Chapter # 11 Heat and Thermodynamics 1st Year Physics MCQ's - MS Education Network

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Chapter # 11 Heat and Thermodynamics, Chapter # 11 Heat and Thermodynamics mcqs

11.01 scientific theory of Gases

The relation for temperature of a gas is given by:

(A) 12KT12KT

(C) 23KT23KT

(D) 32KT32KT

 

A device primarily based upon the physical science property of matter is called:

 

Heat is kind of:

According to Charles law:

(A) VTVT

(B) VnVn

(C) P1TP1T

(D) P1VP1V

 

The ideal gas law is:

The SI unit of product of pressure and volume is:

The value of Boltzman’s constant is:

(A) 1.38×10−23JK1.38×10−23JK

(B) 1.38×1023JK1.38×1023JK

(C) 1.38×10−23Jmol.K1.38×10−23Jmol.K

(D) 1.38×1023Jmol.K1.38×1023Jmol.K

 

 

At constant temperature, if pressure of a given mass of gas is halved, then its volume becomes:

S.I unit of pressure of gas is:

(A) Nm−2Nm−2

(B) N.mN.m

(C) N2mN2m

(D) N2mN2m

 

For gas, the P.E. related to its molecules is:

(C) 12kx2012kx02

(D) 12kx012kx0

 

At constant temperature and pressure, if volume of given mass of a gas is doubled then:

(B) 1414 of original

(C) 1212 of original

 

Temperature of a gas is redoubled from 27oC to 127oC. The quantitative relation of its mean K.E will be:

Boltzman constant, universal universal gas constant Avogadro's number ar referred to as as:

(A) K=RNAK=RNA

(B) K=NAKK=NAK

(C) R=KNAR=KNA

(D) R=NAKR=NAK

 

Boltzman constant “k” has same unit as:

(A) Temperature

(B) Energy

(C) Entropy

(D) Perssure

 

If the temperature of a gas is constant then <12mv2><12mv2> of the molecules of gas can be:

(A) Constant

(B) Zero

(C) Increase

(D) Decrease

 

The mean K.E.  of gas is at :

(A) 0 oC

(B) -273 oC

(C) 100 K

(D) 100 oC

 

The ideal gas law is given within the form:

(A) PV=nR/T

(B) PT=NRV

(C) PV=nRT

(D) TV=nRP

 

The average change of location K.E. of a molecule of a gas at temperature T is proportional to:

(C) T−−√T

(D) T2

 

The unit of pressure of gas is:

(A) 1 Nm-2

(B) One Pascal

(C) One atmosphere

(D) All of the higher than

 

At constant temperature, if the quantity of the given mass of gas is doubled, then the density of gas becomes:

(A) Double

(B) 1/4 of the initial worth

(C) 1/2 of the initial worth

(D) Remains constant

 

General gas law or general gas equation comes from:

(A) Boyle’s law

(B) Charle’s law

(C) Avogadro’s law

(D) All of the higher than

 

Solid ice, Liquid water and water vapours consist in equilibrium at a temperature:

(A) 273 K

(B) 273.16 K

(C) 273o C

(D) 100o C

 

 

11.02 Internal Energy

The internal energy of a bit of lead once crushed by hammer will:

At that of the subsequent temperature, a body has most internal energy:

(A) −273oC−273oC

(B) 0K0K

 

 

 

In physical science system internal energy decreases by 100J and 100J of labor done on the system then heat lost can be:    

A substance gas molecules has:

(A) Temperature

(B) Pressure

(C) Path

(D) Final and initial state

 

The add of all the energies of molecules is understood as:

(A) Elastic P.E.

(B) K.E.

(C) Internal energy

(D) attraction P.E.

 

Average translation K.E of a gas molecule is:

(A) 12KT12KT

(C) 23KT23KT

(D) 32KT32KT

 

11.03 Work and warmth

The value of molar universal gas constant ‘R’ in SI units is:

(A) 83.10 J mol-1 K-1

(B) 0.83 J mol-1 K-1

(C) 8.89 J mol-1 K-1

(D) 8.31 J mol-1 K-1

 

If P = Pressure ; V = Volume of a gas PΔVPΔV represents:

(A) Work

(B) Density

(C) Power

(D) Temperature

 

11.04 initial Law of physical science

The first law of physical science for associate equal method is:

First Law of physical science for associate activity are going to be written as:

(A) Q = W

(B) Q = -W

(C) W=−ΔUW=−ΔU

(D) W=ΔUW=ΔU

 

In case of activity first law of physical science is written as:

(A) W=ΔuW=Δu

(B) W = Q

(C) W=Q−ΔuW=Q−Δu

(D) W=−ΔuW=−Δu

 

If the temperature of a system is unbroken constant, the method is named.

(A) activity

(B) Isochoric method

(C) equal method

(D) Isobaric method

 

Which remains constant in associate adiabatic process?

(A) Volume

(B) Pressure

(C) Entropy

 

Cloud formation in atmosphere is associate example of:

(A) equal method

(C) activity

 

Entropy remains constant in:

(A) equal method

(C) Isochoric method

 

The modification in internal energy is outlined as:

(A) Q - W

 

The work worn out isochoric method is:

(A) Constant

(B) Variable

(C) Zero

(D) depend on condition

 

In thermodynanicsprocess, the equqtion W=−ΔUW=−ΔU reprersents.

(A) equal expension

(B) equal compression

(C) adiabatic expension

(D)Adiabatic compression

 

 

A system will 700 Joules of labor and at an equivalent time its internal energy will increase to four hundred Joules, heat provided by the supply is:

(A) 700 Joules

(B) four hundred Joules

(C) 1100 Joules

(D) three hundred Joules

 

If internal energy decreases by three hundred J and a hundred and twenty J of labor is finished on the system then heat can be:

(A) 420 J

(B) 320 J

(C) 400 J

(D) 300 J

 

If TH = T1 = 327o and metallic element = T2 = twenty seven Co, then potency can be:

(A) 50 %

(B) 52 %

(C) 100 %

(D) Zero

 

 

11.05 Molar Specific Heats of a Gas

The distinction between Cp and Cv is equal to:

(A) Plank's constant

 

For a matter gas Cv = 5R25R2 then gamma  for this gas is:

(A) 5757

(B) 435435

(C) 7575

(D) 354354

 

A gas that strictly obeys the gas law below all conditions of temperature and pressure is called:

(A) Real gas

(B) gas

(C) Permanent gas

(D) chemical element

 

The ideal gas equation PV = RT  holds sensible for:

(A) Any volume of the gas

(B) One metric capacity unit of gas

(C) One metric weight unit of gas

(D) One mole of the gas

 

The amount of warmth needed to lift the temperature of 1 mole of a substance through 1o K is called:

(A) heat energy

(B) Heat capability

(C) Heat of fusion

(D) Molar heat energy

 

The temperature of a system remains constant in:

(A) action

(B) Isobaric method

(C) equal method

(D) action

 

11.06 Reversible and Irreversible Processes

Which is associate example of irreversible process?

 

11.07 engine

Number of spark plugs required within the ICE are:

The potency of ICE is about:

(A) twenty five you have to half-hour

(B) thirty five you have to forty nada

(C) forty you have to fifty nada

(D) fifty you have to hour

 

For operating of warmth engine, there should be:

(A) a supply

(B) a sink

(C) either of those

(D) each of those

 

A system during which there's no transfer of mass and energy across the boundary is called:

(A) associate open system

(B) A closed system

(C) associate isolated system

(D) an ideal system

 

The highest potency of a engine whose lower temperature is 17o C and extreme temperature of 200o C is:

(A) 70 %

(B) 100 %

(C) 35 %

(D) 38 %

 

11.09 Carnot|Nicolas Leonard Sadi Carnot|physicist} Engine and Carnot s Theorem

The potency of warmth engine whose sink is at 17OC and supply at 200OC is:

(A) 38%

(B) 65%

(C) 80%

(D) 90%

 

An ideal reversible engine has:

(A) 100% potency

(B) 80%

 

An ideal engine will solely be 100% economical if its cold temperature reservoir is at:

(A) 0 K

(B) 0oC0oC

(D) 100oC100oC


The heat energy of fusion of ice is:

(A) 3.36×105JK−13.36×105JK−1

(B) 3.36×106JK−13.36×106JK−1

(C) 3.36×107JK−13.36×107JK−1

(D) 3.36×108JK−13.36×108JK−1

 

The potency of a Carnot engine is 100% if temperature of sink T2 is:

(A) 0oC0oC

(C) 0oF0oF

 

A Carnot engine has associate potency of fifty once its sink temperature is at 27oC. The temperature of source:

(A) 273oC273oC

(B) 300oC300oC

(C) 327oC327oC

(D) 373oC373oC

 

If engine absorbs four hundred J and rejects two hundred J energy, its potency can be:

 If the temperature of sink is abnsoluite zero then the potency of warmth engine engine ought to be:

(A) 100%

(B) 50%

(C) Zero

(D) infinite

 

Efficiency of a engine operating between twenty seven oC and 327 oC can be:

(A) 50 %

(B) 90 %

(C) 40 %

(D) 61 %

 

The potency of a Carnot engine operating between higher and lower temperature T1 and T2 severally is given by:

(A) η=T2−T1T1η=T2−T1T1

(B) η=T1−T2T1η=T1−T2T1

(C) η=T2T1−T2η=T2T1−T2

(D) η=T1T1−T2η=T1T1−T2

 

The potency of a Carnot engine is:

(A) Infinite

(B) Zero

(C) but one

(D) bigger than one

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