Heat and Thermodynamics — Page 6 | JEE Physics

Q51

Two thermally insulated vessels

1

and

2

are filled with air at temperatures

\left( {{T_1},{T_2}} \right),

volume

\left( {{V_1},{V_2}} \right)

and pressure

\left( {{P_1},{P_2}} \right)

respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

A

{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)/\left( {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}} \right)

B

\left( {{T_1} + {T_2}} \right)/2

C

{{T_1} + {T_2}}

D

{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)/\left( {{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}} \right)

Correct Answer

Option A

Solution

Here

Q=0

and

W=0.

Therefore from first law of thermodynamics

\Delta U = Q + W = 0

$\therefore$ Internal energy of the system with partition $=$ Internal energy of the system without partition.

{n_1}{C_v}\,{T_1} + {n_2}\,{C_v}{T_2} = \left( {{n_1} + {n_2}} \right){C_v}\,T

$\therefore$

T = {{{n_1}{T_1} + {n_2}T{}_2} \over {{n_1} + {n_2}}}

But

{n_1} = {{{P_1}{V_1}} \over {R{T_1}}}

and

{n_2} = {{{P_2}{V_2}} \over {R{T_2}}}

$\therefore$

T = {{{{{P_1}{V_1}} \over {R{T_1}}} \times {T_1} + {{{P_2}{V_2}} \over {R{T_2}}} \times {T_2}} \over {{{{P_1}{V_1}} \over {R{T_1}}} + {{{P_2}{V_2}} \over {R{T_2}}}}}

= {{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)} \over {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}

Q52

Which of the following statements is correct for any thermodynamic system ?

A The change in entropy can never be zero

B Internal energy and entropy and state functions

C The internal energy changes in all processes

D The work done in an adiabatic process is always zero,

Correct Answer

Option B

Solution

Internal energy and entropy are state function, they do not depend upon path taken.

Q53

A gaseous mixture consists of

16

g

of helium and

16

g

of oxygen. The ratio

{{Cp} \over {{C_v}}}

of the mixture is

A

1.62

B

1.59

C

1.54

D

1.4

Correct Answer

Option A

Solution

{{{n_1} + {n_2}} \over {r - 1}} = {{{n_1}} \over {{r_1} - 1}} + {{{n_2}} \over {{r_2} - 1}}

{{{{16} \over 4} + {{16} \over {32}}} \over {r - 1}} = {{16/4} \over {{5 \over 3} - 1}} + {{16/32} \over {1.4 - 1}}

$\therefore$

\gamma = 1.62

Q54

The work of

146

kJ

is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by

{7^ \circ }C.

The gas is

\left( {R = 8.3J\,\,mo{l^{ - 1}}\,{K^{ - 1}}} \right)

A diatomic

B triatomic

C a mixture of monoatomic and diatomic

D monoatomic

Correct Answer

Option A

Solution

W = {{nR\Delta T} \over {1 - \gamma }} \Rightarrow - 146000

= {{1000 \times 8.3 \times 7} \over {1 - \gamma }}

or

1 - \gamma = - {{58.1} \over {146}} \Rightarrow \gamma

= 1 + {{58.1} \over {146}} = 1.4

Hence the gas is diatomic.

Q55

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature

{T_0},

while Box contains one mole of helium at temperature

\left( {{7 \over 3}} \right){T_0}.

The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases,

{T_f}

in terms of

{T_0}

is

A

{T_f} = {3 \over 7}{T_0}

B

{T_f} = {7 \over 3}{T_0}

C

{T_f} = {3 \over 2}{T_0}

D

{T_f} = {5 \over 2}{T_0}

Correct Answer

Option C

Solution

Heat lost by He $=$ Heat gained by

{N_2}

{n_1}C{v_1}\Delta {T_1} = {n_2}C{v_2}\Delta {T_2}

{3 \over 2}R\left[ {{7 \over 3}{T_0} - {T_f}} \right]

= {5 \over 2}R\left[ {{T_f} - {T_0}} \right] \Rightarrow {T_f}

= {3 \over 2}{T_0}

Q56

A Carnot engine, having an efficiency of

\eta = 1/10

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

100

J

B

99

J

C

90

J

D

1

J

Correct Answer

Option C

Solution

The efficiency

\left( \eta \right)

of a Carnot engine and the coefficient of performance

\left( \beta \right)

of a refrigerator are related as

\beta = {{1 - \eta } \over \eta }

\,\,\,\,\,\,

Here,

\eta = {1 \over {10}}

\,\,\,\,\,\,

$\therefore$

\beta = {{1 - {1 \over {10}}} \over {\left( {{1 \over {10}}} \right)}} = 9.

Also, Coefficient of performance

\left( \beta \right)

is given by

\beta = {{{Q_2}} \over W},

where

{Q_2}

is the energy absorbed from the reservoir. or,

9 = {{{Q_2}} \over {10}}

$\therefore$

{Q_2} = 90\,J.

Q57

The speed of sound in oxygen

\left( {{O_2}} \right)

at a certain temperature is

460\,\,m{s^{ - 1}}.

The speed of sound in helium

(He)

at the same temperature will be (assume both gases to be ideal)

A

1421\,\,m{s^{ - 1}}

B

500\,\,m{s^{ - 1}}

C

650\,\,m{s^{ - 1}}

D

300\,\,m{s^{ - 1}}

Correct Answer

Option A

Solution

The speed of sound in a gas is given by

v = \sqrt {{{\gamma RT} \over M}}

$\therefore$

{{{v_{{O_2}}}} \over {{v_{He}}}} = \sqrt {{{{\gamma _{{O_2}}}} \over {{M_{{O_2}}}}} \times {{{M_{He}}} \over {{\gamma _{He}}}}}

= \sqrt {{{1.4} \over {32}} \times {4 \over {1.67}}} = 0.3237

$\therefore$

{v_{He}} = {{{v_{{O_2}}}} \over {0.3237}}

= {{460} \over {0.3237}}

= 1421\,m/s

Q58

n moles of an ideal gas with constant volume heat capcity CV undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is :

A

{{nR} \over {{C_V} - nR}}

B

{{4nR} \over {{C_V} - nR}}

C

{{4nR} \over {{C_V} + nR}}

D

{{nR} \over {{C_V} + nR}}

Correct Answer

Option D

Solution

w = nR

\Delta

T

\Delta

H = (Cv + nR)

\Delta

T

{\omega \over {\Delta H}} = {{nR} \over {{C_v} + nR}}

Q59

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume

{V_1}

and contains ideal gas at pressure

{P_1}

and temperature

{T_1}

. The other chamber has volume

{V_2}

and contains ideal gas at pressure

{P_2}

and temperature

{T_2}

. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

A

{{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)} \over {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}

B

{{{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}} \over {{P_1}{V_1} + {P_2}{V_2}}}

C

{{{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}} \over {{P_1}{V_1} + {P_2}{V_2}}}

D

{{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)} \over {{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}}}

Correct Answer

Option A

Solution

Since, no work is done and system is thermally insulated from surrounding.

Therefore, total internal energy is constant, that is, $U=U_1+U_2$ Assuming gas have same degree of freedom, we have

\left(n_1+n_2\right) C_{\mathrm{V}} T=n_1 C_{\mathrm{V}} T_1+n_2 C_{\mathrm{V}} T_2

Therefore,

T=\frac{n_1 R T_1+n_2 R T_2}{R n_1+n_2 R}=\frac{\left(P_1 V_1+P_2 V_2\right)}{\frac{P_1 V_1}{T_1}+\frac{P_2 V_2}{T_2}}=\frac{\left(P_1 V_1+P_2 V_2\right) T_1 T_2}{\left(P_1 V_1 T_2+P_2 V_2 T_1\right)}

Q60

A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from

V

to

32

V

, the efficiency of the engine is

A

0.5

B

0.75

C

0.99

D

0.25

Correct Answer

Option B

Solution

{T_1}{V^{\gamma - 1}} = {T_2}{\left( {32V} \right)^{\gamma - 1}}

\Rightarrow {T_1} = {\left( {32} \right)^{\gamma - 1}}.{T_2}

For diatomic gas,

\gamma = {7 \over 5}

$\therefore$

\gamma - 1 = {2 \over 5}

$\therefore$

{T_1} = {\left( {32} \right)^{{2 \over 5}}}.{T_2} \Rightarrow {T_1} = 4{T_2}

Now, efficiency

= 1 - {{{T_2}} \over {{T_1}}}

= 1 - {{{T_2}} \over {4{T_2}}}

= 1 - {1 \over 4}

= {3 \over 4}

= 0.75.