The perfect gas goes from a state A to another state B by absorbing $8 \times {10^5}$J of heat and doing $6.5 \times {10^5}{\rm{J}}$of external work. It is now transferred between the same two states in another process in which it absorbs ${10^5}J$ of heat in the second process. Then

Work done on the gas is $0.5 \times {10^5}{\rm{J}}$

Work done on the gas is ${10^5}J$

A gas expands under constant pressure P from volume ${V_1}$ to ${V_2}$. The work done by the gas is

$P\left( {{V_1} - {V_2}} \right)$

$P\left( {V_1^\gamma - V_2^\gamma } \right)$

$P\frac{{{V_1}{V_2}}}{{{V_2} - {V_1}}}$

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NEET • PHYSICS • THERMODYNAMICS

Thermodynamics practice · Medium

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Showing 1-30 of 126 questions

1 Medium

1 of 126

Analyzing +4 | -1

The work done during expansion of a gas is independent of pressure if the expansion takes place

View answer and explanation

Correct option: D

In none of the above cases

The correct answer is D because the work done during the expansion of a gas depends on the pressure, regardless of whether the process is isothermal or adiabatic.

2 Medium

2 of 126

Remembering +4 | -1

The

p-T

relation for an adiabatic expansion is

View answer and explanation

Correct option: D

p^{1-\gamma} T^\gamma=

constant

Option D is correct because the correct p-T relation for an adiabatic process is given by

p^{1-\gamma}T^{\gamma} = constant

3 Medium

3 of 126

Application +4 | -1

A Carnot engine has efficiency

\frac{1}{4}

. Efficiency becomes

\frac{1}{3}

when temperature of sink is decreased by 50 K (keeping source fixed). What is the original temperature of the sink?

View answer and explanation

Correct option: D

450 K

Option D (450 K) is correct because it is the original temperature of the sink calculated from the Carnot engine efficiency equations.

4 Medium

4 of 126

Understanding +4 | -1

The

SI

unit of mechanical equivalent of heat is

View answer and explanation

Correct option: B

Joule/Calorie

Option B is correct because the mechanical equivalent of heat (J) relates the amount of work (in Joules) to the amount of heat (in Calories), so its unit is Joule/Calorie.

5 Medium

5 of 126

Remembering +4 | -1

Which of the following is/are thermodynamic coordinate?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because volume, pressure, and temperature are all state variables that define the thermodynamic state of a system.

6 Medium

6 of 126

Application +4 | -1

In the certain process, 400 cal of heat are supplied to a system and at the same time 105 J of mechanical work was done on the system. The increase in its internal energy is

View answer and explanation

Correct option: D

425 cal

Option D (425 cal) is correct because it is the increase in internal energy calculated using the first law of thermodynamics:

\Delta U = Q - W

, where W is negative when work is done on the system.

7 Medium

7 of 126

Understanding +4 | -1

A thermally insulated container is divided into two parts by a screen. In one part the pressure and temperature are

P

and

T

for an ideal gas filled. In the second part it is vacuum. If now a small hole is created in the screen, then the temperature of the gas will

View answer and explanation

Correct option: C

Remain same

Option C (Remain same) is correct because the process is a free expansion into a vacuum, which is an irreversible adiabatic process with no work done and no heat exchange, resulting in no change in temperature for an ideal gas.

8 Medium

8 of 126

Analyzing +4 | -1

Assertion: PV value of gas can increase on increasing the pressure.
Reason: Pressure always increases on compression.

View answer and explanation

Correct option: C

If assertion is True but the Reason is False

Option C (If assertion is True but the Reason is False) is correct because PV can increase with pressure, but pressure doesn't always increase on compression.

9 Medium

9 of 126

Understanding +4 | -1

A gas undergoes an adiabatic change. Its specific heat in the process is

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because in an adiabatic process, there is no heat transfer (Q=0), so the specific heat (c = Q/mΔT) is zero.

10 Medium

10 of 126

Application +4 | -1

The perfect gas goes from a state A to another state B by absorbing

8 \times {10^5}

J of heat and doing

6.5 \times {10^5}\mathrm{J}

of external work. It is now transferred between the same two states in another process in which it absorbs

{10^5}J

of heat in the second process. Then

View answer and explanation

Correct option: A

Work done on the gas is

0.5 \times {10^5}\mathrm{J}

Work done on the gas is

0.5 \times {10^5}\mathrm{J}

because the change in internal energy is path-independent, and the difference in heat absorbed corresponds to the difference in work done.

11 Medium

11 of 126

Understanding +4 | -1

The internal energy of the gas increases in

View answer and explanation

Correct option: B

Adiabatic compression

Option B (Adiabatic compression) is correct because in adiabatic compression, work is done on the gas, increasing its internal energy since no heat is exchanged.

12 Medium

12 of 126

Remembering +4 | -1

The adiabatic Bulk modulus of a perfect gas at pressure is given by

View answer and explanation

Correct option: D

\gamma P

Option D (

\gamma P

) is correct because the adiabatic bulk modulus of a perfect gas is given by the formula

\gamma P

, where

\gamma

is the adiabatic index and P is the pressure.

13 Medium

13 of 126

Understanding +4 | -1

Which of the following is unique function of initial and final states?

View answer and explanation

Correct option: C

Option C (dU) is correct because dU represents the change in internal energy, which is a state function and depends only on the initial and final states.

14 Medium

14 of 126

Remembering +4 | -1

The thermodynamic process, in which internal energy of the system remains constant is

View answer and explanation

Correct option: C

Isothermal

Option C (Isothermal) is correct because an isothermal process is defined as a process occurring at constant temperature, implying constant internal energy for an ideal gas.

15 Medium

15 of 126

Understanding +4 | -1

It is difficult to cook rice in an open vessel by boiling it a high altitudes because of

View answer and explanation

Correct option: C

Low boiling point and low pressure

Option C (Low boiling point and low pressure) is correct because at high altitudes, the atmospheric pressure is lower, which reduces the boiling point of water, making it difficult to cook rice.

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16 Medium

16 of 126

Remembering +4 | -1

The Davison – Germer experiment is direct evidence of

View answer and explanation

Correct option: B

Wave nature of electron

Option B (Wave nature of electron) is correct because the Davisson-Germer experiment demonstrated the diffraction of electrons, confirming their wave-like properties.

17 Medium

17 of 126

Application +4 | -1

{C_V} = 4.96cal/mole K

,then increase in internal energy when temperature of 2 moles of this gas is increased from

340K

342K

View answer and explanation

Correct option: B

19.84cal

Option B (

19.84cal

) is correct because it is the increase in internal energy calculated using the formula

\Delta U = nC_V \Delta T

18 Medium

18 of 126

Understanding +4 | -1

When heat is given to a gas in an isobaric process, then

View answer and explanation

Correct option: C

Both 1 and 2

Option C (Both 1 and 2) is correct because in an isobaric process, heat added does both work and increases internal energy.

19 Medium

19 of 126

Understanding +4 | -1

dU+dW=0 is valid for

View answer and explanation

Correct option: D

Isochoric process

Option D (Isochoric process) is correct because in an isochoric process, the volume is constant, so no work is done (dW=0), and thus dU+dW=dU=0 if the process is also isothermal.

20 Medium

20 of 126

Remembering +4 | -1

Two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other.

View answer and explanation

Correct option: D

Zeroth law of thermodynamics

Option D (Zeroth law of thermodynamics) is correct because it describes the transitive property of thermal equilibrium.

21 Medium

21 of 126

Analyzing +4 | -1

n_1

and

n_2

are the moles of two ideal gases, whose thermodynamics constants, [ratio of

\left(\frac{C_p}{C_v}\right)

] are

\gamma_1

and

\gamma_2

respectively, are mixed.

\left(\frac{C_p}{C_v}\right)

for the mixture is-

View answer and explanation

Correct option: D

\frac{n_1 \gamma_1\left(\gamma_2-1\right)+n_2 \gamma_2\left(\gamma_1-1\right)}{n_1\left(\gamma_2-1\right)+n_2\left(\gamma_1-1\right)}

Option D is correct because it represents the correct formula for calculating the ratio of specific heats for a mixture of ideal gases based on their individual ratios and mole fractions.

22 Medium

22 of 126

Remembering +4 | -1

Which property of the system does not change in an adiabatic process

View answer and explanation

Correct option: D

Total amount of heat

Option D is correct because, by definition, an adiabatic process is one in which there is no heat exchange between the system and its surroundings, meaning the total amount of heat remains constant.

23 Medium

23 of 126

Analyzing +4 | -1

The following sets of values for

{C_V}

and

{C_P}

of a gas has been reported by different students. The units are

cal/g

mole

K.

Which of these sets is most reliable

View answer and explanation

Correct option: A

{C_V} = 3,\;{C_P} = 5

Option A (

{C_V} = 3,\;{C_P} = 5

) is the most reliable because the difference between

C_P

and

C_V

should be approximately equal to R (approximately 2 cal/mol-K).

24 Medium

24 of 126

Application +4 | -1

Find the change in the entropy in the following process

100g

of ice at

{0^ \circ }C

melts when dropped in a bucket of water at

{50^ \circ }C

(Assume temperature of water does not change)

View answer and explanation

Correct option: B

+ 4.5cal/K

The change in entropy is

+ 4.5cal/K

because it is calculated by dividing the heat absorbed by the ice during melting by the temperature at which melting occurs.

25 Medium

25 of 126

Remembering +4 | -1

which of the following is the unit of specific heat

View answer and explanation

Correct option: B

J/kg{\;^0}C

Option B (

J/kg{\;^0}C

) is correct because specific heat is defined as the amount of heat required to raise the temperature of 1 kg of a substance by 1 degree Celsius.

26 Medium

26 of 126

Remembering +4 | -1

The first law of thermodynamics is concerned with the conservation of

View answer and explanation

Correct option: B

Energy

Option B is correct because the first law of thermodynamics is a statement of the conservation of energy, stating that energy cannot be created or destroyed, only transformed.

27 Medium

27 of 126

Remembering +4 | -1

First law of thermodynamics is based on

View answer and explanation

Correct option: B

Law of conservation of energy

Option B (Law of conservation of energy) is correct because the first law of thermodynamics is a specific application of the law of conservation of energy.

28 Medium

28 of 126

Understanding +4 | -1

A gas expands under constant pressure P from volume

{V_1}

{V_2}

. The work done by the gas is

View answer and explanation

Correct option: A

P\left( {{V_2} - {V_1}} \right)

Option A is correct because the work done by a gas expanding at constant pressure is given by the formula

W = P(V_2 - V_1)

, where P is the constant pressure, V2 is the final volume, and V1 is the initial volume.

29 Medium

29 of 126

Understanding +4 | -1

The change of entropy, when heat is absorbed by the gas is:

View answer and explanation

Correct option: A

Positive

When heat is absorbed by a gas, the entropy increases because the gas molecules have more energy and can move in more disordered ways. Therefore, the change in entropy is positive.

30 Medium

30 of 126

Remembering +4 | -1

The specific heat of an ideal gas varies with temperature

T

View answer and explanation

Correct option: D

{T^0}

Option D is correct because the specific heat of an ideal gas is constant and does not vary with temperature, hence it is proportional to

T^0

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Key Takeaways

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