Q : 1. Choose the correct.

(a) Cyclic process  is an example of reversible process

(b) Quasi - static is an essential condition for reversibility.

(c) Second law of therm o dynamics follow law of conservation of energy.

(d) If a system comes back to it is original state, it means that Δ U 0.

Answer Option : (B)

Answer Explanation : => Reversibility means that you have to trace the exact path in reverse direction and cyclic process means that system must reach at  its initial state, may be from any path.

=> Quasi static and non dissipation of energy is required conditions for reversibility.

=> I^st law follow law to conservation of energy.

=> For cyclic process ΔU=0.

Q : 2.

Draw the graph between P - T for the given therm o dynamic in T - V graph.

(a)
   

(b)
   

(c)
   

(d)
   

Answer Option : (A)

Answer Explanation : T - V graph shows that process is isochoric. Means PV = URT 

P T

Q : 3. Choose the correct

(a) The value of U of an ideal gas is zero for isothermal expansion.

(b) If temperature of a system is increased ΔU must be non zero.

(c) If temperature of the system decreased, the heat energy of the system must decreased.

(d) For an adiabatic process ΔT = 0.

Answer Option : (B)

Answer Explanation : => ΔU is zero for Isothermal expansion of gas not U.

=> If temperature of the system is increased, the K.E. of the molecules of the system is increased, so the internal energy U of the system must increase.

=> If temperature is decreased, internal energy in transition. Heart energy is not a state variable.

=> For adiabatic process ΔQ=0 , ΔT0.

Q : 4.

If the P-V graph it is shown that a system is taken from state A to state B from two paths BA and BCA. If tells that

(a) Work done is path independent.

(b) work done is more if we choose path A C B.

(c) Energy released is more if we choose path AB.

(d)  ΔU is path dependent

Answer Option : (C)

Answer Explanation : Work done is path independent as you can observe that work done by the gas for both process is different.

Here volume is decreasing

So work done on gas is positive and done by gas is negative.

From first law of therm o dynamics
ΔQ = ΔU + ΔW -------(1)

ΔU is same for both paths.

Here Δw -ve
               ΔQ -ve

From the graph |Δw1|>|Δw2|
ΔW₁ work done by the gas for path AB.
ΔW₂ work done by the gas for path ACB.

In the above situation work is done on the system, internal energy of the system increased and heat energy is released to surrounding.

In the above situation work is done on the system, internal energy of the system increased and heat energy is released to surrounding.

From equation (1)
As ΔU is same for both paths,

ΔQ_{1 =} ΔU + ΔW₁
       ΔQ_{2 =} ΔU + ΔW₂

Here ΔW  is negative, ΔW₁ is more negative than ΔW₂ it meas ΔQ₁ is negative than ΔQ₂ . Negative sign with ΔQ just shows that energy is released.

|ΔQ₁ | > |ΔQ₂ |

If means released is more if we choose path AB.

ΔU is path independent

Q : 5. In a cyclic process

(a) Work done must e negative.

(b) Work done must be positive

(c) Work done is equal to (-ΔQ).

(d) Work done is equal to (+ΔQ).

Answer Option : (D)

Answer Explanation : For cyclic process, ΔU=0.

ΔQ= ΔU + ΔW

So ΔQ = ΔW

Q : 6. A cannot engine has efficiency 1/5. Efficiency become 1/3 when temperature of sink is decreased by 50k. what is the temperature of sink?

(a) 350k

(b) 375k

(c) 325k

(d) 300k

Answer Option : (D)

Answer Explanation :



Q : 7. If in a thermo dynamic process ΔQ+ve and ΔW -ve

(a) ΔU -ve

(b) ΔQ + ΔW = ΔU

(c) |ΔQ| + |ΔW| = |ΔU|

(d) None of the above

Answer Option : (C)

Answer Explanation : ΔQ = ΔU + ΔW

If ΔQ +ve   and   ΔW-ve

It means heat energy is given to system and work done is on the system.
 
So internal energy of the system is increased due to ΔQ and ΔW.

So ΔU +ve

|ΔQ| + |ΔW| = |ΔU|

Q : 8. Choose the correct option/options

(a) A system has a certain amount of heat.

(b) A system has a certain amount of work.

(c) A gas has a certain amount of internal energy.

(d) A certain amount of heat is supplied to the system.

Answer Option : (C) , (D)

Answer Explanation : Heat energy is the energy in transition and internal energy is a state variable.

Q : 9. Choose the correct option/ options

(a) In an ideal gas, the internal energy is just the sum of kinetic energies of all the molecules.

(b) In an ideal gas, the internal energy is sum of potential energies kinetic energies of all the molecules.

(c) For isothermal expansion of an ideal gas, ΔQ = ΔU + ΔW

(d) For isothermal expansion of an ideal gas ΔQ = ΔW.

Answer Option : (A) , (D)

Answer Explanation : In an ideal gas the interaction b/w the molecules is negligible , So internal energy of an ideal gas depend only an K.E of the molecules.

For isothermal expansion ΔT =0, So ΔU =0.
It means ΔQ=ΔW

Q : 10. Choose the correct option/ options

(a) PV = nRT is valid for isothermal process only.

(b) PV = Constant is valid for isothermal process only.

(c) PV = nRT is applicable for adiabatic process

(d) PV^r = Constant is valid for adiabatic process only

Answer Option : (B) , (C)

Answer Explanation : PV = nRT , is an equation which shows relation b/w state variables, it is known as equation of state. It is applicable for any type of process.

PV=Constant  for isothermal process.

PV^r = Constant  Condition for adiabatic process.

Q : 11. In an isothermal process.

(a) Work done by the system means less of internal energy of the system.

(b) Work done by the system means positive work done by the system

(c) Work done by the system means negative work done by enviorement

(d) Work done on the system means, negative work done by the system.

Answer Option : (B), (C) and (D)

Answer Explanation : Work done by the system Positive work done by the system and negative work done by the environment.

Work done on the system Positive work done by environment and negative work done by system.

Q : 12.If ideal gas is heated in an container, while keeping T constant, it will push the piston up and the given heat energy is used to do work

(a) According to IInd Law of thermodynamics, ΔQ=ΔU+W, and ΔU cannot be zero.

(b) According to I^st law of thermodynamics, ΔU must be zero.

(c) According to II^nd law of thermodynamics, ΔQ =( W + change in internal energy of container and piston

(d) ΔQ = ΔU + W is not related to IInd law of thermodynamics.

Answer Option : (B), (C) and (D)

Answer Explanation : Hint Go for Ist and IInd  laws of thermodynamics.

Q : 13. Thermodynamic equilibrium means

(a) The temperature of the system and surrounding must have same values.

(b) The temperature of the system and surrounding may have same values.

(c) The volume of the two system in thermal equilibrium must have same values.

(d) The volume of the two systems in thermal equilibrium may have same values.

Answer Option : (A) , (D)

Answer Explanation : For thermal equilibrium two conditions are necessary (i) the temperature of system and surrounding must be equal and (ii) The macroscopic variables of both system and surrounding must bot change with time.

Q : 14.


From the above P-V graph for ideal gas

If the system goes from A to B from path ADB,

(a) ΔU must be zero.

(b) ΔUcan not be zero.

(c) Work done is continuously decreasing

(d) ΔU in this case will be equal to ΔU when we take system from path ACB.

Answer Option : (B)
Answer Explanation : For isothermal process  and it is represented by graph ACB.  Means ΔU=0.

It means for graph ADB ΔU  0  b/c it does not show an isothermal process.

Work done is continuously increasing.
 
Q : 15.
From the above P-V graph for ideal gas

The relation b/w ΔQ₁ and ΔQ₂  in the  situation above is

(a) ΔQ₁ < ΔQ₂

(b) ΔQ₁ > ΔQ₂

(c) ΔQ₁ = ΔQ₂

(d) Cannot say from the given information.

Answer Option : (B)

Answer Explanation : We Know ΔQ₁ = ΔU + ΔW

For ADB       ΔQ₁ = ΔU₁ + ΔW₁ -----------(1)

For ACB      ΔQ₂ =0 + ΔW₂  ----------------(2)
 
ΔU1 must be positive and ΔW₁ > ΔW₂ ,So ΔQ₁ >ΔQ₂ 

Q : 16.Match The Following:


Answer Option : A:R|B:P|C:Q|D:P:T

Q : 17. Choose the correct option/options

(a) Q is a extensive state variable.

(b) ΔQ is a extensive state variable.

(c) Work is a intensive state variable.

(d) U ( internal energy) is a extensive state variable.

Answer Option : (B), (D)

Answer Explanation : Q and W(work) both are not variables.

They are used to change internal energy of system.

U is a extensive state variable b/c it gets halved when we divided system in to the halves.

ΔQm , so it is also an extensive state variable(?).

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