Heat and Thermodynamics — Page 11 | JEE Physics

Q101

An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true? (A) the mean free path of the molecules decreases. (B) the mean collision time between the molecules decreases. (C) the mean free path remains unchanged. (D) the mean collision time remains unchanged.

A (C) and (D)

B (A) and (D)

C (B) and (C)

D (A) and (B)

Correct Answer

Option C

Solution

The mean free path of molecules of an ideal gas is given as: $\lambda$ =

{V \over {\sqrt 2 \pi {d^2}N}}

where : V = Volume of container N = No of molecules Mean free path is independent of temperature hence with increasing temp since volume of container does not change (closed container), so mean free path is unchanged.

Average collision time = $\lambda$ Vav and Vav $\propto$

\sqrt T

$\therefore$ Average collision time $\propto$

{1 \over {\sqrt T }}

Hence with increase in temperature the average collision time decreases.

Q102

A calorimeter of water equivalent 20 g contains 180 g of water at 25oC. ‘m’ grams of steam at 100oC is mixed in it till the temperature of the mixure is 31oC. The value of ‘m’ is close to : (Latent heat of water = 540 cal g–1, specific heat of water = 1 cal g–1 oC–1)

A 2.6

B 2

C 4

D 3.2

Correct Answer

Option B

Solution

Given Temp of mixture = 31°C 180×1×(31-25) + 20×(31-25) = m×540 + m×1×(100-31) $\Rightarrow$ 180×6 + 20×6 = 540m + 100 m - 31m $\Rightarrow$ 1080 + 120 = 640 m - 31m $\Rightarrow$ 1200 = 609m $\Rightarrow$ m =

{{1200} \over {609}}

= 1.97

Q103

Number of molecules in a volume of 4 cm3 of a perfect monoatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to? (Given, mean kinetic energy of a molecule (at T) is 4

\times

10–14 erg, g = 980 cm/s2, density of mercury = 13.6 g/cm3)

A 5.8

\times

1018

B 4.0

\times

1016

C 5.8

\times

1016

D 4.0

\times

1018

Correct Answer

Option D

Solution

E = {3 \over 2}kT \Rightarrow \left( {T = {{2E} \over {3k}}} \right),

Also

\,PV = NkT

P = \rho gh,\,V = 4c{m^3}

$\therefore$ (

\rho gh

)V =

Nk \times {{2E} \over {3k}}

$\Rightarrow$

13.6 \times {10^3} \times 9.8 \times 2 \times {10^{ - 2}} \times 4 \times {10^{ - 6}}

= Nk \times {{2E} \over {3k}} = {{N \times 2} \over 3} \times 4 \times {10^{ - 14}} \times {10^{-7}}

$\Rightarrow$

N = {{13.6 \times 19.6 \times 4 \times {{10}^{ - 5}} \times 3 \times 10} \over 8}

$\Rightarrow$

N = 399.84 \times {10^{16}}

= 3.99 \times {10^{18}}

$\Rightarrow$

N = 4 \times {10^{18}}

Q104

In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be n times the initial pressure. The value of n is :

A 128

B 32

C 326

D

{1 \over {32}}

Correct Answer

Option A

Solution

In adiabatic process PV $\gamma$ = constant $\Rightarrow$

P{\left( {{m \over \rho }} \right)^\gamma }

= constant As mass is constant $\therefore$ P $\propto$

{{\rho ^\gamma }}

$\Rightarrow$

{{{P_f}} \over {{P_i}}} = {\left( {{{{\rho _f}} \over {{\rho _i}}}} \right)^\gamma }

=

{\left( {32} \right)^{{7 \over 5}}}

= 27 = 128 [ For diatomic gas $\gamma$ =

{{7 \over 5}}

]

Q105

Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom.The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of

\gamma \left( { = {{{C_p}} \over {{C_v}}}} \right)

are given, respectively by:

A U =

{5 \over 2}RT

and

\gamma = {7 \over 5}

B U = 5RT and

\gamma = {6 \over 5}

C U = 5RT and

\gamma = {7 \over 5}

D U =

{5 \over 2}RT

and

\gamma = {6 \over 5}

Correct Answer

Option A

Solution

Total degree of freedom (f) = 3 + 2 = 5 U =

{{nfRT} \over 2}

=

{{5RT} \over 2}

$\gamma$ =

{{{C_p}} \over {{C_v}}}

=

1 + {2 \over f}

=

1 + {2 \over 5}

=

{7 \over 5}

Q106

A carnot engine having an efficiency of

{1 \over {10}}

is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is :

A 90 J

B 99 J

C 1 J

D 100 J

Correct Answer

Option A

Solution

n =

1 - {{{T_C}} \over {{T_H}}}

$\Rightarrow$

{1 \over {10}}

=

1 - {{{T_C}} \over {{T_H}}}

$\Rightarrow$

{{{T_C}} \over {{T_H}}} = {9 \over {10}}

As for carnot engine T $\propto$ Q $\therefore$

{{{Q_C}} \over {{Q_H}}} = {9 \over {10}}

$\Rightarrow$

{{Q_H} = {{10} \over 9}{Q_C}}

As W = Q - QC $\Rightarrow$ 10 =

{{{10} \over 9}{Q_C}}

- QC $\Rightarrow$ QC = 90 J

Q107

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

\gamma

= 1.40, the work done by air is : (31.4 = 4.6555) [Take air to be an ideal gas]

A 60.7 J

B 100.8 J

C 90.5 J

D 48 J

Correct Answer

Option C

Solution

{P_1}V_1^\gamma = {P_2}V_2^\gamma

$\Rightarrow$ P2 = P1

{\left[ {{{{V_1}} \over {{V_2}}}} \right]^\gamma }

= 105 $\times$

{\left[ {{1 \over 3}} \right]^{1.4}}

Work done =

{{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}

=

{{{{10}^5} \times {{10}^{ - 3}} - {{{{10}^5}} \over {{3^{1.4}}}} \times 3 \times {{10}^{ - 3}}} \over {1.4 - 1}}

= 88.7 J $\approx$ 90.5 J

Q108

Two moles of an ideal gas with

{{{C_P}} \over {{C_V}}} = {5 \over 3}

are mixed with 3 moles of another ideal gas with

{{{C_P}} \over {{C_V}}} = {4 \over 3}

. The value of

{{{C_P}} \over {{C_V}}}

for the mixture is :

A 1.50

B 1.45

C 1.47

D 1.42

Correct Answer

Option D

Solution

Cp =

{{{n_1}{C_{{p_1}}} + {n_2}{C_{{p_2}}}} \over {{n_1} + {n_2}}}

Cv =

{{{n_1}{C_{{V_1}}} + {n_2}{C_{{V_2}}}} \over {{n_1} + {n_2}}}

$\gamma$ mix =

{{{C_p}} \over {{C_v}}}

=

{{2 \times {5 \over 2}R + 3 \times {8 \over 2}R} \over {2 \times {3 \over 2}R + 3 \times {6 \over 2}R}}

=

{{5 + 12} \over {3 + 9}}

= 1.42

Q109

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperature, T1 and T2 . The temperature of the hot reservoir of the first engine is T1 and the temperature of the cold reservoir of the second engine is T2 . T is temperature of the sink of first engine which is also the source for the second which is also the source for the second engine. How is T related to T1 and T2 . If both engines perform equal amount of work?

A

T = {{2{T_1}{T_2}} \over {{T_1} + {T_2}}}

B

T = \sqrt {{T_1}{T_2}}

C

T = {{{T_1} + {T_2}} \over 2}

D T = 0

Correct Answer

Option C

Solution

QH : Heat input to 1st engine QL : Heat rejected from 1st engine Q'L : Heat rejected from 2nd engine T : Lower temperature of first engine Work done by 1st engine = work done by 2nd engine QH – QL = QL – Q'L $\Rightarrow$ 2QL = QH + Q'L $\Rightarrow$ 2 =

{{{Q_H}} \over {{Q_L}}} + {{Q{'_L}} \over {{Q_L}}}

$\Rightarrow$ 2 =

{{{T_1}} \over T} + {{{T_2}} \over T}

$\Rightarrow$

T = {{{T_1} + {T_2}} \over 2}

Q110

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from

{\tau _1}

to

{\tau _2}

. If

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

for this gas then a good estimate for

{{{\tau _2}} \over {{\tau _1}}}

is given by :

A

{\left( 2 \right)^{{{1 + \gamma } \over 2}}}

B 2

C

{\left( {{1 \over 2}} \right)^{{{1 + \gamma } \over 2}}}

D

{\left( {{1 \over 2}} \right)^\gamma }

Correct Answer

Option A

Solution

$\tau$ $\propto$

{V \over {\sqrt T }}

....(1) Also we know, PV $\gamma$ = k We know, PV = nRT $\Rightarrow$ P $\propto$

{T \over V}

$\therefore$

\left( {{T \over V}} \right)

V $\gamma$ = k $\Rightarrow$ TV $\gamma$ - 1 = k $\Rightarrow$ T $\propto$ V1 - $\gamma$ Using this value in equation (1) $\tau$ $\propto$

{V \over {{V^{{{1 - \gamma } \over 2}}}}}

$\Rightarrow$ $\tau$ $\propto$

{{V^{1 - {{1 - \gamma } \over 2}}}}

$\Rightarrow$ $\tau$ $\propto$

{{V^{{{\gamma + 1} \over 2}}}}

$\therefore$

{{{\tau _2}} \over {{\tau _1}}}

=

{\left( {{{2V} \over V}} \right)^{^{{{\gamma + 1} \over 2}}}}

=

{\left( 2 \right)^{{{1 + \gamma } \over 2}}}