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**MULTIPLE CHOICE QUESTIONS I**

**Q.1. An ideal gas undergoes four different processes from the same initial state (Figure). Four processes are adiabatic, isothermal, isobaric and isochoric. Out of 1, 2, 3 and 4 which one is adiabatic.**

**(a) 4(b) 3(c) 2(d) 1Ans.** (c)

4 is isobaric process, 1 is isochoric. Out of 3 and 2, 3 has the smaller slope (magnitude) hence is isothermal. Remaining process 2 is adiabatic.

(a) 0.25 kg

(b) 2.25 kg

(c) 0.05 kg

(d) 0.20 kg

Ans.

580 x 10

1 cal will produce sweat = 1/580 x 10

14.5 x 10

= 145/5800 kg per minute = 0.025 kg per min

**Out of the following diagrams (Figures), which represents the T-P diagram?**

** **

**(a) (iv)(b) (ii)(c) (iii)(d) (i)Ans.** (c)

According to P-V diagram at constant temperature, P increases as V decreases. So, it is Boyle’s law in options (iii) and (iv). If P increases at constant temperature, volume V decreases. As in (iii) T-P diagram, P is smaller at 2 and larger at 1, which tallies with option (c).

**The amount of work done by the gas is(a) 6P _{o}V_{o }(b) -2 P_{o}V_{o} (c) +2 P_{o}V_{o }(d) +4 P_{o}V_{o}Ans.** (b)

Work (∆W): Work can be defined as the energy that is transferred from one body to the other owing to a force that acts between them.

If P be the pressure of the gas in the cylinder, then force exerted by the gas on the piston of the cylinder F= PA.

In a small displacement of piston through dx, work done by the gas dW = F.dx = PA dx = PdV

∴ Total amount of work done

In P-V diagram or indicator diagram, the area under P-V curve represents work done.

W = area under P-V diagram

According to the P-V diagram given in the question,

Work done in the process ABCD = Area of rectangle ABCDA

= AB x BC = (3V_{0} - V_{0}) x (2P_{0} - P_{0})

= 2V_{0} x P_{0} = 2P_{0}V_{0}

Since, the cyclic process is anti-clockwise, work done by the gas is negative, i.e., - 2P_{0}V_{0}. Hence there is a net compression in the gas.**Important point:** In a cyclic process work dine is

1. positive if the cycle is clockwise.

2. negative if the cycle is anticlockwise.**Q.5. Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is(a) 2 ^{γ-1}(b) (1/2)^{γ-1}(c) (d) Ans.** (a)

According to the P-V diagram shown for the container A (which is going through isothermal process) and for container B (which is going through adiabatic process).

Both the process involves compression of the gas.

(i) Isothermal compression (gas A) during 1 → 2)

P_{1}V_{1} = P_{2}V_{2}

⇒ P_{0}(2V_{0})^{γ} = P_{2}(V_{0})^{γ}

⇒ P_{0}(2V_{0}) = P_{2}(V_{0})

(ii) Adiabatic compression, (gas B) (during 1 → 2)

P_{1}V_{1}^{γ} = P_{2}V_{2}^{γ}

⇒ P_{0}(2V_{0})^{γ} = P_{2}(V_{0})^{γ}

⇒

Hence = Ratio of final pressure =

where, γ is ratio of specific heat capacities for the gas.**Q.6. Three copper blocks of masses M _{1}, M_{2} and M_{3} kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at T_{1}, T_{2}, T_{3} (T_{1} > T_{2} > T_{3} ). Assuming there is no heat loss to the surroundings, the equilibrium temprature T is (s is specific heat of copper)** (b)

(a)

(b)

(c)

(d)

Ans.

According to question, since there is no net loss to the surroundings and the equilibrium temperature of the system is T.

Let us assume that T

Heat lost by M

⇒ M

(where s is the specific heat of the copper material)

⇒ T[M

⇒

**MULTIPLE CHOICE QUESTIONS II**

**Q.7. Which of the processes described below are irreversible?(a) The increase in temprature of an iron rod by hammering it.(b) A gas in a small cantainer at a temprature T** (a, b, d)

The conditions for reversibility are:

- There must be complete absence of dissipative forces such as friction, viscosity, electric resistance etc.
- The direct and reverse processes must take place infinitely slowly.
- The temperature of the system must not differ appreciably from its surroundings.

**Irreversible process**: Any process which is not reversible exactly is an irreversible process. All natural processes such as conduction, radiation, radioactive decay etc. are irreversible. All practical processes such as free expansion, Joule-Thomson expansion, electrical heating of a wire are also irreversible.

(a) In this case internal energy of the rod is increased from external work done by hammer which in turn increases its temperature. So, the process cannot be retraced itself.

(b) In this process energy in the form of heat is transferred to the gas in the small container by big reservoir at temperature T_{2}.

(c) In a quasi-static isothermal expansion, the gas is ideal, this process is reversible because the cylinder is fitted with frictionless piston.

(d) As the weight is added to the cylinder arrangement in the form of external pressure hence, it cannot be reversed back itself. **Q.8. An ideal gas undergoes isothermal process from some initial state i to final state f. Choose the correct alternatives.(a) DU = 0(b) DQ= 0(c) DQ = DU(d) DQ = DWAns. **(a, d)

According to it heat given to a system (∆Q) is equal to the sum of increase in its internal energy (AIT) and the work done (AW) by the system against the surroundings.

∆Q = ∆U + ∆W

According to the first law of thermodynamics. ∆AQ = ∆U + ∆Wbut

∆U ∝ ∆T

∆U = 0 [As ∆T = 0]

∆Q = ∆W, i.e., heat supplied in an isothermal change is used to do work against external surrounding.

If the work is done on the system then equal amount of heat energy will be liberated by the system.

**(a) Change in internal energy is same in IV and III cases, but not in I and II.(b) Change in internal energy is same in all the four cases.(c) Work done is maximum in case I(d) Work done is minimum in case II.Ans.** (b, c)

**Internal energy (U):** Internal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration.

The energy due to molecular motion is called internal kinetic energy U_{K }and that due to molecular configuration is called internal potential energyUp.

i.e., Total internal energy U= U_{K}+ U_{P}(i) For an ideal gas, as there is no molecular attraction U_{P} = 0

i.e., internal energy of an ideal gas is totally kinetic and is given by

U = U_{K} = 3/2 RT

and change in internal energy ΔU = 3/2 μR

(ii) In case of gases whatever be the process

(iii) Change in internal energy in a cyclic process is always zero as for cyclic process U_{f} = U_{i}

So ΔU = U_{f} - U_{i} = 0

Change in internal energy does not depend on the path of the process. So it is called a point function, i.e., it depends only on the initial and final states (A and B) of the system, i.e., ΔU = Uf - U_{i}

Hence internal energy is same for all four paths I, II, III and IV.

The work done by an ideal gas is equal to the area bounded between P-V curve.

Work done from A to B, ΔW_{A→}_{B} = Area under the P-V curve which is maximum for the path I.**Q.10. Consider a cycle followed by an engine (Fig.)1 to 2 is isothermal2 to 3 is adiabatic3 to 1 is adiabaticSuch a process does not exist because**

**(a) Heat is completely converted to mechanical energy in such a process, which is not possible.(b) Mechanical energy is completely converted to heat in this process,which is not possible.(c) Curves representing two adiabatic processes don’t intersect.(d) Curves representing an adiabatic process and an isothermal process don’t intersect.Ans.** (a, c)

(a) The given process is a cyclic process, i.e. it returns to the original state 1. And change in internal energy in a cyclic process is always zero as for cyclic process U

Hence, total heat is completely converted to mechanical energy. Such a process is not possible by second law of thermodynamics.

(c) Here, two curves are intersecting, when the gas expands adiabatically from 2 to 3. It is not possible to return to the same state without being heat supplied, hence the process 3 to 1 cannot be adiabatic. So, we conclude that such a process does not exist because curves representing two adiabatic processes do not intersect.

If W > 0, then possibilities are:

**(a) Q _{1} > Q_{2} > 0**(a, c)

(b) Q_{2} > Q_{1} > 0

(c) Q_{2} < Q_{1} < 0

(d) Q_{1} < 0, Q_{2} > 0

Ans.

Refrigerator or Heat Pump:

A refrigerator or heat pump is basically a heat engine run in reverse direction. It essentially consists of three parts:

**Sink:** At lower temperature T_{2}.

The working substance takes heat Q_{2} from a sink (contents of refrigerator) at lower temperature, has a net amount of work done W on it by an external agent (usually compressor of refrigerator) and gives out a larger amount of heat Q_{1}, to a hot body at temperature T_{1} (usually atmosphere). Thus, it transfers heat from a cold body to a hot body at the expense of mechanical energy supplied to it by an external agent. The cold body is thus cooled more and more.

We know that the diagram represents the working of a refrigerator. So, we can write

Q1 = W + Q_{2}

According to the problem, W > 0, then

⇒ W = Q_{1} - Q_{2} > 0

So there are two possibilities:

(a) If both Q_{1} and Q_{2} are positive,

⇒ Q_{1} > Q_{2} > 0

(c) If both Q_{1} and Q_{2} are negative,

Q_{2} < Q_{1} < 0

**VERY SHORT ANSWER TYPE QUESTIONS**

**Q.1. Can a system be heated and its temperature remains constant?****Ans. **If the system does work against the surroundings so that it compensates for the heat supplied, the temperature can remain constant.

It is given that ΔT = 0 ⇒ ΔU = 0

∴ ΔQ = ΔU + ΔW

⇒ ΔQ = ΔW So heat supplied to the system is utilized in expansion system is isothermal.**Q.2. A system goes from P to Q by two different paths in the P-V diagram as shown in Figure. Heat given to the system in path 1 is 1000 J. The work done by the system along path 1 is more than path 2 by 100 J. What is the heat exchanged by the system in path 2?**

**Ans.** According to the first law of thermodynamics,

∆Q = AU + ∆W. Let us apply this for each path.

For path 1: Heat given Q_{1} = +1000 J

Let work done for path 1 = W_{1} .

For path 2:

Work done (W_{2}) = (W_{I }–* 100)* J

Heat given Q_{2} – ?

As change in internal energy between two states for different path is same.

∆ U = Q_{i }- W_{1} = Q_{2 }- W_{2}1000 *– *W_{! }= Q_{2 }- (W_{1} – 100)

⇒ Q_{2 }= 1000 - 100 = 900 J **Q.3. If a refrigerator’s door is kept open, will the room become cool or hot? Explain.Ans.** A refrigerator is a heat engine it extracts heat from low temperature reservoir and transfer it to high temperature. If a refrigerator’s door is kept open, then room will become hot, because then refrigerator exhaust more heat into the room than earlier. In this way, temperature of the room increases and room becomes hot. No refrigerator is efficient. Thus it exhaust more heat into the room than it extract from it. Thus, a room cannot be cooled by keeping the door of a refrigerator open.

Ans.

dQ = dU + dW

As dQ = 0 (adiabatic process)

so dU = -dW

In compression, work is done on the system So, dW = -ve

⇒ dU = + ve

So internal energy of the gas increases, i.e. its temperature increases.

Ans.

Pressure (P) ∝Temperature (T). Therefore, pressure of gas increases

**SHORT ANSWER TYPE QUESTIONS**

**Q.1. Consider a Carnot’s cycle operating between T _{1} = 500 K and T_{2 }= 300K producing 1 k J of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.**

Temperature of source or reservoir = T

Temperature of sink = T

1000J/x = 1 - 0.6

1000/x = 0.4

x = 1000/0.4 = 2500 J.

Ans.

According to the problem, height of the stairs = h = 10 m

Work done to burn 5 kg of fat

= (5 kg)(7000 x 10

= 147 x 10

Work done towards burning of fat in one trip (up and down the stairs)

(as only half the work done while coming down is useful in burning fat)

∴ Number of times, the person has to go up and down the stairs (no. of trips required)

Ans.

For just before and after an stroke, we can write

As ΔV << V, so by using binomial approximation we get

⇒

Hence, work done is increasing the pressure from P

Ans.

Efficiency of refrigerator’s 50% of perfect engine

∴ Efficiency of refrigerator = 50% of 1 = .5

Net efficiency = η' = 0.5 x 0.1 = 0.05

∴ Coefficient of performance

Q

= 19 x 1 kW = 19 kJ/s

Ans.

i.e.,

A perfect refrigerator is the one which transfers heat from cold to hot body without doing work.

i.e., W = 0 so that Q

According to the problem, coefficient of performance (ω) = 5

T

Coefficient of performance (ω) =

(a) the expansion takes place at constant temperature.

(b) the expansion takes place at constant pressure.

Plot the P-V diagram for each case. In which of the two cases, is the work done by the gas more?

Ans.

In case (i) P_{i} V_{i} = P_{f} V_{f} ; therefore process is isothermal. Work done = area under the PV curve so work done is more when the gas expands at constant pressure.

**LONG ANSWER TYPE QUESTIONS**

**Q.1. Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in Figure. **

**(a) Find the work done when the gas is taken from state 1 to state 2.****(b) What is the ratio of temperature T _{1}/T_{2}, if V_{2} = 2V_{1}?**

Work done by the gas (Let PV

Then, the change in internal energy

**A to B : volume constantB to C : adiabaticC to D : volume constantD to A : adiabaticV**

(a) A to B

(b) C to D

(c)

Similarly.

Similarly,

Now

Similarly, P

Totat work done = W

(d) Heat supplied during process A, B

dQ

**AB : constant volumeBC : constant pressureCD : adiabaticDA : constant pressureAns. **

(a): For A → B, dV = 0

So,

dW = 0

By 1^{st} law of thermodynamics

dQ = dU + dW = dU + 0

∴ dQ = dU

(∵ dQ = nC_{v}dT)

n = 1; C_{v} = 3/2 R

So,

∴ Heat exchange [to system]

(b) For B to C, ΔP = 0 n = 1

dQ = dU + dW = C_{v}(dT) + P_{B}dV

V_{A} = V_{B} and P_{B} = P_{C}

∴

= 5/2 P_{B}V_{C} - 5/2P_{B}V_{A}

dQ_{2} = 5/2P_{B}[V_{C} - V_{A}]

(c) For diagram C → D, adiabatic change

∴ dQ_{3} = 0 (No exchange of heat)

(d) For diagram D → A, ΔP = 0 Compression of gas from volume V_{D} to V_{A} at constant pressure hence heat exchange similar to part (b) i.e. Heat exchange**Q.4. Consider that an ideal gas (n moles) is expanding in a process given by P = f (V), which passes through a point (V _{0}, P_{0}). Show that the gas is absorbing heat at (P_{0}, V_{0}) if the slope of the curve P = f (V) is larger than the slope of the adiabat passing through (P_{0}, V_{0}).**Slope of P = f (V), curve at (V

Ans.

Slope of adiabat at (V

Now heat absorbed in the process P = f (V)

dQ = dV + dW

= nC

Since T = (1/nR ) PV = (1/ nR ) V f (V)

dT = (1/nR ) [f (V) + V f′ (V)] dV

Thus

Heat is absorbed when dQ/dV > 0 when gas expands, that is when

**(a) What is the initial pressure of the system?(b) What is the final pressure of the system?(c) Using the first law of thermodynamics, write down a relation between Q, P** (a): It is considered that piston is mass less and piston is balanced by atmospheric pressure (P

(b) On supply heat Q. Volume of gas increases from V

So increase in volume = V

If displacement of piston is x then volume increase in cylinder)

= Area of base x height = A

A

Force exerted by spring

As the piston is of unit area of cross-section ∴ A = 1

Force due to spring = K(V

Final total pressure on gas P

(c) By Ist law of thermodynamics dQ = dU + dW

dU = C

T = final temperature of gas

T

n = 1

W.D. by gas = p. dV + increase in PE of spring

It is required relation.

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