# 11th Class Physics Chapter 11 Heat and Thermondyanmics Short Questions Answer

## 11th Class Physics Chapter 11 Heat and Thermondyanmics Short Questions Answer

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^{2}) = v

^{2}]is also a positive, therefore the average of the square of the velocities will not be zero.

(i) Some part of heat is used doing the external work to move the piston up against the constant atmospheric pressure.

(ii) The other part of heat is used to increase the internal energy and temperature.

If the same gas is heated at constant volume, no external work is done to expend the gas. The total heat supplied is used to increase the the internal energy and temperature of the gas. This shows that more heat is required to heat the gas at constant pressure than at constant volume for the same rise of temperature. So we conclude that specific heat at constant pressure is greater than the specific heat at constant volume i.e. C

_{P}> C

_{v}

**Solution:-**

From kinetic theory of gases, we have the relation

P = 1/3/ N

_{0}<mv

^{2}>

Where N

_{0}= No of molecules in a unit volume

m = mass of one molecule

N

_{0}m = mass of unit volume of the gas. But mass per unite volume is called density of gas

or N

_{0}m = P

Hence P = 1.3 (N

_{0}) <mv

^{2}> = 1/3 N

_{0}m<v

^{2}>

Or P = 1/3 p <v

^{2}>

Hence P = <v

^{2}> = 3P/p

**Solution-
** Temperature in centigrade = T

_{c}= 37

^{0}C

Temperature in Fahrenheit= T

_{f}= ?

Temperature in Kelvin = T

_{k}= ?

Using the formula to convert Centigrade into Fahrenheit

T

_{f}= 9/5 T

_{c}+ 32

Putting the value, we get

T

_{f}= 9/5*37+32= 333/5+32

or T

_{f}= 66.6.+32=98.6

^{0}F

T

_{f}= 98.6

^{0}F

(ii) Formula to convert centigrade into Fahrenheit

T

_{k}= 37+273=310k

T

_{k }= 310 k ans.

It is customary to define molar specific heat of a gas in two following wats.

(i) Molar specific heat at constant volume.

(ii) Molar specific heat at constant pressure.

**(i) Molar specific heat at constant volume (C _{v})
**It is defined as the amount of heat required to raise the temperature of on mole of a gas through l K at constant volume, It is denoted by ‘C

_{v})

Mathematically it is expressed as Δ

C

_{v}= Q

_{v}/ ΔT

Where ‘Q

_{v}’ is the amount of heat supplied at constant volume and It is customary to define molar specific heat of a gas in two following wats.

(i) Molar specific heat at constant volume.

(ii) Molar specific heat at constant pressure.

**(i) Molar specific heat at constant volume (C**

It is defined as the amount of heat required to raise the temperature of on mole of a gas through l K at constant volume, It is denoted by ‘C

_{v})_{v})

Mathematically it is expressed as Δ

C

_{v}= Q

_{v}/ ΔT

Where ‘Q

_{v}’ is the amount of heat supplied at constant volume and T is rise of temperature.

**(ii) Molar specific heat at constant pressure (C**

_{p})It is defined as the amount of heat required to raise the temperature of one mole of a gas through l K at constant pressure. It is denoted by C

_{p}.

Mathematically, it is written as

Q

_{p}= C

_{p}Δ

_{T }= or C

_{p}= Q

_{p}/ ΔT

Where Q

_{p}is the amount of heat supplied at constant pressure. It is customary to define molar specific heat of a gas in two following wats.

(i) Molar specific heat at constant volume.

(ii) Molar specific heat at constant pressure.

**(i) Molar specific heat at constant volume (C**

It is defined as the amount of heat required to raise the temperature of on mole of a gas through l K at constant volume, It is denoted by ‘C

_{v})_{v})

Mathematically it is expressed as Δ

C

_{v}= Q

_{v}/ ΔT

Where ‘Q

_{v}’ is the amount of heat supplied at constant volume and ΔT is rise of temperature.

**(ii) Molar specific heat at constant pressure (C**

_{p})It is defined as the amount of heat required to raise the temperature of one mole of a gas through l K at constant pressure. It is denoted by C

_{p}.

Mathematically, it is written as

Q

_{p}= C

_{p}Δ

_{T }= or C

_{p}= Q

_{p}/ ΔT

Where Q

_{p}is the amount of heat supplied at constant pressure.ΔT is rise of temperature.

**(ii) Molar specific heat at constant pressure (C**

_{p})It is defined as the amount of heat required to raise the temperature of one mole of a gas through l K at constant pressure. It is denoted by C

_{p}.

Mathematically, it is written as

Q

_{p}= C

_{p}Δ

_{T }= or C

_{p}= Q

_{p}/ ΔT

Where Q

_{p}is the amount of heat supplied at constant pressure.

^{2}> where p is the density of gas?

P = 1/3 N

_{0}m <v

^{2}>……………..(1)

we have

N

_{0}m = Number of molecules in unit volume of gas *mass of the gas molecules.

= Mass of unit volume of the gas

= Density of the gas

(Because density= mass/volume)

Therefore, the equation (1) become

P = 1/3 p<v

^{2}>

P= 2/3 N

_{0}<1/2 mv

^{2}>

But N

_{0}= N/V = No. of molecules / Volume

P = 2/3 N/V<1/2mv

^{2}>

Or PV= 2/3N<1/2 mv

^{2}>…………………………….(1)

If average translational K.E. <1/2mv

^{2}> remains constant, the temperature of the gas will remain constant and 2/3 N is also constant.

So equation (1) becomes

PV = constant

Which is the required equation of Boyle’s law.

P = 2/3 N

_{0}<1/2mv

^{2}>

But N

_{0}= N/V = No. of molecules/volume

P = 2/3 N/V<1/2mc

^{2}>

or V = 2/3 N/P<1/2mv

^{2}>

Since <1/2mv

^{2}>αT

V/T = 2/3 N/P…………………………………(1)

If pressure ‘P’ is kept constant, the light hand side of equ. (1) becomes constant

Hence, V/T = constant

This is required equation of Charle’s law.

It is expressed by the formula

Boltzmann constant = R = R/N

_{0}

Knowledge the values,

Molar gas constant = R = 8.314 jmol

^{-1}k

^{-1}

Number of molecules in a mole = Avogadro’s N

_{0}= 6.02*10

^{33}

K = 8.31/6.02*10

^{23}= 1.38*10

^{-23}Jk

^{–}

ɳ = out put /input = Net work done/Heat energy absorbed

= ΔW/Q

_{1}= Q

_{1}-Q

_{2}/Q

_{1}

or ɳ = 1- Q

_{2}/Q

_{1}

Since T

_{2}/T

_{1}= Q

_{2}/Q

_{1}

ɳ = 1- T

_{2}/ T

_{1 }Where T

_{1}and T

_{2}are the temperature of source and sink respectively.

Have Any other short question related to this 11th Class 11th Class Physics Chapter 11 Heat and Thermodynamics please write down in comment section.

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