Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

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(c) For the curve 1 volume is constant so it is isochoric process. But in curve 2 and 3 curve 2 is steeper so 2 is adiabatic and 3 is isothermal.

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process 1 isochoric and process 2 is isothermal .

Since, work done = area under P-V curve . here area under the pV curve 1 is more . so work done is more when the gas expands in isochoric process.

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Coefficient of $\beta =5$

T₁= 27+273=300K

Coefficient of performance $\beta =\frac{T_2}{300-T_2}$

1500-5T₂=T₂

6T₂=1500

T₂= 250K

T₂= 250-273=-23^oC

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Temperature of the source is 27⁰C

T₁= 27+273= 300K

T₂= -3+273= 270K

Efficiency of heat engine = 1-T₂/T₁= 1-270/300=1/10

Efficiency of refrigerator  is 50% of a perfect engine

= 0.5 $*\eta$ = 1/20

Coefficient of performance of the refrigerator $\beta =\frac{Q_2}{W}=\frac{1-{\eta }^{\text{'}}}{\eta }$

= $\frac{1-1/20}{1/20}=\frac{19/20}{1/20}=19$

Q₂= $\beta W$ =19W

= 19 $*$ 1KW=19KW= 19kJ/s

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For adiabatic change process we know

P₁V₁^y= P₂V₂^y

P (V+ $?V$ )^y = (P+ $?P$ )V^y

PV^y (1+ $\frac{?V}{V}$ ) ^y=p (1+ $\frac{?P}{P}$ )V^y

PV^y (1+ $\frac{?V}{V}$ ) $\approx$ PV^y (1+ $\frac{?V}{V}$ )

Y $\frac{?V}{V}=\frac{?P}{P}$

dV= $\frac{1}{Y}\frac{V}{p}dP$

hence work done increasing the pressure from P₁ to P₂

W= ${\int }_{p1}^{p2}pdV={\int }_{p2}^{p1}p\frac{1}{Y}\frac{V}{p}dp$

= $\frac{V}{Y}\left(P_2-P_1\right)$

W= $\frac{P_2-P_1}{Y}V$

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Height of stairs h= 10m

Energy produced by burning 1 kg of fat = 7000Kcal

Energy produced by burning 5kg of fat = 5 $*7000=35000Kcal$

Energy utilised in going up and down one time

= mgh + $\frac{1}{2}mgh$ = $\frac{3}{2}mgh$

= $\frac{3}{2}*60*10*10$

= 9000J= 9000/4.2=3000/1.4cal

Number of times, the person has to go up and down the stairs

= $\frac{35*{10}^6}{\frac{3000}{1.4}}=\frac{3.5*1.4*{10}^6}{3000}$  = 16.3 $*{10}^3$ times

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Temperature of the source T₁= 500K and sink T₂= 300K

Work done W= 1000J

Efficiency of Carnot engine = 1-T₂/T₁= 1-300/500= 200/500= 2/5

Efficiency = W/Q₁

So Q₁= W/efficiency = 1000 $\left(\frac{5}{2}\right)=2500J\mathit{}$

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During driving temperature of the gas increases while volume remains constant. So according charle's law, at constant volume V.

Pressure is directly proportional to temperature. Therefore pressure of gas increases.

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Yes during adiabatic compression the temperature of a gas increases while no heat is

In adiabatic compression dQ=0

From the first law of thermodynamics dU= dQ-dW

dU=-dW

in compression work is done on the gas i.e work done is negative

dU=positive

hence internal energy of the gas increases due to which its temperature increases.

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If a refrigerator's doors is kept open, then room will become hot, because amount of heat removed would be less than the amount of heat released in the room.