Kinetic Theory Test

Result

Kinetic Theory Test - 46

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If the potential energy of a gas molecule is $$U=\dfrac{M}{r^6}-\dfrac{N}{r^{12}}, M$$ and $$N$$ being positive constants, then the potential energy at equilibrium must be

A
Zero

B
$$NM^2/4$$

C
$$MN^2/4$$

D
$$M^2/4N$$

Solution

Given in question,
$$U=\dfrac{M}{r^6}-\dfrac{N}{r^{12}}$$
$$\therefore F=\dfrac{-du}{dr}=\dfrac{-d}{dr}\begin{pmatrix}\dfrac{M}{r^6}-\dfrac{N}{r^{12}}\end{pmatrix}$$
$$=-\begin{pmatrix}\dfrac{-6M}{r^7}+\dfrac{12N}{r^{13}}\end{pmatrix}=\begin{pmatrix}\dfrac{6M}{r^7}-\dfrac{12N}{r^{13}}\end{pmatrix}$$
For equilibrium position, Force $$F=0$$
$$\therefore\dfrac{6M}{r^7}=\dfrac{12N}{r^{13}}$$ or $$r^6= \dfrac{2N}{M}$$
Hence, $$U=\dfrac{M}{(2N/M)}-\dfrac{N}{(2N/M)^2}=\dfrac{M^2}{4N}$$
So the potential energy at equilibrium must be $$U=\dfrac{M^2}{4N}$$
Question 2
1 / -0

One mole of an ideal monoatomic gas is heated at a constant pressure from $$0^oC$$ to $$100^oC$$. Then the change in the internal energy of the gas is (Given $$R = 8.32 {Jmol}^{-1}K^{-1}$$)

A
$$0.83 \times 10^3 J$$

B
$$4.6 \times 10^3 J$$

C
$$2.08 \times 10^3 J$$

D
$$1.25 \times 10^3 J$$

Solution

At constant pressure, the change in internal energy of system=$$nC_v\Delta T$$
$$=n\dfrac{3}{2}R\Delta T$$
$$=1.25\times 10^3J$$
Question 3
1 / -0

The number of degrees of freedom for a rigid diatomic molecule is ____________.

A
$$3$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

Number of degree of freedom $$d = 3N-1$$
where $$N$$ is the number of atoms in a molecules
In diatomic molecules, $$N =2$$
$$\implies$$ $$d = 3(2)-1 =5$$
Hence diatomic molecule has $$5$$ degrees of freedom ($$3$$ translational and $$2$$ rotational).
Question 4
1 / -0

In a thermodynamic process, helium gas obeys the law, $$T\ P^{-2/5} =$$ constant. The heat is given to $$n$$ moles of $$He$$ in order to raise the temperature from $$T$$ to $$2T$$ is :

A
$$8 RT$$

B
$$4 RT$$

C
$$16 RT$$

D
$$Zero$$

Solution

As the helium gas obeys $$TP^{-2/5}=$$ constant which is a characteristic of the adiabatic reversible process, the process is an adiabatic reversible process for which $$q=0$$.

Hence, option D is correct.
Question 5
1 / -0

During an experiment an ideal gas is found to obey an additional law $$V{p}^{2}=$$ constant. The gas is initially at temperature $$T$$ and volume $$V$$, when it expands to volume $$2V$$, the resulting temperature is $$T_2$$:

A
$$\cfrac { T }{ 2 } $$

B
$$2T$$

C
$$\sqrt { 2 } T $$

D
$$\cfrac { T }{ \sqrt { 2 } } $$

Solution

Ideal gas: $$VP^2=$$ constant
Again $$PV =nRT$$ from equation of state,
Hence, $$VP\times P=$$ constant i.e. $$nRT\times P=$$ constant
Again, $$P=\dfrac{nRT}{V}$$

$$\therefore$$ $$\dfrac{(nRT)^2}{V}=$$ constant

$$\dfrac{T^2}{V}=$$ constant

Thus volume $$V$$ when expanded to $$2V$$, temperature $$T_2$$

$$T_2=\sqrt{\dfrac{2v}{v}}=\sqrt{2}T_1$$
Question 6
1 / -0

If a gas mixture contains 2 moles of $$O_2$$ and 4 moles of Ar at temperature $$T$$, then what will be the total energy of the system (neglecting all vibrational modes)

A
11 RT

B
15 RT

C
8 RT

D
RT

Solution

Given, At a temperature $$T$$,
Number of moles of $$O_2 =2 moles$$
Number of moles of $$Ar=4moles$$
We Know,
Total energy of the gases,
$$U=2\begin{pmatrix}\dfrac{n_1}{2}RT\end{pmatrix}+4\begin{pmatrix}\dfrac{n_2}{2}RT\end{pmatrix}$$
For $$O_2, n_1 = 5$$ and for $$Ar, n_2=3$$
$$\therefore U=2\begin{pmatrix}\dfrac{5}{2}RT\end{pmatrix}+4\begin{pmatrix}\dfrac{3}{2}RT\end{pmatrix}=11RT$$
Question 7
1 / -0

$$4.48L$$ of an ideal gas at STP requires $$12cal$$ to raise the temperature by $${15}^{o}C$$ at constant volume. The $${C}_{P}$$ of the gas is ______ cal.

A
2

B
4

C
6

D
8

Solution

Ideal gas at STP :-$$ 1 mole \longrightarrow 22.4 l $$
$$? \longrightarrow 4.48l $$
$$x= \frac {4.48}{22.4} = 0.2$$
$$C_v = \frac { Q }{n \triangle T } = \frac{12}{0.2*15}$$ $$=4$$
$$C_p - C_v = R$$
$$R = 1.987 Cal K^-1 Mol ^-1$$
therefore; $$Cp = R + Cv = 2 + 4$$
$$=6$$
Question 8
1 / -0

The average kinetic energy of thermal neutron is of the order of :
(Boltzmann's constant $${ k }_{ B }=8\times { 10 }^{ -5 }{ eV }/{ K }$$)

A
$$0.03 eV$$

B
$$3 eV$$

C
$$3 keV$$

D
$$3 MeV$$

Solution

Thermal neutron, any free neutron (one that is not bound within an atomic nucleus) that has an average energy of motion (kinetic energy) corresponding to the average energy of the particles of the ambient materials.
Average Kinetic Energy=kT=0.03eV
Question 9
1 / -0

The speed of sound in hydrogen at NTP 1270 m/s. Then, the speed in a mixture of hydrogen and oxygen in the ratio 4 : 1 by volume will be :

A
317 m/s

B
635 m/s

C
830 m/s

D
950 m/s

Solution

The volume of hydrogen and oxygen in a mixture is 4 : 1. So let V be the volume of oxygen. The volume of hydrogen will be 4V if $$ \rho_m $$ be the density of mixture, then
$$ \rho_m = \dfrac {4V \times 1 + V \times 1 }{5V} = 4 $$
$$ V \infty \left( \dfrac {1}{\rho} \right)^{1/2} $$
Velocity in mixture $$ = \dfrac {1270}{(4)^{3/2}} = 635 m/s $$
Question 10
1 / -0

In Mayer's realation
$${ C }_{ P }-{ C }_{ V }=R$$
'$$R$$' stands for:

A
translational kinetic energy of $$1$$ mol of gas

B
rotational kinetic energy of $$1$$ mol gas

C
vibratonal kinetic energy of $$1$$ mol gas

D
work done to increase the temperature of $$1$$ mol gas by one degree

Solution

$$PV=RT...(i)$$ (For 1 mol)
$$P(V+\Delta V)=R(T+1)...(ii)$$
From eqs. i an dii
$$P\Delta V=R$$
$$w=R$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Kinetic Theory Test - 46

Result

Kinetic Theory Test - 46

Score

-

Rank

-

SHARING IS CARING

Question 1

Zero

$$NM^2/4$$

$$MN^2/4$$

$$M^2/4N$$

Solution

Question 2

$$0.83 \times 10^3 J$$

$$4.6 \times 10^3 J$$

$$2.08 \times 10^3 J$$

$$1.25 \times 10^3 J$$

Solution

Question 3

$$3$$

$$5$$

$$6$$

$$7$$

Solution

Question 4

$$8 RT$$

$$4 RT$$

$$16 RT$$

$$Zero$$

Solution

Question 5

$$\cfrac { T }{ 2 } $$

$$2T$$

$$\sqrt { 2 } T $$

$$\cfrac { T }{ \sqrt { 2 } } $$

Solution

Question 6

11 RT

15 RT

8 RT

RT

Solution

Question 7

2

4

6

8

Solution

Question 8

$$0.03 eV$$

$$3 eV$$

$$3 keV$$

$$3 MeV$$

Solution

Question 9

317 m/s

635 m/s

830 m/s

950 m/s

Solution

Question 10

translational kinetic energy of $$1$$ mol of gas

rotational kinetic energy of $$1$$ mol gas

vibratonal kinetic energy of $$1$$ mol gas

work done to increase the temperature of $$1$$ mol gas by one degree

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.