Kinetic Theory Test

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Kinetic Theory Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of adiabatic to isothermal elasticity of diatomic gas is:-

A
$$1.67$$

B
$$1.4$$

C
$$1.33$$

D
$$1.27$$

Solution

The ratio of adiabatic to isothermal elasticity of diatomic gas is,

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

$$=\dfrac{C_v+R}{C_v}$$

$$=\dfrac{\frac{7R}{2}}{\frac{5R}{2}}$$

$$=\dfrac{7}{5}=1.4$$
Question 2
1 / -0

A vessel contains 28 gm of $$N-2$$ and 32 gm of $$O_2$$ at temperature T = 1800 K and pressure 2 atm pressure it $$N_2$$ dissociates $$30%$$ and $$O_2$$ dissociates $$50%$$ if temperature remains constant.

A
2 atm

B
1 atm

C
2.8 atm

D
1.4 atm
Question 3
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An ideal gas has an initial volume $$V$$ and pressure $$P$$. In doubling its volume the minimum work done will be in (of the given processes):

A
Isobaric process

B
Isothermal process

C
Adiabatic process

D
Same in all given processes

Solution

The isothermal case delivers more work because it has heat supplied to it, some fraction of which is converted to work. But for a given increment in volume, the adiabatic case begins to deliver more work. In the adiabatic case, as the volume fraction approaches infinity, the specific volume is made smaller by the fact that the temperature must decrease, but this is not the case for the isothermal expansion.
Answer B:
Question 4
1 / -0

The velocities of sound in an ideal gas at temperature $$T_1$$ and $$T_2 k$$ are found to be $$V_1$$ and $$V_2$$ respectively.If the r.m.s. velocities of the molecules of the same gas at the same temperatures $$T_1$$ and $$T_2$$ are $$v_1$$ and $$v_2$$ respectively then

A
$$v_2=v_1(\dfrac{V_1}{V_2})$$

B
$$v_2=v_1(\dfrac{V_2}{V_1})$$

C
$$v_2=v_1\sqrt{\dfrac{V_2}{V_1}}$$

D
$$v_2=v_1\sqrt{\dfrac{V_1}{V_2}}$$

Solution
Question 5
1 / -0

A gas undergoes a process such that $$P \propto \dfrac{1}{T}$$. If the molar heat capacity for this process is $$24.93 \,J/mol \,K$$, then what is the degree of freedom of the molecules of the gas?

A
$$8$$

B
$$4$$

C
$$2$$

D
$$6$$

Solution
Question 6
1 / -0

A vessel contains a non-linear triatomic gas. If $$50$$% of gas dissociate into individual atom, then find new value of degree of freedom by ignoring the vibrational mode and any further dissociation

A
$$2.15$$

B
$$3.75$$

C
$$5.25$$

D
$$6.35$$

Solution

Let's assume we have $$1$$ mole of triatomic gas
$$\therefore 3Na$$ is present
So, $$0.5$$ moles= $$1.5 Na$$ atoms
$$1$$ part of $$0.5$$ moles remains untouched
Degree of dissociation= $$0.5 \times 6=3$$
Degree of freedom for $$0.5Na= 1.5 \times 0.5=0.75$$
Total=$$3+0.75=3.75$$
Question 7
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At which of the following temperature would the molecules of a gas have twice the average kinetic energy they have at $${20}^{o}C$$

A
$${40}^{o}C$$

B
$${80}^{o}C$$

C
$${313}^{o}C$$

D
$${586}^{o}C$$

Solution

$$\textbf{Step 1 - Average KE of gas molecule initial and final}$$
Given, $$E_{2} = 2E_{1}$$ and $$T_{1} = 20^{\circ}C$$

$$E_{1} = \dfrac {3}{2} KT_{1}$$ $$....(1)$$

Similarly,

$$E_{2} = \dfrac {3}{2} KT_{2}$$ $$....(2)$$

$$\textbf{Step 2 - Calculation of final temperature }$$
Dividing equation (1) with (2)
$$\dfrac {E_{1}}{E_{2}} = \dfrac {\dfrac {3}{2} KT_{1}}{\dfrac {3}{2} KT_{2}} = \dfrac {T_{1}}{T_{2}}$$

$$\dfrac {E_{1}}{2E_{1}} = \dfrac {(273 + 20)K}{T_{2}}$$

$$\Rightarrow T_{2} = 586\ K$$

In $$^{\circ}C$$
$$T_{2} = 586 - 273$$
$$= 313^{\circ}C$$

Correct option : C
Question 8
1 / -0

The lowest pressure(the best Vaccum) that can be created in laboratory at 27 degree is $$10^{-11} $$ mm of Hg. At this pressure, the number of ideal gass molecules per $$cm^{3}$$ will be

A
$$3.22 \times 10 ^{12} $$

B
$$1.61 \times 10 ^{12} $$

C
$$3.21 \times 10 ^{6} $$

D
$$3.22 \times 10 ^{5} $$

Solution
Question 9
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Which graph shows how the average kinetic energy of the particles varies with absolute temperature for an ideal gas?

A

B

C

D
Question 10
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Select the incorrect relation. (Where symbols have their usual meanings)

A
$${ C }_{ p }=\dfrac { \gamma R }{ \gamma -1 }$$

B
$${ C }_{ p }-{ C }_{ v }=R$$

C
$$\Delta U=\dfrac { { P }_{ f }{ V }_{ f }-{ P }_{ i }{ V }_{ i } }{ 1-\gamma }$$

D
$${ C }_{ v }=\dfrac { R }{ \gamma -1 }$$

Solution

Solution:- (C) $$\Delta{U} = \cfrac{{P}_{f}{V}_{f} - {P}_{i}{V}_{i}}{1 - \gamma}$$
As we know that,
$$\Delta{U} = n {C}_{v} \Delta{T}$$
Whereas,
$$n =$$ No. of moles
$${C}_{v} =$$ Heat capacity at constant volume $$= \cfrac{R}{\gamma - 1}$$
$$\Delta{T} =$$ Change in temperature $$= {T}_{f} - {T}_{i}$$
$$\therefore \Delta{U} = n \cfrac{R}{\gamma - 1} \left( {T}_{f} - {T}_{i} \right)$$
$$\Rightarrow \Delta{U} = \cfrac{nR{T}_{f} - nR{T}_{i}}{\gamma - 1} ..... \left( 1 \right)$$
Now from ideal gas equation,
$$PV = nRT$$
For the same no. of moles of a gas,
$$nR{T}_{f} = {P}_{f}{V}_{f}$$
$$nR{T}_{i} = {P}_{i}{V}_{i}$$
Substituting these values in $${eq}^{n} \left( 1 \right)$$, we have
$$\Delta{U} = \cfrac{{P}_{f}{V}_{f} - {P}_{i}{V}_{i}}{\gamma - 1}$$
Hence the given option $$\left( C \right) \Delta{U} = \cfrac{{P}_{f}{V}_{f} - {P}_{i}{V}_{i}}{1 - \gamma}$$ is incorrect.