles of oxygen gas (molecules taken
to be rigid) as an ideal gas. Its CP/CV value
will be :
A.
23/15
B.
67/45
C.
40/27
D.
19/13
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
A carnot engine having an efficiency of ${1 \over {10}}$ is
being used as a refrigerator. If the work done
on the refrigerator is 10 J, the amount of
heat absorbed from the reservoir at lower
temperature is :
A.
90 J
B.
99 J
C.
1 J
D.
100 J
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
A thermodynamic cycle xyzx is shown on a V-T diagram.
The P-V diagram that best describes this cycle
is :
(Diagrams are schematic and not to scale)
The P-V diagram that best describes this cycle
is :(Diagrams are schematic and not to scale)
A.
B.
C.
D.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
The plot that depicts the behavior of the mean
free time t (time between two successive
collisions) for the molecules of an ideal gas, as
a function of temperature (T), qualitatively, is:
(Graphs are schematic and not drawn to scale)
(Graphs are schematic and not drawn to scale)
A.
B.
C.
D.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other
engine to produce work) between temperature, T1
and T2
. The temperature of the hot reservoir of
the first engine is T1
and the temperature of the cold reservoir of the second engine is T2
. T is
temperature of the sink of first engine which is also the source for the second which is also the
source for the second engine. How is T related to T1
and T2
. If both engines perform equal amount
of work?
A.
$T = {{2{T_1}{T_2}} \over {{T_1} + {T_2}}}$
B.
$T = \sqrt {{T_1}{T_2}} $
C.
$T = {{{T_1} + {T_2}} \over 2}$
D.
T = 0
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean
collision time between the gas molecule changes from ${\tau _1}$
to ${\tau _2}$
. If ${{{C_p}} \over {{C_v}}} = \gamma $ for this gas then a good
estimate for ${{{\tau _2}} \over {{\tau _1}}}$
is given by :
A.
${\left( 2 \right)^{{{1 + \gamma } \over 2}}}$
B.
2
C.
${\left( {{1 \over 2}} \right)^{{{1 + \gamma } \over 2}}}$
D.
${\left( {{1 \over 2}} \right)^\gamma }$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
Two moles of an ideal gas with ${{{C_P}} \over {{C_V}}} = {5 \over 3}$
are mixed with 3 moles of another ideal gas
with ${{{C_P}} \over {{C_V}}} = {4 \over 3}$. The value of ${{{C_P}} \over {{C_V}}}$ for the
mixture is :
A.
1.50
B.
1.45
C.
1.47
D.
1.42
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
A litre of dry air at STP expands adiabatically to a volume of 3 litres. If $\gamma $ = 1.40, the work done by air is : (31.4 = 4.6555) [Take air to be an ideal gas]
A.
60.7 J
B.
100.8 J
C.
90.5 J
D.
48 J
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Evening Slot
An engine operates by taking a monoatomic ideal gas through the cycle shown in the figure. The
percentage efficiency of the engine is close to __________.
Correct Answer: 19
Explanation:
% efficiency of carnot engine $\eta $ = ${W \over {{Q_{in}}}} \times 100$
Work = Area of ABCD = (2P0)(V0)
Qin = QAB + QBC
QAB = isochoric process
= nCV(TB - TA)
= 1 $ \times $ ${3 \over 2}R\left( {{T_B} - {T_A}} \right)$
= ${3 \over 2}R\left( {{{3{P_0}{V_0}} \over R} - {{{P_0}{V_0}} \over R}} \right)$
= 3P0V0
QBC = isobaric process
= nCP$\Delta $T
= $1 \times {5 \over 2}R$(TC - TB)
= ${5 \over 2}R\left( {{{6{P_0}{V_0}} \over R} - {{3{P_0}{V_0}} \over R}} \right)$
= 7.5 P0V0
$ \therefore $ $\eta $ = ${{{2{P_0}{V_0}} \over {3{P_0}{V_0} + 7.5{P_0}{V_0}}}}$ $ \times $ 100
$ \simeq $ 19%
Work = Area of ABCD = (2P0)(V0)
Qin = QAB + QBC
QAB = isochoric process
= nCV(TB - TA)
= 1 $ \times $ ${3 \over 2}R\left( {{T_B} - {T_A}} \right)$
= ${3 \over 2}R\left( {{{3{P_0}{V_0}} \over R} - {{{P_0}{V_0}} \over R}} \right)$
= 3P0V0
QBC = isobaric process
= nCP$\Delta $T
= $1 \times {5 \over 2}R$(TC - TB)
= ${5 \over 2}R\left( {{{6{P_0}{V_0}} \over R} - {{3{P_0}{V_0}} \over R}} \right)$
= 7.5 P0V0
$ \therefore $ $\eta $ = ${{{2{P_0}{V_0}} \over {3{P_0}{V_0} + 7.5{P_0}{V_0}}}}$ $ \times $ 100
$ \simeq $ 19%
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Morning Slot
Initially a gas of diatomic molecules is contained in a cylinder of volume V1
at a pressure P1
and
temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in
number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a
volume 2V1
is given by P2
. The ratio ${{{P_2}} \over {{P_1}}}$
is ________.
Correct Answer: 5
Explanation:
We know, PV = nRT
$ \therefore $ P1V1 = nR (250)
and P2(2V1) = ${{5n} \over 4}R \times \left( {2000} \right)$
By Dividing
${{{P_1}} \over {2{P_2}}}$ = ${{4 \times 250} \over {5 \times 2000}}$
$ \Rightarrow $ ${{{P_1}} \over {{P_2}}} = {1 \over 5}$
$ \Rightarrow $ ${{{P_2}} \over {{P_1}}} = 5$
$ \therefore $ P1V1 = nR (250)
and P2(2V1) = ${{5n} \over 4}R \times \left( {2000} \right)$
By Dividing
${{{P_1}} \over {2{P_2}}}$ = ${{4 \times 250} \over {5 \times 2000}}$
$ \Rightarrow $ ${{{P_1}} \over {{P_2}}} = {1 \over 5}$
$ \Rightarrow $ ${{{P_2}} \over {{P_1}}} = 5$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 5th September Evening Slot
Nitrogen gas is at 300oC temperature. The
temperature (in K) at which the rms speed of a
H2 molecule would be equal to the rms speed
of a nitrogen molecule, is _______.
(Molar mass of N2 gas 28 g).
(Molar mass of N2 gas 28 g).
Correct Answer: 40TO41
Explanation:
Vrms = $\sqrt {{{3RT} \over M}} $
VN2 = $\sqrt {{{3R(573)} \over 28}} $
VH2 = $\sqrt {{{3RT} \over 2}} $
Given, VN2 = VH2
$\sqrt {{{3RT} \over 2}} $ = $\sqrt {{{3R(573)} \over 28}} $
$ \Rightarrow $ ${T \over 2} = {{573} \over {28}}$
$ \Rightarrow $ T = 41 K
VN2 = $\sqrt {{{3R(573)} \over 28}} $
VH2 = $\sqrt {{{3RT} \over 2}} $
Given, VN2 = VH2
$\sqrt {{{3RT} \over 2}} $ = $\sqrt {{{3R(573)} \over 28}} $
$ \Rightarrow $ ${T \over 2} = {{573} \over {28}}$
$ \Rightarrow $ T = 41 K
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Evening Slot
The change in the magnitude of the volume of an ideal gas when a small additional pressure $\Delta $P is
applied at a constant temperature, is the same as the change when the temperature is reduced by
a small quantity $\Delta $T at constant pressure. The initial temperature and pressure of the gas were 300
K and 2 atm. respectively.
If |$\Delta $T| = C|$\Delta $P| then value of C in (K/atm.) is _________.
If |$\Delta $T| = C|$\Delta $P| then value of C in (K/atm.) is _________.
Correct Answer: 150
Explanation:
We know, $PV = nRT$
$ \therefore $ $P\Delta V + V\Delta P = 0$ (for constant temp.)
and $P\Delta V$ = $nR\Delta T$ (for constant pressure)
$\Delta T = {{P\Delta V} \over {nR}}$
$\Delta P = - {{P\Delta V} \over V}$ ($\Delta V$ is same in both cases)
${{\Delta T} \over {\Delta P}} = {{P\Delta V} \over {nR}}{V \over { - P\Delta V}} = {{ - V} \over {nR}} = - {T \over P}$
[As PV = nRT
$ \Rightarrow $ $ {{V \over {nR}} = {T \over P}} $]
$ \therefore $ $\left| {{{\Delta T} \over {\Delta P}}} \right| = \left| {{{ - 300} \over 2}} \right| = 150$
$ \therefore $ $P\Delta V + V\Delta P = 0$ (for constant temp.)
and $P\Delta V$ = $nR\Delta T$ (for constant pressure)
$\Delta T = {{P\Delta V} \over {nR}}$
$\Delta P = - {{P\Delta V} \over V}$ ($\Delta V$ is same in both cases)
${{\Delta T} \over {\Delta P}} = {{P\Delta V} \over {nR}}{V \over { - P\Delta V}} = {{ - V} \over {nR}} = - {T \over P}$
[As PV = nRT
$ \Rightarrow $ $ {{V \over {nR}} = {T \over P}} $]
$ \therefore $ $\left| {{{\Delta T} \over {\Delta P}}} \right| = \left| {{{ - 300} \over 2}} \right| = 150$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Morning Slot
A closed vessel contains 0.1 mole of a monoatomic ideal gas at 200 K. If 0.05 mole of the same gas
at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close
to _______.
Correct Answer: 266
Explanation:
As work done on gas and heat supplied to the gas are zero,
$ \therefore $ Total internal energy of gases remain same.
u1 + u2 = u1' + u2'
We know, $\Delta $U = nCv$\Delta $T
$ \therefore $ $\left( {0.1 \times {{3R} \over 2} \times 200} \right) + \left( {0.05 \times {{3R} \over 2} \times 400} \right)$ = $\left( {0.15 \times {{3R} \over 2} \times {T_f}} \right)$
$ \Rightarrow $ (20 + 20) = 0.15 Tf
$ \Rightarrow $ Tf = 266.67
$ \therefore $ Total internal energy of gases remain same.
u1 + u2 = u1' + u2'
We know, $\Delta $U = nCv$\Delta $T
$ \therefore $ $\left( {0.1 \times {{3R} \over 2} \times 200} \right) + \left( {0.05 \times {{3R} \over 2} \times 400} \right)$ = $\left( {0.15 \times {{3R} \over 2} \times {T_f}} \right)$
$ \Rightarrow $ (20 + 20) = 0.15 Tf
$ \Rightarrow $ Tf = 266.67