Thermodynamics

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New answer posted•a year ago
S
a year ago

This is a multiple choice answer as classified in NCERT Exemplar

(b) Let us assume that T1, T2T3

According to questions there is no loss of heat in the surroundings

Heat lost by M3 = heat gained by M1+ heat gained by M2

M3s (T3-T)= M1s (T-T1)+M2s (T-T2)

T [M1+M2+M3]= M3T3+M1T1+M2T2

T = M 1 T 1 + M 2 T 2 + M 3 T 3 M 1 + M 2 + M 3

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New answer posted•a year ago
S
a year ago

This is a multiple choice answer as classified in NCERT Exemplar

(a) Container A is isothermal and container B is adiabatic

For isothermal process P1V1=P2V2

Po (2Vo)= P2 (Vo)

P2= 2Po

for adiabatic process

P1V1y= P2V2y

Po (2Vo)y=P2 (Vo)y

P2= ( 2 V o V o )yPo= 2yPo

Hence ratio of final pressure = 2 γ P o 2 P o = 2 γ - 1

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New answer posted•a year ago
S
a year ago

This is a multiple choice answer as classified in NCERT Exemplar

(b) Work done ABCD = area of rectangle ABCDA

= AB × B C = (3Vo-Vo) × (2po-po)

= 2V0 × Po= 2poVo

And work done by the gas =- 2poVo

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New answer posted•a year ago
S
a year ago

This is a multiple choice answer as classified in NCERT Exemplar

(c) As we know PV =constant

Hence we can say that gas is going through an isothermal process.

Clearly from the graph that between process 1 and 2 temperature is constant and the gas expands and pressure decreases. So density of 2 is less

...Read more

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New answer posted•a year ago
S
a year ago

This is a multiple choice answer as classified in NCERT Exemplar

(a) Amount of sweat evaporated /minute = s w e a t p r o d u c e d m i n u t e n u m b e r o f c a l o r i e s r e q u i r e d f o r e v a p o r a t i o n k g

 = 14.5 × 10 3 580 × 10 3 = 0.25 k g

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New answer posted•a year ago
S
a year ago

This is a multiple choice answer as classified in NCERT Exemplar

(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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New answer posted•a year ago
S
a year ago
This is a short answer type question as classified in NCERT Exemplar

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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New answer posted•a year ago
S
a year ago

This is a short answer type question as classified in NCERT Exemplar

Coefficient of β = 5

T1= 27+273=300K

Coefficient of performance β = T 2 300 - T 2

1500-5T2=T2

6T2=1500

T2= 250K

T2= 250-273=-23oC

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New answer posted•a year ago
S
a year ago

This is a short answer type question as classified in NCERT Exemplar

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 × η = 1/20

Coefficient of performance of the refrigerat

...Read more

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New answer posted•a year ago
S
a year ago

This is a short answer type question as classified in NCERT Exemplar

For adiabatic change process we know

P1V1y= P2V2y

P (V+ ? V )y = (P+ ? P )Vy

PVy (1+ ? V V ) y=p (1+ ? P P )Vy

PVy (1+ ? V V ) ≈ PVy (1+ ? V V )

Y ? V V = ? P P

dV= 1 Y V p d P

hence work done increasing the pressure from P1 to P2

W= ∫ p 1 p 2 p d V = ∫ p 2 p 1 p 1 Y V p d p

= V Y P 2 - P 1

W= P 2 - P 1 Y V

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