Heat and Thermodynamics — Page 5 | NEET Physics

Q41

The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is

-

20 o C, the temperature of the surroundings to which it rejects heat is

A 11 o C

B 21 o C

C 31 o C

D 41 o C

Correct Answer

Option C

Solution

Coefficient of performance,

Cop = {{{T_2}} \over {{T_1} - {T_2}}}

5 = {{273 - 20} \over {{T_1} - \left( {273 - 20} \right)}} = {{253} \over {{T_1} - 253}}

5T 1 – (5 × 253) = 253 5T 1 = 253 + (5 × 253) = 1518 $\therefore$

{T_1} = {{1518} \over 5} = 303.6

$\Rightarrow$ T 1 = 303.6 – 273 = 30.6 $\cong$ 31°C

Q42

An ideal gas is compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas ?

A Isochoric

B Isothermal

C Adiabatic

D Isobaric

Correct Answer

Option C

Solution

The P-V diagram of an ideal gas compressed from its initial volume

{V_0}

to

{{{V_0}} \over 2}

by several processes is shown in the figure.

Work done on the gas = Area under P–V curve As area under the P–V curve is maximum for adiabatic process, so work done on the gas is maximum for adiabatic process.

Q43

Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is

A 2

B

{1 \over 2}

C

{2 \over 3}

D

{3 \over 4}

Correct Answer

Option D

Solution

From PV = nRT

{P_A} = {{{\rho _A}{M_A}} \over {RT}}

and

{P_B} = {{{\rho _B}{M_B}} \over {RT}}

From question,

{{{P_A}} \over {{P_B}}} = {{{\rho _A}} \over {{\rho _B}}}{{{M_A}} \over {{M_B}}} = 2{{{M_A}} \over {{M_B}}} = {3 \over 2}

$\therefore$

{{{M_A}} \over {{M_B}}} = {3 \over 4}

Q44

One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure. The change in internal energy of the gas during the transition is

A 20 J

B

-

12 kJ

C 20 kJ

D

-

20 kJ

Correct Answer

Option D

Solution

Change in internal energy from A $\to$ B

\Delta U = {f \over 2}nR\Delta T = {f \over 2}nR\left( {{T_f} - {T_i}} \right)

= {5 \over 2}\left\{ {{P_f}{V_f} - {P_i}{V_i}} \right\}

(As gas is diatomic $\therefore$ f = 5)

= {5 \over 2}\left\{ {2 \times {{10}^3} \times 6 - 5 \times {{10}^3} \times 4} \right\}

= {5 \over 2}\left\{ {12 - 20} \right\} \times {10^3}J = 5 \times \left( { - 4} \right) \times {10^3}J

\therefore \Delta U = - 20\,KJ

Q45

The ratio of the specific heats

{{{C_p}} \over {{C_v}}} = \gamma

in terms of degrees of freedom (n) is given by

A

\left( {1 + {2 \over n}} \right)

B

\left( {1 + {n \over 2}} \right)

C

\left( {1 + {1 \over n}} \right)

D

\left( {1 + {n \over 3}} \right)

Correct Answer

Option A

Solution

For n degrees of freedom,

{C_v} = {n \over 2}R

Also,

{C_p} - {C_v} = R

{C_p} = {C_v} + R = {n \over 2}R + R = \left( {{n \over 2} + 1} \right)R

\gamma = {{{C_p}} \over {{C_v}}} = {{\left( {{n \over 2} + 1} \right)R} \over {{n \over 2}R}} = {{n + 2} \over n}

$\therefore$

\gamma = 1 + {2 \over n}

Q46

Figure below shows two paths that may be taken by a gas to go from a state A to a state C. In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be

A 460 J

B 300 J

C 380 J

D 500 J

Correct Answer

Option A

Solution

Considering the cyclic process ABCA Q cyclic = W = area of

\Delta

ABC

{Q_{AB}} + {Q_{BC}} + {Q_{CA}} = {1 \over 2} \times BC \times AB

400 + 100 + {Q_{CA}} = {1 \over 2} \times \left( {2 \times {{10}^{ - 3}}} \right) \times 4 \times {10^4}

400 + 100 - {Q_{AC}} = 40

Hence, Q AC = 460 J

Q47

A Carnot engine, having an efficiency of as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

A 90 J

B 1 J

C 100 J

D 99 J

Correct Answer

Option A

Solution

Given, efficiency of engine,

\eta = {1 \over {10}}

work done on system W = 10J Coefficient of performance of refrigerator

\beta = {{{Q_2}} \over W} = {{1 - \eta } \over \eta } = {{1 - {1 \over {10}}} \over {{1 \over {10}}}} = {{{9 \over {10}}} \over {{1 \over {10}}}} = 9

Energy absorbed from reservoir

{Q_2} = \beta w

{Q_2} = 9 \times 10 = 90\,J

Q48

A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is

A P 0 V 0

B 2P 0 V 0

C

{{{P_0}{V_0}} \over 2}

D zero

Correct Answer

Option D

Solution

In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise.

As is clear from figure,

{W_{AEDA}} = + area\,of\,\Delta AED = + {1 \over 2}{P_0}{V_0}

{W_{BCEB}} = - area\,of\Delta BCE = - {1 \over 2}{P_0}{V_0}

The net work done by the system is

{W_{net}} = {W_{AEDA}} + {W_{BCEB}}

= + {1 \over 2}{P_0}{V_0} - {1 \over 2}{P_0}{V_0} = 0

Q49

A monatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically to a volume 16V. The final pressure of the gas is (Take

\gamma

= 5/3)

A 64P

B 32P

C P/64

D 16P

Correct Answer

Option C

Solution

First, isothermal expansion

PV = P'\left( {2V} \right);P' = {P \over 2}

Then, adiabatic expansion

P'{\left( {2V} \right)^\gamma } = {P_f}{\left( {16V} \right)^\gamma }

(For adiabatic process, PV $\gamma$ = constant)

{P \over 2}\left( {2V} \right) = {P_f}{\left( {16V} \right)^{5/3}}

{P_f} = {P \over 2}{\left( {{{2V} \over {16V}}} \right)^{5/3}} = {P \over 2}{\left( {{1 \over 8}} \right)^{5/3}}

= {P \over 2}{\left( {{1 \over {{2^3}}}} \right)^{5/3}} = {P \over 2}\left( {{1 \over {{2^5}}}} \right) = {P \over {64}}

Q50

The mean free path of molecules of a gas, (radius r) is inversely proportional to

A r 3

B r 2

C r

D

\sqrt r

Correct Answer

Option B

Solution

Mean free path

{\lambda _m} = {1 \over {\sqrt 2 \pi {d^2}n}}

where d = diameter of molecule and d = 2r

\therefore {\lambda _m} \propto {1 \over {{r^2}}}