EASY

Agniveer Vayu

IMPORTANT

Earn 100

Assertion: The average degree of freedom per molecule for a gas is $6$ . If gas performs $25 J$ of work when it expands at a constant pressure then the heat absorbed by the gas is $100 J$ .

Reason: First law of thermodynamics is $d Q = ∆ U + d W$ where $∆ U = n C_{v} d T$ .

100% studentsanswered this correctly

Important Questions on Thermodynamics

EASY

Agniveer Vayu

IMPORTANT

Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 12 - Thermodynamics>Exercise>Q 2

A gas can be taken from

A

B

via two different processes

ACB

and

ADB

When path

ACB

is used

60 J

of heat flows into the system and

30 J

of work is done by the system. If the path

ADB

is used then work done by the system is

10 J

, the heat flows into the system in the path

ADB

is:

EASY

Agniveer Vayu

IMPORTANT

Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 12 - Thermodynamics>Exercise>Q 3

An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states are

Q₁ = 6000 J,
Q₂ = -5500 J;
Q₃ = -3000 J;
Q₄ = 3500 J
W₁ = 2500 J;
W₂ = -1000 J;
W₃ = -1200 J;
W₄ = x J.

The ratio of the net work done by the gas to the total heat absorbed by the gas is n. The values of x and η respectively are

EASY

Agniveer Vayu

IMPORTANT

Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 12 - Thermodynamics>Exercise>Q 4

For path

A B C

, Heat given to the system is

60 J

and work done by the system is

30 J

. For path

A D C

, work done by the system is

10 J

. The heat given to the system for path

A D C

EASY

Agniveer Vayu

IMPORTANT

Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 12 - Thermodynamics>Exercise>Q 5

Five moles of an ideal monoatomic gas with an initial temperature of

150 ° C

expand and in the process absorb

1500 J

of heat and does

2500 J

of work. The final temperature of the gas in

° C

is (ideal gas constant,

R = 8.314 J K^{- 1} {mol}^{- 1}

EASY

Agniveer Vayu

IMPORTANT

Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 12 - Thermodynamics>Exercise>Q 6

An ideal gas with adiabatic exponent

(γ = 1.5)

undergoes a process in which work done by the gas is same as increase in internal energy of the gas. The molar heat capacity of gas for the process is

EASY

Agniveer Vayu

IMPORTANT

Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 12 - Thermodynamics>Exercise>Q 7

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

EASY

Agniveer Vayu

IMPORTANT

Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 12 - Thermodynamics>Exercise>Q 8

In the thermodynamical process, the pressure of a fixed mass of gas is changed in such a manner that the gas releases

20 J

of heat and

8 J

of work is done on the gas. If the internal energy of the gas was

30 J

, then the final internal energy will be

EASY

Agniveer Vayu

IMPORTANT

Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 12 - Thermodynamics>Exercise>Q 1

The temperature of a silver bar rises by

10 ° C

when it absorbs

1.23 kJ

of energy by heat. The mass of the bar is

525 g

. Determine the specific heat of silver.

Assertion: The average degree of freedom per molecule for a gas is 6. If gas performs 25 J of work when it expands at a constant pressure then the heat absorbed by the gas is 100 J. Reason: First law of thermodynamics is dQ=∆U+dW where ∆U=nCv dT .

Important Questions on Thermodynamics

Learn and Prepare for any exam you want !

Assertion: The average degree of freedom per molecule for a gas is $6$ . If gas performs $25 J$ of work when it expands at a constant pressure then the heat absorbed by the gas is $100 J$ .

Reason: First law of thermodynamics is $d Q = ∆ U + d W$ where $∆ U = n C_{v} d T$ .