Kinetic Theory Test

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Kinetic Theory Test - 74

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Weekly Quiz Competition

Question 1
1 / -0

2 moles of an ideal gas A ($$ {C}_{P} $$ $$= 4R$$) and 4 moles of an ideal gas B ($$ {C}_{V} $$ $$= \dfrac {3R}{2}$$) are taken together in a container and allowed to expand reversibly and adiabatically from 49L to 64L, starting from an initial temperature of $$ {47}^{0} $$C. The final temperature of the gas is :

A
$$4{}^{0}C$$

B
$$5{}^{0}C$$

C
$$7{}^{0}C$$

D
$$8{}^{0}C$$

Solution

$$ {\gamma}_{min} = \dfrac {{n}_{A}{C}_{{P}_{A}} + {n}_{B}{C}_{{P}_{B}}}{{n}_{A}{C}_{{V}_{A}} + {n}_{B}{C}_{{V}_{B}}} = \dfrac {2 \times 4R + 4 \times \dfrac {5R}{2}}{2 \times 3R + 4 \times \dfrac{3R}{2}}$$$$= \dfrac {18R}{12R} = 1.5 $$

Now,

$$ {T}_{1}{V}_{1}^{\gamma - 1} = {T}_{2}{V}_{2}^{\gamma - 1} $$

$$ 320\times{(49)}^{0.5} = {T}_{2}\times {(64)}^{0.5} $$

$$ {T}_{2} = \dfrac {320 \times 7}{8} =280 K = {7}^{0}C $$
Question 2
1 / -0

The ratio of translational and rotational kinetic energies at 100 K temperature is 3 : 2. Then the internal energy of one mole gas at that temperature is $$[R = 8.3 J/mol-K]$$

A
$$1175J$$

B
$$1037.5 J$$

C
$$2075 J$$

D
$$4150J$$

Solution

According to law of equipartition of energy, energies
equally distributed among its degree of freedom,
Let translational and rotational degree of freedom be $$f_1$$ and $$f_2$$.
Therefore $$\dfrac{K_T}{K_R}=\dfrac{3}{2}$$
and $$K_T+K_R=U$$
Hence the ratio of translational to rotational degrees of freedom is $$3:2$$. Since translational degrees of freedom is 3, the rotational degrees of freedom must be 2.
Internal energy $$U=1\times (f_1+f_2)\times \dfrac{1}{2}RT$$
$$=\dfrac{5}{2}RT=2075J$$
Question 3
1 / -0

A monoatomic ideal gas ($$ {C}_{V} = \dfrac {3}{2} R$$) is allowed to expand adiabatically and reversibly from initial volume of 8L at 300 K to a volume of $$ {V}_{2} $$ at 250 K. $$ {V}_{2} $$ is:
(Given $${(4.8)}^{1/2}$$ = 2.2)

A
10.5 L

B
23 L

C
8.5 L

D
50.5 L

Solution

$$ T{V}^{\gamma - 1} = constant $$
$$ \gamma = \dfrac{5}{3} $$
Therefore,$$ \gamma - 1 = \dfrac{2}{3} $$
$$ 300 \times {(8)}^{2/3} = 250 \times {({V}_{2})}^{2/3} \implies {({V}_{2})}^{2/3} = 4.8 $$
$$ \implies {V}_{2} = {(4.8)}^{3/2} \approx 4.8 \times 2.2 = 10.5 L $$
Question 4
1 / -0

On heating 128 g of oxygen gas from 0$$^{\circ}$$C to 100$$^{\circ}$$C, $$C_V$$ and $$C_P$$ on an average are 5 and 7 cal mol$$^{-1}$$ degree$$^{-1}$$, the value of $$\Delta$$U and $$\Delta$$H are respectively :

A
2800 cal, 2000 cal

B
2000 cal, 2800 cal

C
280 cal, 200 cal

D
none of the above

Solution

128 g of oxygen (molecular weight = 32) corresponds to 4 moles.

The enthalpy change is $$\Delta H = n C_P \Delta T=4 \times 7\times 100=2800$$ cal.

The change in internal energy is $$\Delta U = n C_V \Delta T=4\times 5\times 100=2000$$ cal.
Question 5
1 / -0

$$S_1$$: At extermely high temperature degree of freedom of monatomic gas become 5. due to activation of vibrational degree of freedom.
$$S_2$$: For a uniformly charged solid non-conducting sphere potential is maximum at it's centre.
$$S_3$$: Breaking stress of wire depends on its corss-sectional Area.
$$S_4$$: Only time varying magnetic field produces induced electric field which have lines of force.

A
FFFT

B
TTTF

C
FTFT

D
TFTF

Solution

$${S}_{1}$$: monatomic gases can never have degree of freedom more than 3.
false
$${S}_{2}$$:uniformly charged solid non-conducting sphere's potential will be maximum at it's centre.(shown in image)
true
$${S}_{3}$$:Breaking stress of wire doesn't depends on its corss-sectional Area but breaking load of wire depends on the area of cross section of wire.
false
$${S}_{4}$$:Only time varying magnetic field can induce electric field which have lines of force.
true
answer is C.

Question 6

1 / -0

The heat capacity of a certain amount of a particular gas at constant pressure is greater than that at constant volume by 29.1 J/K. Match the items given in Column I with the items given in Column II.

List 1	List 2
If the gas is monatomic, heat capacity at constant volume	131 J/K
If the gas monatomic, heat capacity at constant pressure	43.7 J/K
If the gas is rigid diatomic, heat capacity at constant pressure	72.7 J/K
If the gas is vibrating diatomic, heat capacity at constant pressure	102 J/K

A$$\rightarrow$$2, B$$\rightarrow$$3, C$$\rightarrow$$4, D$$\rightarrow$$1

B$$\rightarrow$$1, C$$\rightarrow$$3, D$$\rightarrow$$4, A$$\rightarrow$$1

C$$\rightarrow$$1, D$$\rightarrow$$3, A$$\rightarrow$$4, B$$\rightarrow$$1

D$$\rightarrow$$1, A$$\rightarrow$$3, B$$\rightarrow$$4, C$$\rightarrow$$1

Solution

Given:
$$C_p-C_v= 29.1$$ ....(1)
For monoatomic gas, $$\frac{C_p}{C_v}=\frac{5}{3}$$ ...(2)
Substituting for $$C_p$$ in (1), we get $$\frac{2}{3}C_v= 29.1$$
$$\therefore C_v= 43.7 \:J\:K^{-1}$$

$$C_p= 43.7 +29.1 =72.9 \:J\:K^{-1}$$
If the gas is diatomic $$C_v=\frac{5}{7}C_p$$
$$ \therefore \frac{2}{7}C_p= 29.1$$
$$\therefore CP=102 \:J\:K^{-1}$$

If the gas is vibrating di-atomic $$C_v= \frac{7}{9}C_p$$
$$ \therefore \frac{2}{9}C_p= 29.1$$
$$\therefore C_p=131 \:J\:K^{-1}$$

Question 7
1 / -0

Consider a classroom of dimensions $$(5 \times 10 \times 3)\ $$m$$^3$$ at temperature $$20^{o}$$C and pressure $$1$$ atm. There are $$50$$ people in the room, each losing energy at the average of $$150$$ watts. Assuming that the walls, ceiling, floor, and furniture are perfectly insulated and none of them is absorbing heat. How much time will be needed for raising the temperature of air in the room to the body temperature (37$$^{o}C$$)? [For air $$C_p = \dfrac{7}{2} R$$ and neglect the loss of air to the outside as the temperature rises]

A
$$422$$ sec

B
$$411.3$$ sec

C
$$421.1$$ sec

D
$$413.1$$ sec

Solution

The volume of air present in the room is $$ 5 \times 10 \times 3 = 150 \ m^3 = 150 \times 10^3$$ $$litre$$ ($$\because\ 1\ m^3=1000\ L$$)

The number of moles of air is calculated using the ideal gas equation.

$$n= \dfrac{PV}{RT} = \dfrac{1 \times 150 \times 10^3}{0.0821 \times 293} = 6.236 \times 10^3$$

For 1 mole of air, $$\left ( \dfrac{\partial H}{\partial T} \right )_P = C_p = \dfrac{\Delta H }{\Delta T}$$

For n moles of air, $$\therefore \Delta H = n \times C_p \times \Delta T$$

Substitute values in the above expression,
$$ \Delta H= 6.236 \times 10^3 \times \dfrac 72 \times 8.314 \times (310 - 293)= 3.085 \times 10^6 J$$

Thus, the heat need to increase the temperature of the room to $$37^{\circ}$$C is $$=3.085 \times 10^6 J$$

The heat released by $$50$$ people is $$=150 \times 50\ J/sec=7500 \ J/sec$$. This is the heat provided in 1 second.

$$\therefore$$ The time required to provide $$3.085 \times 10^6 J$$ heat is $$\displaystyle \dfrac{1 \times 3.085 \times 10^6}{7500} = 411.3 $$ sec.

Hence, the correct option is $$\text{B}$$
Question 8
1 / -0

At what temperature does the mean kinetic energy of hydrogen atoms increase to such an extent that they will escape out of the gravitational field of earth forever?

A
$$10075^o$$K

B
$$10000^o$$K

C
$$12075^o$$K

D
$$20000^o$$K

Solution

When the velocity of the hydrogen atoms become equal to escape velocity of earth, then they will be able to escape from earth.
Escape velocity of Earth=11200m/s
RMS velocity of Hydrogen atom=$$\sqrt{\dfrac{3 \times RT} {M}}$$
$$R=8.314JK^{-1}mol^{-1}$$
$$M=0.002Kg$$
Substituting we get
$$T=10075K$$
Question 9
1 / -0

$$1$$ mole of an ideal gas $$A\ ({ C }_{ v,m }=3R)$$ and $$2$$ moles of an ideal gas $$B\ \left( { C }_{ v,m }=\cfrac { 3 }{ 2 } R \right)$$ are taken in a container and expanded reversibly and adiabatically from $$1$$ litre to $$4$$ litres starting from initial temperature of $$320K$$. $$\Delta E$$ or $$\Delta U$$ for the process is:

A
$$-240R$$

B
$$240R$$

C
$$480R$$

D
$$-960R$$
Question 10
1 / -0

In a certain gas $$\displaystyle \frac{2}{5}$$th of the energy of molecules is associated with the rotation of molecules and the rest of it is associated with the motion of the centre of mass. The average translation energy of one such molecule, when the temperature is $$27^{\circ}C$$ is given by $$x\times 10^{-23}\ J$$,then find $$x$$?

A
621

B
623

C
6.21

D
62.1

Solution

Since two fifth energy is associated with the rotation, thus there are five degrees of freedom out of which two are rotational and three translational.

Hence, $${ E }_{ trans. }=\dfrac { 3 }{ 2 } kT=\dfrac { 3 }{ 2 } (1.38\times { 10 }^{ -23 })(300)=6.21\times { 10 }^{ -21 }J$$

Answer is $$621\times { 10 }^{ -23 }J$$

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Kinetic Theory Test - 74

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Kinetic Theory Test - 74

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Question 1

$$4{}^{0}C$$

$$5{}^{0}C$$

$$7{}^{0}C$$

$$8{}^{0}C$$

Solution

Question 2

$$1175J$$

$$1037.5 J$$

$$2075 J$$

$$4150J$$

Solution

Question 3

10.5 L

23 L

8.5 L

50.5 L

Solution

Question 4

2800 cal, 2000 cal

2000 cal, 2800 cal

280 cal, 200 cal

none of the above

Solution

Question 5

FFFT

TTTF

FTFT

TFTF

Solution

Question 6

A$$\rightarrow$$2, B$$\rightarrow$$3, C$$\rightarrow$$4, D$$\rightarrow$$1

B$$\rightarrow$$1, C$$\rightarrow$$3, D$$\rightarrow$$4, A$$\rightarrow$$1

C$$\rightarrow$$1, D$$\rightarrow$$3, A$$\rightarrow$$4, B$$\rightarrow$$1

D$$\rightarrow$$1, A$$\rightarrow$$3, B$$\rightarrow$$4, C$$\rightarrow$$1

Solution

Question 7

$$422$$ sec

$$411.3$$ sec

$$421.1$$ sec

$$413.1$$ sec

Solution

Question 8

$$10075^o$$K

$$10000^o$$K

$$12075^o$$K

$$20000^o$$K

Solution

Question 9

$$-240R$$

$$240R$$

$$480R$$

$$-960R$$

Question 10

621

623

6.21

62.1

Solution

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