Thermodynamics Online Objective Test | Thermodynamics Online Objective Test

# Thermodynamics # CBSE 11TH PHYSICS Prepare / Learn

Q1. Kelvin-Planck statement of Second Law of Thermodynamics states that

A. Processes are possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work.

B. No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work.

C. No process is possible whose sole result is the absorption of heat from a reservoir and the partial conversion of the heat into work.

D. Processes are not possible whose sole result is the absorption of heat from a reservoir and the partial conversion of the heat into work.

Answer : Option B
Explaination / Solution:

it is impossible to devise a cyclically operating device, the sole effect of which is to absorb energy in the form of heat from a single thermal reservoir and to deliver an equivalent amount of work. This implies that it is impossible to build a heat engine that has 100% thermal efficiency.

# Thermodynamics # CBSE 11TH PHYSICS Prepare / Learn

Q2. Clausius statement of Second Law of Thermodynamics states that

A. Many processes are possible whose sole result is the transfer of heat from a colder object to a hotter object

B. No process is possible whose sole result is the transfer of work from a hotter object to a hotter object

C. No process is possible whose sole result is the transfer of heat from a hotter object to a colder object

D. No process is possible whose sole result is the transfer of heat from a colder object to a hotter object

Answer : Option D
Explaination / Solution:

It is impossible for a self acting machine unaided by any external agency to transfer heat from a cold to a hot reservior. In other words,heat by itself cannot pass from a colder to a hotter body.

# Thermodynamics # CBSE 11TH PHYSICS Prepare / Learn

Q3. The spontaneous processes of nature are

A. dynamic

B. irreversible

C. reversible

D. static

Answer : Option B
Explaination / Solution:

Spontaneous processes are irreversible because they can be reversed only by taking a different path to get back to their original state. A reversible process can take the same path to return to its original state.

# Thermodynamics # CBSE 11TH PHYSICS Prepare / Learn

Q4. A thermodynamic process (state i →state f) is reversible

A. if the process can be turned back such that both the system and the surroundings return to their original states, with lots of change in the universe.

B. if the process can be turned back such that both the system and the surroundings return to different states, with lots of change in the universe.

C. if the process can be turned back such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe.

D. if the process can be turned back such that both the system and the surroundings return to their different states, with no other change anywhere else in the universe.

Answer : Option C
Explaination / Solution:

Irreversible process is one which can be reversed in such a way that all changes taking place in the direct process are exactly replaced in the inverse order and opposite sense and no changes are left behind.

# Thermodynamics # CBSE 11TH PHYSICS Prepare / Learn

Q5. The Carnot cycle consists of

A. Two isobaric processes connected by two adiabatic processes

B. Two isothermal processes connected by two isobaric processes

C. Two isothermal processes connected by two adiabatic processes

D. Two isochoric processes connected by two adiabatic processes

Answer : Option C
Explaination / Solution:

The Carnot cycle consists of the following four processes:

A reversible isothermal gas expansion process
A reversible adiabatic gas expansion process
A reversible isothermal gas compression process
A reversible adiabatic gas compression process

# Thermodynamics # CBSE 11TH PHYSICS Prepare / Learn

Q6. A geyser heats water flowing at the rate of 3.0 liters per minute from 27

^{\circ}

C to 77

^{\circ}

C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0

\times

1 0^{4}

J/g?

A. 14 g per min

B. 16 g per min

C. 12 g per min

D. 18 g per min

Answer : Option B
Explaination / Solution:

$Q = m s Δ T = 3000 \times 4.2 \times 50$

m = mass of water

s = specific heat of water

heat of combustion is 4.0 $\times$ $1 0^{4}$ J/g

rate of consumption of the fuel = Q / Heat of combustion

$= \frac{3000 \times 4.2 \times 50}{4 \times 10^{4}} = 16 g m / min$

# Thermodynamics # CBSE 11TH PHYSICS Prepare / Learn

Q7. What amount of heat must be supplied to 2.0

\times

1 0^{- 2}

kg of nitrogen (at room temperature) to raise its temperature by 45

^{\circ}

C at constant pressure? (Molecular mass of

N_{2}

= 28; R = 8.3 J

m o l^{- 1}

K^{- 1}

A. 934 J

B. 904

C. 954

D. 974

Answer : Option A
Explaination / Solution:

Q=nCvΔT=2×10−2×10−328×72×8.3×45=934Jn=wtingmMolecularwtCv=72RforDiatomicGas

# Thermodynamics # CBSE 11TH PHYSICS Prepare / Learn

Q8. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

A. 2.64

B. 2.84

C. 3.04

D. 2.44

Answer : Option A
Explaination / Solution:

PVγ=ConstPVγ=Pf(V2)γforH2,γ=1.4Pf=P21.4=2.64P

# Thermodynamics # CBSE 11TH PHYSICS Prepare / Learn

Q9. In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case?

A. 17.2 J

B. 17.6 J

C. 16.8 J

D. 17.9 J

Answer : Option C
Explaination / Solution:

$Δ Q = Δ U + W$

In adiabatic process $Δ Q = 0$

$Δ U = - W$

Given that W = -22.3J

$Δ U = 22.3 J$

in second process $Δ Q = 9.35 \times 4.18 = 39.083 J$

$Δ Q = Δ U + W$

$\begin{aligned} Δ Q = 9.35 \times 4.18 = 39.083 J \\ Δ Q = Δ U + W \\ 39.083 = 22.3 + W \\ W = 16.8 J \end{aligned}$

# Thermodynamics # CBSE 11TH PHYSICS Prepare / Learn

Q10. Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. What is the final pressure of the gas in A and B?

A. 0.5 atm

B. 0.6 atm

C. 0.55 atm

D. 0.45 atm

Answer : Option A
Explaination / Solution:

PV = Pf.2V

Pf = V / 2V

Pf = 0.5P

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