One mole of an ideal monoatomic gas is heated in a constant pressure of one atmosphere from $0^\circ \,C$ to $100^\circ \,C$. Then the change in the internal energy is

$12.48 \times {10^2}{\rm{ joules}}$

$8.32 \times {10^2}{\rm{ joules}}$

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NEET • PHYSICS • THERMODYNAMICS

Thermodynamics practice · Hard

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1 Hard

1 of 17

Understanding +4 | -1

In an adiabatic process, which of the following
statements is true?

View answer and explanation

Correct option: C

The molar heat capacity is zero

Option C (The molar heat capacity is zero) is correct because in an adiabatic process, there is no heat transfer (Q=0), and molar heat capacity is defined as C = Q/nΔT, so C=0.

2 Hard

2 of 17

Application +4 | -1

One mole of an ideal diatomic gas undergoes a thermodynamic process, in which its molar heat capacity varies directly proportional to temperature as

{C}=\alpha{T}

, where

\alpha

is a positive constant. Work done by the gas when it is heated from initial temperature

T_0

to a final temperature

3 T_0

will be

View answer and explanation

Correct option: C

\left(4 \alpha {T}_0-5 {R}\right) {T}_0

Option C (

\left(4 \alpha {T}_0-5 {R}\right) {T}_0

) is correct because it represents the work done calculated by integrating the heat capacity over the temperature range and applying the first law of thermodynamics.

3 Hard

3 of 17

Application +4 | -1

One mole of an ideal monoatomic gas is heated in a constant pressure of one atmosphere from

0^\circ \,C

100^\circ \,C

. Then the change in the internal energy is

View answer and explanation

Correct option: C

12.48 \times {10^2}\mathrm{ joules}

The change in the internal energy is

12.48 \times {10^2}\mathrm{ joules}

because it is calculated using the formula

\Delta U = nC_v\Delta T

for a monoatomic gas.

4 Hard

4 of 17

Understanding +4 | -1

Identify the characteristics of an adiabatic process in a monoatomic gas.

(A) Internal energy is constant.

(B) Work done in the process is equal to the charge in internal energy.

(C) The product of temperature and volume is a constant.

(D) The product of pressure and volume is a constant.

(E) The work done to change the temperature from

\mathrm{T}_1

\mathrm{T}_2

is proportional to

\left(\mathrm{T}_2-\mathrm{T}_1\right)

Choose the correct answer from the options given below:

View answer and explanation

Correct option: C

(B), (E) only

Option C ((B), (E) only) is correct because in an adiabatic process, work done equals the negative change in internal energy, and work done is proportional to the temperature difference.

5 Hard

5 of 17

Application +4 | -1

In an adiabatic process where pressure is increased by

\frac{2}{3}\% \mathrm{if}\frac{{{C_p}}}{{{C_v}}} = \frac{3}{2},

then the volume decreases by about

View answer and explanation

Correct option: A

\frac{4}{9}\%

Option A (

\frac{4}{9}\%

) is correct because it is derived from the adiabatic process equation relating pressure and volume changes.

6 Hard

6 of 17

Analyzing +4 | -1

Consider the efficiency of carnot's engine is given by

\eta=\frac{\alpha \beta}{\sin \theta} \log e \frac{\beta x}{k T}

, where

\alpha

and

\beta

are constants. If

T

is temperature,

k

is Boltzmann constant,

\theta

is angular displacement and

x

has the dimensions of length. Then, choose the incorrect option

View answer and explanation

Correct option: D

Dimensions of

\alpha

is same as that of

\beta

Option D is the incorrect option because, based on dimensional analysis of the given equation, the dimensions of

\alpha

and

\beta

are not the same.

7 Hard

7 of 17

Application +4 | -1

A litre of dry air STP expands adiabatically to a volume of 3 L. If

\mathrm{\gamma } = 1.40,

the work done by air is

\mathrm{(}{\mathrm{3}^{\mathrm{1}\mathrm{.4}}}\mathrm{ = 4}\mathrm{.6555)}

[Take, air to be an ideal gas]

View answer and explanation

Correct option: B

90.5 J

Option B (90.5 J) is correct because the work done by the air is calculated using the adiabatic process equation and the ideal gas law.

8 Hard

8 of 17

Application +4 | -1

N

moles of non-linear polyatomic gas (degree of freedom 6) is mixed with 2 moles of monoatomic gas. The resultant mixture has molar specific heat equal to that of a diatomic gas, then

N

View answer and explanation

Correct option: D

Option D (4) is correct because the number of moles N is calculated by equating the molar specific heat of the mixture to that of a diatomic gas.

9 Hard

9 of 17

Application +4 | -1

Work done in rising temperature of five mole of Helium adiabatically through

\theta^{\circ} {C}

{W}

. Then the work done in an isobaric process of the sample for same change of temperature is

View answer and explanation

Correct option: D

-\frac{2 W}{3}

Option D (

-\frac{2 W}{3}

) is correct because the ratio of work done in the isobaric process to the adiabatic process is -2/3 based on the specific heat capacities.

10 Hard

10 of 17

Application +4 | -1

A diatomic gas

(\gamma=1.4)

does

100 \mathrm{~J}

of work in an isobaric expansion. The heat given to the gas is:

View answer and explanation

Correct option: A

350 \mathrm{~J}

Option A (

350 \mathrm{~J}

) is correct because it is calculated using the relationship between heat, work, and the adiabatic index for an isobaric process.

11 Hard

11 of 17

Application +4 | -1

One mole of an ideal monatomic gas is taken from state

\mathrm{A}

to state B through the process

P=(3 / 2) T^{1 / 2}

. It is found that its temperature increases by

100 \mathrm{~K}

in this process. Now it is taken from state B to C through a process for which internal energy is related to volume as

U_{\text {int }}=(1 / 2) V^{1 / 2}

. Find the total work performed by the gas (in joule), if it is given that volume at

\mathrm{B}

100 \mathrm{~m}^3

and at C it is

1600 \mathrm{~m}^3

. (Use

R=8.3 \mathrm{~J} / \mathrm{mol} \mathrm{K}

View answer and explanation

Correct option: A

435

The total work performed by the gas is 435 J. The work done in process A to B is calculated using the given relation P = (3/2)T^(1/2) and the ideal gas law. The work done in process B to C is calculated using the relation between internal energy and volume.

12 Hard

12 of 17

Application +4 | -1

A man of 60 Kg gains 1000 Cal of heat by eating 5 mangoes. His efficiency is 56%. The height to which he can Jump by using this energy is

View answer and explanation

Correct option: A

4 m

Option A (4 m) is correct because it is the height calculated using the man's efficiency and the energy gained from eating mangoes.

13 Hard

13 of 17

Application +4 | -1

A monoatomic gas

(\gamma = 5/3)

is suddenly compressed to 1/8 of its original volume adiabatically, then the pressure of the gas, will change to

View answer and explanation

Correct option: D

32 times its initial pressure

Option D (32 times its initial pressure) is correct because the pressure changes by a factor of 32 when the volume is adiabatically compressed to 1/8 of its original volume with γ = 5/3.

14 Hard

14 of 17

Application +4 | -1

One mole of an ideal gas with

\gamma = 1.4,

is adiabatically compressed so that its temperature rises from

27^\circ C

35^\circ C,

. The change in the internal energy of the gas is

\left( {R = 8.3J/mol.\,k} \right)

View answer and explanation

Correct option: B

166 J

Option B (166 J) is correct because it is the change in internal energy calculated using the given temperature change and the number of moles.

15 Hard

15 of 17

Application +4 | -1

A bullet of mass

m \mathrm{~kg}

is moving with velocity

K^2 \, \mathrm{m} / \mathrm{s}

collides against a vertical wall, consequently half of its kinetic energy is converted into heat. If whole heat is acquired by the bullet, the rise in temperature

\text{(in }{ }^{\circ} \mathrm{C})

of bullet is

\text{(specific heat for bullet }=s \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C})

View answer and explanation

Correct option: B

\frac{K^4}{4 s}

Option B (

\frac{K^4}{4 s}

) is correct because it is the rise in temperature calculated from the kinetic energy converted to heat.

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16 Hard

16 of 17

Application +4 | -1

The specific heat at constant pressure of a real gas obeying

\mathrm{PV}^2=\mathrm{RT}

equation is :

View answer and explanation

Correct option: D

\mathrm{C}_{\mathrm{V}}+\frac{\mathrm{R}}{2 \mathrm{V}}

Option D (

\mathrm{C}_{\mathrm{V}}+\frac{\mathrm{R}}{2 \mathrm{V}}

) is correct because it accounts for the additional term arising from the equation of state

\mathrm{PV}^2=\mathrm{RT}

when calculating

C_p

17 Hard

17 of 17

Application +4 | -1

Two heat engines are operating in such a way that the heat rejected by the first engine is used as the heat input of the second. If both the engines are

40 \%

efficient, then the overall efficiency of the system is

View answer and explanation

Correct option: A

0.64

Option A (0.64) is correct because the overall efficiency is the product of the individual efficiencies when engines are operating in series.

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Key Takeaways

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