Description

This mock test of First Law Of Thermodynamics MCQ Level - 2 (Part - 2) for Physics helps you for every Physics entrance exam.
This contains 10 Multiple Choice Questions for Physics First Law Of Thermodynamics MCQ Level - 2 (Part - 2) (mcq) to study with solutions a complete question bank.
The solved questions answers in this First Law Of Thermodynamics MCQ Level - 2 (Part - 2) quiz give you a good mix of easy questions and tough questions. Physics
students definitely take this First Law Of Thermodynamics MCQ Level - 2 (Part - 2) exercise for a better result in the exam. You can find other First Law Of Thermodynamics MCQ Level - 2 (Part - 2) extra questions,
long questions & short questions for Physics on EduRev as well by searching above.

QUESTION: 1

Two moles of an ideal monoatomic gas is taken through a cycle ABCA as shown in the p-T diagram. During the process AB, Pressure and temperature of the gas vary such that pT = constant. If T_{1} = 300K then the work done on the gas in the process AB is

Select one:

Solution:

Number of moles, n = 2, T_{1}= 300 K, T_{2} = 2T_{1} = 600 K

During process A → B

pT = constant

or p^{2}V = constant = k [∴ pV = nRT]

p^{2}V = k

= 2 × 2 × R × [300 – 600] = -1200 R

The correct answer is: –1200 R

QUESTION: 2

One mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are 10^{5}N/m and 6L respectively. The final volume of the gas is 2L, molar specific heat of the gas at constant volume is 3R/2. The total work done is

Select one:

Solution:

In adiabatic process

Further,

or

= 6.24 × 10^{5} N/m^{2}

New york done in adiabatic process is given by

= –972 J

The correct answer is: -972 J

QUESTION: 3

Two identical containers A and B with frictionless pistons contains the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is m_{A} and that in B is m_{B} . The gas in each cylinder is now allowed to expand isothermally to the final volume 2V. The changes in the pressure in A and B are found to be Δp and 1.5 Δp respectively. Then

Select one:

Solution:

Process is isothermal. Therefore, T = constant. volume is increasing, therefore, pressure will decrease.

In chamber A, as Pressure decrease,

So,

...(i)

Similarly, in chamber B→

...(ii)

From equation (1) and (2)

or 2m_{A} = 2m_{A}

The correct answer is: 3 m_{A} = 2 m_{B}

QUESTION: 4

A cyclic process ABCA shown in the V-T diagram is performed with a constant mass of an ideal gas. Show the same process on a P-V diagram

Select one:

Solution:

Process A → B

From the graph,

From ideal gas equation, pV = nRT

⇒ p = constant

⇒ A → B, is isobaric process

So, in process A → B, when volume increases, pressure remains constant.

So, pV graph will be straight line parallel to V–axis as p = constant

Process B → C

From the graph, V = constant

From ideal gas equation, pV = nRT

⇒ pV = nRT

[V is constant]

But from the graph T is decreasing, so p also decrease.

But V remain constant.

pV graph will be straight line parallel to p-axis as V = constant

Process C → A

From the graph,

T = constant

From ideal gas equation, pV = nRT

⇒pV = nRT [T is constant]

But from the graph, V decreases, so p should increase.

so, pV graph will be rectangular hyperbola as

The correct answer is:

QUESTION: 5

An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66V and its temperature falls to T/2. How many degree of freedom do the gas molecules have?

Solution:

In adiabatic process = constant

Here,

We know

Here f = degree of freedom

So, putting the value of

f = 5

The correct answer is: 5

QUESTION: 6

An ideal gas has specific heat at constant pressure The gas is kept in a closed vessel of volume 0.0083^{3}m , at a temperature of 300K and a pressure of 1.6 × 10^{6}N/m^{2}. An amount of 2.49 × 10^{4}J of heat energy is supplied to the gas.

The final pressure and temperature of the gas are

Select one:

Solution:

Vessel is closed. Therefore, ΔW = 0

From first law of thermodynamics and ideal gas equation.

Q = ΔU + W

and pV = nRT

∴ W = 0

⇒ Q = ΔU

Also, Δ = nC_{V}ΔT

or,

Substituting the values, we have

ΔT = 375K

T_{f} = ΔT + T = 675K

= 3.6 × 10^{6}N/m^{2}

The correct answer is:

QUESTION: 7

A monoatomic ideal gas, initially at temperature T_{1}, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T_{2 }by releasing the piston suddenly. If L_{1} and L_{2} are the lengths of the gas column before and after expansion respectively, then T_{1} /T_{2} is given by

Select one:

Solution:

During adiabatic expansion, we known

For a monoatomic gas,

(A = Area of cross-section of piston)

The correct answer is: (L_{2} /L_{1})^{2/3}

QUESTION: 8

p-V plots for two gas during adiabatic process are shown in the figure. Plots 1 and 2 should correspond respectively to

Select one:

Solution:

In adiabatic process

Slope of p-V graph,

slope (with negative sign)

From the given graph,

(slope)_{2} > (slope)_{1}

∴

Therefore, 1 should corresponding to O_{2 } and 2 should correspond to He ( = 1.67)

The correct answer is: He and O_{2}

QUESTION: 9

Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The pistons of A is free to move, while that of B is kept fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is

Select one:

Solution:

A is free to move, therefore, heat will be supplied at constant pressure

∴ ...(i)

B is held fixed, therefore, heat will be supplied at constant volume.

∴ ...(ii)

But dQ_{A} = dQ_{B} (given)

∴

[ = 1.4 (diatomic)] (dT_{A} = 30 K)

= (1.4) (30 K)

dT_{B} = 42 K

The correct answer is: 42 K

QUESTION: 10

An ideal gas is taken through the cycle A → B → C → A a shown in the figure. If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process C → A is

Solution:

From first law of thermodynamics

Q = W + ΔU

ΔU = 0 (process ABCA is cyclic)

= 5 – 10 – 0 = –5 J

The correct answer is: –5 J

### First law of thermodynamics - Thermodynamics

Video | 13:24 min

### First Law of Thermodynamics

Video | 11:27 min

### First law of thermodynamics

Video | 24:27 min

### First Law of Thermodynamics

Video | 04:09 min

- First Law Of Thermodynamics MCQ Level - 2 (Part - 2)
Test | 10 questions | 45 min

- First Law Of Thermodynamics MCQ Level â€“ 2 (Part - 1)
Test | 10 questions | 45 min

- First Law Of Thermodynamics MCQ Level â€“ 1 (part - 2)
Test | 10 questions | 30 min

- First Law Of Thermodynamics NAT Level â€“ 2
Test | 10 questions | 45 min

- First Law Of Thermodynamics - 2
Test | 15 questions | 45 min