Kinetic Theory Test

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Kinetic Theory Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

Molecular motion shows itself as.

A
Temperature

B
Internal energy

C
Friction

D
Viscosity

Solution

Molecular mother shows itself as temperature.
And not as internal energy, because internal energy is sum of potential energy and kinetic energy.
Question 2
1 / -0

The kinetic energy of $$1$$g molecule of a gas, at normal temperature and pressure, is?

A
$$0.56\times 10^4$$J

B
$$2.7\times 10^2$$J

C
$$1.3\times 10^2$$J

D
$$3.4\times 10^3$$J

Solution

The kinetic energy of $$1$$g molecule of a gas at temperature
$$T=\dfrac{3}{2}$$ RT$$=\dfrac{3}{2}\times 8.31\times 273=3.4\times 10^3$$J
Question 3
1 / -0

If for a gas $$\dfrac{R}{C_V}=0.67$$, this gas is made up of molecules which are.

A
Monatomic

B
Diatomic

C
Polyatomic

D
Mixture of diatomic and polyatomic molecules

Solution

For a gas, we know $$\dfrac{R}{C_V}=\gamma -1$$
or $$0.67=\gamma -1$$ or, $$\gamma =1.67$$
Hence the gas is monatomic.
Question 4
1 / -0

Pressure depends on distance as $$P = \dfrac{\alpha}{\beta} exp (- \dfrac{\alpha z}{k \theta})$$, where $$\alpha, \beta$$ are constants, z is distance, k is Boltzmann's constant, and $$\theta$$ is temperature. The dimensions of $$\beta$$ are

A
$$M^0L^0T^0$$

B
$$M^{-1}L^{-1}T^{-1}$$

C
$$M^0L^2T^0$$

D
$$M^{-1}L^{2}T^{2}$$

Solution

$$\cfrac{\alpha z}{k\theta}$$ is unitless.
$$Z\to L$$
$$\alpha\to MLT^{-2}$$
$$\cfrac{\alpha}{\beta}=ML^{-1}T{-2}$$ (unit of pressure)
$$\beta \to L^2$$
Question 5
1 / -0

A vessel contains a mixture consisting of m$$_{1}$$ - 7 g of nitrogen (M$$_{1}$$ = 28) and m$$_{2}$$ = 11 g of carbon dioxide (M$$_{2}$$ = 44) at temperature T - 300 K and pressure P$$_{0}$$ = 1 atm. The density of the mixture is

A
$$1.46g\ per\ litre$$

B
$$2.567 g \ per \ litre$$

C
$$3.752 g \ per \ litre$$

D
$$4.572 g \ per \ litre$$

Solution

Let the volume occupied $$=V$$
By Dalton's law of partial pressure
$$\cfrac{P_{nit}}{P_0}=\cfrac{n_{nit}}{n_{nit}+n_{carb}}$$
No. of moles of Nitrogen $$\eta_{nit}=\cfrac{M_1}{M_{nit}}=\cfrac{7}{28}=0.25 mol$$
No. of moles of carbon $$\eta_{carb}=\cfrac{M_2}{M_{carb}}=\cfrac{11}{4}=0.25 mol$$
Thus,
$$P_{nit}=P_0\times\cfrac{0.25}{0.25\times0.25}=P_0/2=0.5atm$$
From ideal gas equation
$$V=\cfrac{nRT}{P}=\cfrac{0.25\times8.314\times290}{0.5\times101325}=0.0119mole$$
Total mixture $$m=(7+11)\times 10^{-11}kg$$
Thus density s $$P=\cfrac{m}{V}=\cfrac{(7+11)\times 10^{-3}}{0.0119}\approx1.46kg/m^3$$
Option A is correct.
Question 6
1 / -0

A vessel of volume V contains a mixture of $$1$$mole of hydrogen and $$1$$ mole of oxygen(both considered as ideal). Let $$f_1(v)dv$$ denote the fraction of molecules with speed between v and $$(v+dv)$$ with $$f_2(v)dv$$, similarly for oxygen. then

A
$$f_1(v)+f_2(v)=f(v)$$ obeys the Maxwell's distribution law

B
$$f_1(v), f_2(v)$$ will obey the Maxwell's distribution law separately

C
Neither $$f_1(v)$$ nor $$f_2(v)$$ will obey the Maxwell's distribution law

D
$$f_2(v)$$ and $$f_1(v)$$ will be the same

Solution

The Maxwell-Boltzmann speed distribution function $$\left(N_v=\dfrac{dN}{dv}\right)$$ depends on the mass of the gas molecule. [Here, dN is the number of molecules with speeds between v and $$(v+dv)$$]. The masses of hydrogen and oxygen molecules are different.
Question 7
1 / -0

A gas mixture consists of $$2$$ moles of oxygen and $$4$$ moles of Argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is?

A
$$4$$RT

B
$$9$$RT

C
$$11$$RT

D
$$15$$RT

Solution

For two moles of diatomic oxygen with no vibrational mode.
$$U_1=2\times \dfrac{5}{2}RT=5RT$$
For four moles of monoatomic Argon,
$$U_2=4\times \dfrac{3}{2}RT=6RT$$
$$\therefore u=u_1+u_1=5RT+6RT=11RT$$
Question 8
1 / -0

A vessel contains a mixture consisting of $${m}_{1}=7kg$$ of nitrogen $$\left( { M }_{ 1 }=28 \right) $$ and $${m}_{2}=11g$$ of carbon dioixide $$\left( { M }_{ 2 }=44 \right) $$ at temeprature $$T=300K$$ and pressure $${ P }_{ 0 }=1\quad atm$$. The density of the mixture is:

A
$$1.446g$$ per litres

B
$$2.567g$$ per litre

C
$$3.752g$$ per litre

D
$$4.572g$$ per litre

Solution

Let V is the volume of the vessel.
Now, let $$p_{1}$$ and $$p_{2}$$ be the partial pressure, then using gas law:
$$p_{1}V = \dfrac{m_1}{M_1}RT\\$$
$$p_{2}V = \dfrac{m_2}{M_2}RT,\ p_{0} = p_{1} + p_{2}\\$$
$$p_{0} = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{V}\\$$
$$V = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{p_{0}}\\$$
$$\because \rho_{mix}=\dfrac{(m_{1} + m_{2})}{V}\\$$
$$\rho_{mix}=\dfrac {(m_1 + m_2)M_1 M_2} {(m_1M_2 + m_2M_1)} \times \dfrac{p_0}{RT}\\$$
Substituting values,
$$\rho_{mix}=\dfrac {(7 + 11) \times 28 \times 44\times 10^{-3}} {(7 \times 44 + 11\times 28))} \times \dfrac{10^{5}}{8.3 \times 300}\\$$
$$= 1.446 \ per \ litre$$
Option A is correct.
Question 9
1 / -0

The average degree of freedom per molecule for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. The heat absorbed by the gas is

A
75 J

B
100 J

C
150 J

D
125 J

Solution

Hint: Use First law of thermodynamics

Step 1: Note all the values given.
It is given that aa gas with degree of freedom per molecule of gas $$f = 6$$ performs $$W = 25J$$ work when it expands st constant Pressure. We know that change in Intenral Energy of gas is given by,
$$∆U = \dfrac{f}{2}R∆T$$
Where $$R$$ is Gas Constant and $$∆T$$ is change in temperature.
Therefore,
$$∆U = 3R∆T$$

Also for Isobaric process, taht is process at constant temperature, Eork done is given by,
$$W = P∆V$$
$$\Rightarrow W = 25J = P∆V$$
$$\Rightarrow W = R∆T = 25J$$ (Since, P∆V = R∆T)

Step 2: Calculate heat absorbed by the gas
By First Law of Thermodynamics, we have
$$ Q = ∆U + W$$
Where $$Q$$ is heat given to gas, ∆U is change in Internal Energy and W is work done by gas. We have
$$Q = 3R∆T + R∆T$$
$$\Rightarrow Q = 4R∆T$$
$$\Rightarrow Q = 4 \times 25J$$
$$\Rightarrow Q = 100J$$

Therefore, heat absorbed by gas is, $$Q = 100J$$
Option B is correct.

Question 10

1 / -0

Match List I with List II and select the correct answer using the codes given the lists.

List I	List II
P. Boltzmann Constant	$$1$$. $$[ML^2T^{-1}]$$
Q. Coefficient of viscosity	$$2$$. $$[ML^{-1}T^{-1}]$$
R. Plank Constant	$$3$$. $$[MLT^{-3}K^{-1}]$$
S. Thermal conducivity	$$4$$. $$[ML^2T^{-2}K^{-1}]$$

P-$$3$$, Q-$$1$$, R-$$2$$, S-$$4$$

P-$$3$$, Q-$$2$$, R-$$1$$, S-$$4$$

P-$$4$$, Q-$$2$$, R-$$1$$, S-$$3$$

P-$$4$$, Q-$$1$$, R-$$2$$, S-$$3$$

Solution

$$E=\cfrac{3}{2}KT\\ Boltzman constant\rightarrow[K]=\cfrac{[ML^2T^{-2}]}{[\theta]}=[ML^2T^{-2}K^{-1}]$$

Coefficient of viscosity of $$\rightarrow \eta=\cfrac{F}{A/dv/dx}=\cfrac{MLT^{-2}}{L^2\times T^{-1}}=[ML^{-1}T^{-1}]$$

Planck's constant $$\rightarrow ML^2T^{-1}$$

Dimension formula of thermal conductivity

Thermal conductivity is measured in $$(wall/m.K)\\=MLT^{-3}K^{-1}$$

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Kinetic Theory Test - 50

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Kinetic Theory Test - 50

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SHARING IS CARING

Question 1

Temperature

Internal energy

Friction

Viscosity

Solution

Question 2

$$0.56\times 10^4$$J

$$2.7\times 10^2$$J

$$1.3\times 10^2$$J

$$3.4\times 10^3$$J

Solution

Question 3

Monatomic

Diatomic

Polyatomic

Mixture of diatomic and polyatomic molecules

Solution

Question 4

$$M^0L^0T^0$$

$$M^{-1}L^{-1}T^{-1}$$

$$M^0L^2T^0$$

$$M^{-1}L^{2}T^{2}$$

Solution

Question 5

$$1.46g\ per\ litre$$

$$2.567 g \ per \ litre$$

$$3.752 g \ per \ litre$$

$$4.572 g \ per \ litre$$

Solution

Question 6

$$f_1(v)+f_2(v)=f(v)$$ obeys the Maxwell's distribution law

$$f_1(v), f_2(v)$$ will obey the Maxwell's distribution law separately

Neither $$f_1(v)$$ nor $$f_2(v)$$ will obey the Maxwell's distribution law

$$f_2(v)$$ and $$f_1(v)$$ will be the same

Solution

Question 7

$$4$$RT

$$9$$RT

$$11$$RT

$$15$$RT

Solution

Question 8

$$1.446g$$ per litres

$$2.567g$$ per litre

$$3.752g$$ per litre

$$4.572g$$ per litre

Solution

Question 9

75 J

100 J

150 J

125 J

Solution

Question 10

P-$$3$$, Q-$$1$$, R-$$2$$, S-$$4$$

P-$$3$$, Q-$$2$$, R-$$1$$, S-$$4$$

P-$$4$$, Q-$$2$$, R-$$1$$, S-$$3$$

P-$$4$$, Q-$$1$$, R-$$2$$, S-$$3$$

Solution

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