Kinetic Theory Test

Result

Kinetic Theory Test - 51

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

An insulated box containing monatomic ideal gas of molar mass M is moving with a uniform speed v. The box suddenly stops and consequently the gas acquires a new temperature. Calculate the change in the temperature of the gas. Neglect heat absorbed by the box.

A
$$\Delta T=2\dfrac{Mv^2}{3R}$$.

B
$$\Delta T=\dfrac{Mv^2}{3R}$$.

C
$$\Delta T=3\dfrac{Mv^2}{3R}$$.

D
$$\Delta T=4\dfrac{Mv^2}{3R}$$.

Solution

An insulated Box containing monoatomic ......... neglect heat absorbed by box
molar mass$$=M$$
speed $$=v$$
Initial energy $$=KE=\dfrac{1}{2}MV^{2}$$
Final energy = Temperature energy (or $$PE$$)
$$=\dfrac{3}{2}RT$$
For ideal conditions, $$\dfrac{1}{2}, mv^{2}=\dfrac{3}{2}RT$$
$$\Delta T=\dfrac{mv^{2}}{3R}$$
Question 2
1 / -0

A gas molecule of mass $$M$$ at the surface of the earth has kinetic energy equivalent to $${0}^{o}C$$. If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than radius of the earth. ($${k}_{B}$$ is Boltzmann constant)

A
zero

B
$$\cfrac { 273{ k }_{ B } }{ 2Mg } $$

C
$$\cfrac { 546{ k }_{ B } }{ 3Mg } $$

D
$$\cfrac { 819{ k }_{ B } }{ 2Mg } \quad $$

Solution
Question 3
1 / -0

An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes $$5.66$$V while its temperature falls to $$T/2$$. How many degrees of freedom do the gas molecules have?

A
7

B
$$5$$.

C
6

D
8

Solution

Adiabatic equation of perfect gas is given as $$TV^{r-1}=$$ constant
$$m=T_{1}V_{1}^{(r-1)}=T_{2}V_{2}^{(r-1)}$$
$$T_{1}=T_{1}V_{1}=V_{1}V_{2}=5.66\ V$$
and $$T_{2}=\dfrac{T}{2}$$
$$TV^{r-1}=\dfrac{T}{2}(5.66\ V)^{r-1}$$
$$2=5.66^{r-1}$$
Taking $$\log$$ on both sides
$$(r-1)\log 5.66=\log 2(r-1)$$
$$r=\dfrac{\log 2}{\log 5.66}=1+0.3010/0.75$$
$$r=1+0.4$$
$$r=1.4$$ for $$r=1.4$$ Agree of freedom $$=5$$
Question 4
1 / -0

A gas performs Q work when it expand at constant pressure. During this process heat absorbed by the gas is 4Q. The average number of degrees of freedom for the gas is:

A
5

B
6

C
4

D
3.5

Solution

$$Given:$$ A gas performs $$Q $$ work when it expand at constant pressure. During this process heat absorbed by the gas is $$4Q.$$
$$Solution:$$ In a isobaric process,
$$U=4 Q-Q=n C_{v} d T=3 Q$$
Work = $$n R dT$$ So, U/Work $$=\mathrm{C}_{\mathrm{v}} / \mathrm{R}$$
$$3=\mathrm{C}_{\mathrm{v}} / \mathrm{R}$$
$$R=C_{p}-C_{v}$$
$$3=\dfrac{C_{v}}{C_{p}-C_{v}}$$
$$\dfrac{C_{p}}{C v}=4 / 3=\gamma$$
$$\gamma=1+\frac{2}{f}$$
where f is degree of freedom $$4 / 3=1+\frac{2}{f}$$
$$f=6$$
$$So, the$$ $$correct$$ $$option: B$$
Question 5
1 / -0

Two monatomic ideal gases $$1$$ and $$2$$ of molecular masses $$m_1$$ and $$m_2$$ respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas $$1$$ to that in gas $$2$$ is given by.

A
$$\sqrt{\dfrac{m_1}{m_2}}$$

B
$$\sqrt{\dfrac{m_2}{m_1}}$$

C
$$\dfrac{m_1}{m_2}$$

D
$$\dfrac{m_2}{m_1}$$

Solution

$$V_{rm}=\sqrt {\dfrac {rRT}{m}}$$
i.e $$vr\dfrac {1}{\sqrt m}$$
which given $$v-1\times \dfrac {1}{\sqrt {m_1}}$$ and $$v_2\times \dfrac {1}{\sqrt {m_1}}$$
$$\dfrac {v_1}{v_2}=\sqrt {\dfrac {m_2}{m_1}}$$
Question 6
1 / -0

An ideal gas is initially at $$P_1$$, $$V_1$$ is expanded to $$P_2, V_2$$ and then compressed adiabatically to the same volume $$V_1$$ and pressure $$P_3$$. If W is the net work done by the gas in the complete process which of the following is true.

A
$$W > 0; P_3 > P_1$$

B
$$W < 0; P_3 > P_1$$

C
$$W > 0; P_3 < P_1$$

D
$$W < 0; P_3 < P_1$$

Solution

Since work is done by the gas as area under the graph is $$-ve$$ ( anticlockwise) ie energy is lost by the gas and $$w<0$$ when $$P_3>P_1$$.
Question 7
1 / -0

One kg of a diatomic gas is at a pressure of $$8\times {10}^{4}N/{m}^{2}$$. The density of gas is $$4kg/{m}^{2}$$. What is the energy of the gas due to its thermal motion?

A
$$6\times {10}^{4}J$$

B
$$7\times {10}^{4}J$$

C
$$3\times {10}^{4}J$$

D
$$5\times {10}^{4}J$$

Solution
Question 8
1 / -0

A gas thermometer measures the temperature from the variation of pressure of a sample of gas. If the pressure measured at the melting point of lead is $$2.20\ times$$ the pressure measured at the triple point of water find the melting point of lead.

A
$$600K$$

B
$$420K$$

C
$$790K$$

D
$$510K$$

Solution

$$\textbf{Correct option: Option (A).}$$

$$\textbf{Explanation:}$$

Given, the pressure at melting point is 2.20 times the pressure at triple point of water.

$${{P}_{MP}}=2.20\times {{P}_{TP}}$$

We know that, Melting point, $$MP=\dfrac{{{P}_{MP}}}{{{P}_{TP}}}\times 273.16K$$

$$\Rightarrow MP=\dfrac{2.20\times {{P}_{TP}}}{{{P}_{TP}}}\times 273.16\approx 600K$$
$$\textbf{So, correct option is (A).}$$
Question 9
1 / -0

Ratio of $$C_P$$ and $$C_V$$ of gas '$$X$$' is 1.4. The number of atoms of the gas '$$X$$' present in $$11.2$$ litres of at NTP will be:

A
$$6.02\times 10^{23}$$

B
$$5.61\times 10^{23}$$

C
$$3.01\times 10^{23}$$

D
$$2.01\times 10^{23}$$

Solution

$$\dfrac{C_p}{C_v} = \gamma = 1.4$$

Hence, the gas is diatomic.
Number of atoms $$=$$ atoms in one molecule of gas $$\times$$ mole$$\times N_A$$
$$= 2 \times \dfrac{V}{22.4} \times N_A$$
$$= 2 \times \dfrac{11.2}{22.4}\times 6.02\times 10^{23} = 6.02\times 10^{23}$$
Question 10
1 / -0

The average thermal energy of a oxygen atom at room temperature $$(27^{o}C)$$

A
$$4.5\times {10}^{-21}{J}$$

B
$$6.2 \times {10}^{-21}{J}$$

C
$$3.4 \times {10}^{-21}{J}$$

D
$$1.8 \times {10}^{-21}{J}$$

Solution

Boltzman's constant is given as $$k=1.38 \times 10^{-23} J/k$$
Average kinetic energy will be $$KE=\dfrac{FkT}{2}$$
Where $$F$$ is degree of freedom which will be $$3$$ because $$O$$ is a mono-atomic molecule.
Temperature, $$T=27+273=300K$$
So energy=$$\dfrac{3 \times 1.38 \times 10^{-23} \times 300}{2}=6.2\times 10^{-21} Joule$$
Option B is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Kinetic Theory Test - 51

Result

Kinetic Theory Test - 51

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\Delta T=2\dfrac{Mv^2}{3R}$$.

$$\Delta T=\dfrac{Mv^2}{3R}$$.

$$\Delta T=3\dfrac{Mv^2}{3R}$$.

$$\Delta T=4\dfrac{Mv^2}{3R}$$.

Solution

Question 2

zero

$$\cfrac { 273{ k }_{ B } }{ 2Mg } $$

$$\cfrac { 546{ k }_{ B } }{ 3Mg } $$

$$\cfrac { 819{ k }_{ B } }{ 2Mg } \quad $$

Solution

Question 3

7

$$5$$.

6

8

Solution

Question 4

5

6

4

3.5

Solution

Question 5

$$\sqrt{\dfrac{m_1}{m_2}}$$

$$\sqrt{\dfrac{m_2}{m_1}}$$

$$\dfrac{m_1}{m_2}$$

$$\dfrac{m_2}{m_1}$$

Solution

Question 6

$$W > 0; P_3 > P_1$$

$$W < 0; P_3 > P_1$$

$$W > 0; P_3 < P_1$$

$$W < 0; P_3 < P_1$$

Solution

Question 7

$$6\times {10}^{4}J$$

$$7\times {10}^{4}J$$

$$3\times {10}^{4}J$$

$$5\times {10}^{4}J$$

Solution

Question 8

$$600K$$

$$420K$$

$$790K$$

$$510K$$

Solution

Question 9

$$6.02\times 10^{23}$$

$$5.61\times 10^{23}$$

$$3.01\times 10^{23}$$

$$2.01\times 10^{23}$$

Solution

Question 10

$$4.5\times {10}^{-21}{J}$$

$$6.2 \times {10}^{-21}{J}$$

$$3.4 \times {10}^{-21}{J}$$

$$1.8 \times {10}^{-21}{J}$$

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.