- Introduction
- Concept of Heat
- P-V Indicator Digram
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- Work in volume changes
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- Internal Energy and first law of thermodynamics
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- Specific heat capacity of an ideal gas
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- Thermodynamic Processes
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- Quasi static Processes
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- Isothermal Process
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- Adiabatic Process
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- Isochoric process
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- Isobaric process
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- Work done in Isothermal process
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- Work done in an Adiabatic process
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- Heat Engine and efficiency
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- Principle of a Refrigerator
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- Second law of thermodynamics
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- Reversibility and irreversibility
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- Carnot's Heat Engine
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- Carnot Theorem
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- Solved Examples

a, ΔT >0

b, ΔU=0

c ΔQ=ΔW

d PV=constants

In an Isothermal Process

Temperature remains constant ΔT =0

Since Internal energy depends on the temperature

ΔU=0

From first law of Thermodynamics

ΔU=ΔQ-ΔW

Since ΔU=0

ΔQ=ΔW

Also PV=nRT

As T is constant

PV= constant

Given that on absolute scale

Triple point of water on scale A = 200 A

Triple point of water on scale B = 350 B

Also, triple point of water on Kelvin scale = 273.16 K

Now temperature on scale A and on scale B is equivalent to 273.16 K on absolute temperature scale.

Thus, value of one degree on absolute scale A = (273.16/200) K

Or,

Value of temperature T

Similarly value of temperature T

Since T

273.16×T

Or, T

Total heat supplied =Workdone + Change in internal energy

So work done=2140-1580=560 J

Let s be the distance moved then

the workdone is given by =Fs

Fs=560

s=560/F

=560/102*10

Calculate the ratio of final pressure to the intial pressure

Calculate the final temperature

Change in internal energy

Calculate the molar specific heat capacity of the process

Given

n=2 T=27°C=300 K ,V

Now PV

P

P

=.5

ALso

T

or T

Change in internal energy=nC

For monoatomic gas C

Substituting all the values

Change in internal energy==-2764J

As in adiabatic process ΔQ=0,molar specific heat capacity=0

An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C.It absorbs 6*10

T

T

Efficiency of the carnot cycle is given by

=1-(T

Now also efficency =Heat supplied from source/Heat absorbed at high temperature

so Heat supplied from source=6*10

Class 11 Maths Class 11 Physics Class 11 Chemistry

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