Practice Start Test

Heat and Thermodynamics

NEET Physics · 97 questions · Page 4 of 10 · Click an option or "Show Solution" to reveal answer

Q31
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is
A 26.8%
B 20%
C 6.25%
D 12.5%
Correct Answer
Option A
Solution

Efficiency of ideal heat engine,

η=(1−T2T1)\eta = \left( {1 - {{{T_2}} \over {{T_1}}}} \right)

Freezing point of water = 0 o C = 273 K Boiling point of water = 100 o C = (100 + 273) K = 373 K ∴\therefore Sink temperature, T 2 = 0 o C = 0 + 273 = 273 K ∴\therefore Source temperature, T 1 = 100 o C = 100 + 273 = 373 K Percentage efficiency, %

η=(1−T2T1)×100\eta = \left( {1 - {{{T_2}} \over {{T_1}}}} \right) \times 100

=

(1−273373)×100\left( {1 - {{273} \over {373}}} \right) \times 100

=

(100373)×100\left( {{{100} \over {373}}} \right) \times 100

= 26.8%

Q32
Thermodynamic processes are indicated in the following diagram. <br><br><img src=https://neetcare-image.cdn.examgoal.net/1qpq2yck7l3pzfc/4c595905-b0aa-4f28-91c6-820e4a471e2a/5b96f890-625e-11ea-bddc-5f310cab17d0/file-1qpq2yck7l3pzfd-354w.jpg style=max-width: 100%;height: auto;display: block;margin: 0 auto; loading=lazy alt=NEET 2017 Physics - Heat and Thermodynamics Question 85 English> <br>Match the following <br><br> <table class=tg> <tbody><tr> <th class=tg-p7ly colspan=2>Column-1</th> <th class=tg-adtk></th> <th class=tg-p7ly colspan=2>Column-2</th> </tr> <tr> <td class=tg-13k7>P.</td> <td class=tg-13k7>Process I</td> <td class=tg-13k7></td> <td class=tg-13k7>A.</td> <td class=tg-13k7>Adiabatic</td> </tr> <tr> <td class=tg-60hs>Q.</td> <td class=tg-60hs>Process II</td> <td class=tg-60hs></td> <td class=tg-60hs>B.</td> <td class=tg-60hs>Isobaric</td> </tr> <tr> <td class=tg-60hs>R.</td> <td class=tg-60hs>Process III</td> <td class=tg-60hs></td> <td class=tg-60hs>C.</td> <td class=tg-60hs>Isochoric</td> </tr> <tr> <td class=tg-60hs>S.</td> <td class=tg-60hs>Process IV</td> <td class=tg-60hs></td> <td class=tg-60hs>D.</td> <td class=tg-60hs>Isothermal</td> </tr> </tbody></table>
A P → \to C, Q → \to A, R → \to D, S → \to B
B P → \to C, Q → \to D, R → \to B, S → \to A
C P → \to D, Q → \to B, R → \to A, S → \to C
D P → \to A, Q → \to C, R → \to D, S → \to B
Correct Answer
Option A
Solution

Process I volume is constant hence, it is isochoric In process IV, pressure is constant hence, it is isobaric.

Q33
A carnot engine having an efficiency of 110{1 \over {10}} as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
A 90 J
B 99 J
C 100 J
D 1 J
Correct Answer
Option A
Solution

Given, efficiency of engine,

η=110\eta = {1 \over {10}}

work done on system W = 10J Coefficient of performance of refrigerator

β=Q2W=1−ηη=1−110110=910110=9\beta = {{{Q_2}} \over W} = {{1 - \eta } \over \eta } = {{1 - {1 \over {10}}} \over {{1 \over {10}}}} = {{{9 \over {10}}} \over {{1 \over {10}}}} = 9

Energy absorbed from reservoir

Q2=βw{Q_2} = \beta w
Q2=9×10=90 J{Q_2} = 9 \times 10 = 90\,J
Q34
A gas mixture consists of 2 moles of O 2 and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is
A 15 RT
B 9 RT
C 11 RT
D 4 RT
Correct Answer
Option C
Solution

Internal energy of the system is given by

U=f2nRTU = {f \over 2}nRT

Degree of freedom F diatomic = 5 f monoatomic = 3 and, number of moles n(O 2 ) = 2 n(Ar) = 4

Utotal=52(2)RT+32(4)RT=11RT{U_{total}} = {5 \over 2}\left( 2 \right)RT + {3 \over 2}\left( 4 \right)RT = 11RT
Q35
A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas ?
A P/(kT)
B Pm/(kT)
C P/(KTV)
D mkT
Correct Answer
Option B
Solution

As PV = nRT

⇒n=PVRT=massmolar mass\Rightarrow n = {{PV} \over {RT}} = {{mass} \over {molar\,mass}}

...(i) Density, ρ\rho =

massvolume=(molar mass)PRT=(mNA)PRT{{mass} \over {volume}} = {{\left( {molar\,mass} \right)P} \over {RT}} = {{\left( {m{N_A}} \right)P} \over {RT}}

[From eqn. (i)]

ρ=mPkT\rho = {{mP} \over {kT}}
(∵R=NAk)\left( \because{R = {N_A}k} \right)
Q36
One mole of an ideal monatomic gas undergoes a process described by the equation PV 3 = constant. The heat capacity of the gas during this process is
A 32{3 \over 2} R
B 52{5 \over 2} R
C 2R
D R
Correct Answer
Option D
Solution

Process described by the equation, PV 3 = constant For a polytropic process, PV α\alpha = constant

C=CV+R1−α=32R+R1−3=RC = {C_V} + {R \over {1 - \alpha }} = {3 \over 2}R + {R \over {1 - 3}} = R
Q37
The temperature inside a refrigerator is t 2 o C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
A t1t1−t2{{{t_1}} \over {{t_1} - {t_2}}}
B t1+273t1−t2{{{t_1} + 273} \over {{t_1} - {t_2}}}
C t2+273t1−t2{{{t_2} + 273} \over {{t_1} - {t_2}}}
D t1+t2t1+273{{{t_1} + {t_2}} \over {{t_1} + 273}}
Correct Answer
Option B
Solution

Temperature inside refrigerator = t 2 °C Room temperature = t 1 °C For refrigerator,

Heat given to high temperature (Q1)Heat taken from lower temperature (Q2)=T1T2{{{\rm{Heat\,given\,to\,high\,temperature}}\,\left( {{Q_1}} \right)} \over {{\rm{Heat\,taken\,from\,lower \,temperature\, }}\left( {{Q_2}} \right)}} = {{{T_1}} \over {{T_2}}}
Q1Q2=t1+273t2+273{{{Q_1}} \over {{Q_2}}} = {{{t_1} + 273} \over {{t_2} + 273}}
⇒Q1Q1−W=t1+273t2+273\Rightarrow {{{Q_1}} \over {{Q_1} - W}} = {{{t_1} + 273} \over {{t_2} + 273}}
⇒1−WQ1=t2+273t1+273\Rightarrow 1 - {W \over {{Q_1}}} = {{{t_2} + 273} \over {{t_1} + 273}}
⇒WQ1=t1−t2t1+273\Rightarrow {W \over {{Q_1}}} = {{{t_1} - {t_2}} \over {{t_1} + 273}}

The amount of heat delivered to the room for each joule of electrical energy (W = 1 J)

Q1=t1+273t1−t2{Q_1} = {{{t_1} + 273} \over {{t_1} - {t_2}}}
Q38
A refrigerator works between 4 o C and 30 o C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take 1 cal = 4.2 Joules)
A 236.5 W
B 2365 W
C 2.365 W
D 23.65 W
Correct Answer
Option A
Solution

Now

COP=LQPower=TLTH−TLCOP = {{{L_Q}} \over {Power}} = {{{T_L}} \over {{T_H} - {T_L}}}
⇒Power=LQ×(TH−TL)TL\Rightarrow Power = {{{L_Q} \times \left( {{T_H} - {T_L}} \right)} \over {{T_L}}}

= (600 × 4.2) × 26/277 = 236.53 W

Q39
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then
A Compressing the gas isothermally or adiabatically will require the same amount of work.
B Which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.
C Compressing the gas isothermally will require more work to be done.
D Compressing the gas through adiabatic process will require more work to be done.
Correct Answer
Option D
Solution

V 1 = V, V 2 = V/2 On P-V diagram, Area under adiabatic curve > Area under isothermal curve.

So compressing the gas through adiabatic process will require more work to be done.

Q40
The molecules of a given mass of a gas have r.m.s. velocity of 2000 m s −-1 at 27 o C and 1.0 × \times 10 5 N m −-2 pressure. When the temperature and pressure of the gas are respectively, 127 o C and 0.05 × \times 10 5 N m −-2 , the r.m.s. velocity of its molecules in m s −-1 is
A 10023{{100\sqrt 2 } \over 3}
B 1003{{100} \over 3}
C 1002100\sqrt 2
D 4003{{400} \over {\sqrt 3 }}
Correct Answer
Option D
Solution

It is observed that the rms velocity of molecule is directly proportional to temperature, so

vrms∝T{v_{rms}} \propto \sqrt T
⇒vrms′=vrmsT′T\Rightarrow v_{rms}^{'} = {v_{rms}}\sqrt {{{T'} \over T}}

Hence,

vrms′=200×400300=400/3v_{rms}^{'} = 200 \times \sqrt {{{400} \over {300}}} = 400/\sqrt 3
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