Heat and Thermodynamics — Page 2 | JEE Physics

Q11

Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20oC to 90oC. Work done by gas is close to – (Gas constant R = 8.31 J/mol.K)

A 581 J

B 73 J

C 146 J

D 291 J

Correct Answer

Option D

Solution

WD = P

\Delta

V = nR

\Delta

T =

{1 \over 2} \times 8.31 \times 70

Q12

A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and the of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is :

A K1 + K2

B

{{{K_1} + 3{K_2}} \over 4}

C

{{{K_1} + {K_2}} \over 2}

D

{{2{K_1} + 3{K_2}} \over 5}

Correct Answer

Option B

Solution

Keq =

{{{K_1}{A_1} + {K_2}{A_2}} \over {{A_1} + {A_2}}}

=

{{{K_1}\left( {\pi {R^2}} \right) + {K_2}\left( {3\pi {R^2}} \right)} \over {4\pi {R^2}}}

=

{{{K_1} + 3{K_2}} \over 4}

Q13

A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030 cal/(g – oC) (1 cal = 4.2 × 107 ergs) close to :

A 87.5 oC

B 83.3 oC

C 38.4 oC

D 119.2 oC

Correct Answer

Option A

Solution

{1 \over 2}m{v^2} \times {1 \over 2} = ms\Delta T

$\Rightarrow$

\Delta T = {{{v^2}} \over {4 \times 5}} = {{{{210}^2}} \over {4 \times 30 \times 4.200}}

= 87.5^\circ C

Q14

To raise the temperature of a certain mass of gas by 50oC at a constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by 100oC at constant volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?

A 6

B 7

C 5

D 3

Correct Answer

Option A

Solution

160 = n{C_p}50

....(i)

240 = n{C_v}100

....(ii) Dividing (i) by (ii), we get

{{160} \over {240}} = {{{C_p}} \over {{C_v}}} \times {1 \over 2}

$\Rightarrow$

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

=

{4 \over 3}

We know,

\gamma = {{{C_p}} \over {{C_v}}} = 1 + {2 \over f} = {4 \over 3}

\Rightarrow f = 6

Q15

In a dilute gas at pressure P and temperature T, the mean time between successive collisions of a molecule varies with T as :

A

\sqrt T

B T

C

{1 \over T}

D

{1 \over {\sqrt T }}

Correct Answer

Option D

Solution

Time (t) =

{V \over {4\pi \sqrt 2 {r^2}vN}}

....(1) Here, v = most probable speed =

\sqrt {{{2RT} \over {\pi M}}}

$\Rightarrow$ v $\propto$

\sqrt T

$\therefore$ From (1), t $\propto$

{1 \over {\sqrt T }}

Q16

A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) of the mixture is :

A 11

B 20

C 15

D 13

Correct Answer

Option C

Solution

U

= {{{f_1}} \over 2}{n_1}RT + {{{f_2}} \over 2}{n_2}RT

= {5 \over 2}\left( {3RT} \right) + {3 \over 2} \times 5RT

$\therefore$ U

= 15RT

Q17

For a gas CP

-

CV = R in a state P and CP

-

CV = 1.10 R in a state Q, TP and TQ are the temperatures in two different states P and Q respectively. Then

A TP = TQ

B TP < TQ

C TP = 0.9 TQ

D TP > TQ

Correct Answer

Option D

Solution

CP $-$ CV = R for ideal gas and gas behaves as ideal gas at high temperature, so TP > TQ

Q18

The parameter that remains the same for molecules of all gases at a given temperature is :

A kinetic energy

B mass

C momentum

D speed

Correct Answer

Option A

Solution

\mathrm{KE}=\frac{\mathrm{f}}{2} \mathrm{kT}

Q19

The specific heat at constant pressure of a real gas obeying

P V^2=R T

equation is:

A R

B

C_V+R

C

C_V+\dfrac{R}{2 V}

D

\dfrac{R}{3}+C_V

Correct Answer

Option C

Solution

\begin{aligned} & \because \quad P V^2=R T \\ & P(2 v d v)+V^2(d P)=R d T \end{aligned}

at

P=

const.

P d v=\frac{R d T}{2 V} \quad \text{... (i)}

Now, for

n=1

\begin{aligned} & d \theta=d v+d w \\ & C_P d T=C_v d T+P d v \quad \text{... (ii)} \end{aligned}

from (i) and (ii)

C_P=C_V+\frac{R}{2 V}

Q20

The kinetic energy of translation of the molecules in 50 g of

\text{CO}_2

gas at 17°C is :

A 4205.5 J

B 3582.7 J

C 3986.3 J

D 4102.8 J

Correct Answer

Option D

Solution

\begin{aligned} & (\mathrm{KE})_{\text{Transsational }}=\left[\frac{3}{2} \mathrm{KT}\right] \times \text{ no. of molecule } \\ & \text{ No. of molecule }=\left[\frac{50}{44} \times 6.023 \times 10^{23}\right] \\ & (\mathrm{KE})_{\text{Transsational }}=4108.644 \mathrm{~J} \end{aligned}