Kinetic Theory Test

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Kinetic Theory Test - 72

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Weekly Quiz Competition

Question 1
1 / -0

The number of translational degrees of freedom for a diatomic gas is

A
$$2$$

B
$$3$$

C
$$5$$

D
$$6$$

Solution
Question 2
1 / -0

On colliding in a closed container the gas molecules

A
Transfer momentum to the walls

B
Momentum becomes zero

C
Move in opposite directions

D
Perform Brownian motion

Solution
Question 3
1 / -0

A thin walled hollow cylinder is rolling down an incline , without slipping. At any instant, without slipping. At any instant the ratio "Rotational K.E : Translational K.E : Total K.E" is

A
$$ 1 : 1 : 2 $$

B
$$ 1 : 2 : 3 $$

C
$$ 1 : 1 : 1 $$

D
$$ 2 : 1 : 3 $$

Solution
Question 4
1 / -0

According to the law of equipartition of energy, the energy associated with each degree of freedom is:

A
$$\frac{1}{3}K_{B}T$$

B
$$\frac{1}{2}K_{B}T$$

C
$$K_{B}T$$

D
$$\frac{3}{2}K_{B}T$$

Solution

According to the law of equipartition of energy, the energy associated with each degree of freedom is $$\frac{1}{2}K_{B}T$$

Hence, Option "B" is the correct answer.
Question 5
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Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of $$N_{2}$$ at temperature $$T_{0}$$, while box B contains one mole of $$H_{2}$$ at temperature 7/3 $$T_{0}$$. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature [Ignore the heat capacity of boxes]. Then the final temperature of the gases $$T_{f}$$ in terms of $$T_{0}$$ is

A
$$T_{f}=\dfrac{5}{2}T_o$$

B
$$T_{f}=\dfrac{3}{7}T_o$$

C
$$T_{f}=\dfrac{5}{3}T_o$$

D
$$T_{f}=\dfrac{3}{2}T_o$$

Solution

According to kinetic theory of gases, the KE of an ideal gas molecule at temp $$T$$ is given by $$\dfrac{1}{2}fk_BT$$ where f is the no of degrees of freedom, $$k_B$$ is the Stefan Boltzman constant and T is the temp in Kelvin.
Both Nitrogen and Hydrogen are diatomic and the no of degrees of freedom $$f=5$$.

Total KE is $$KE=\dfrac{1}{2}fn_1N_Ak_BT_1 + \dfrac{1}{2}fn_2N_Ak_BT_2=\dfrac{1}{2}fN_Ak_B(n_1T_1+n_2T_2)$$ ....(1) where $$N_A$$ is the Avogadro's number.
$$T_1=T_0$$ , $$T_2=\frac{7}{3}T_0$$
$$n_1=n_2=n$$
Let the final temp be $$T_f$$ after both gases are mixed.

$$KE= \dfrac{1}{2}(n_1+n_2)fN_Ak_BT$$ ...(2)

Equating (1) and (2) since total energy is conserved,

$$\dfrac{1}{2}(n_1+n_2)fN_Ak_BT_f = \dfrac{1}{2}fN_Ak_B(n_1T_1+n_2T_2)$$

$$ \therefore T_f=\dfrac{n_1T_1+n_2T_2}{n_1+n_2}=\dfrac{1}{2}\times (T_1+ T_2)=\dfrac{1}{2}\times \dfrac{10}{3}T_0=\dfrac{5}{3}T_0 $$
Question 6
1 / -0

A 90 cm long barometer tube contains some air above the mercury. The reading is 74.5 cm when the true pressure is 76 cm at the temperature $$15^{0} C$$. If the reading is observed to be 75.8 cm on a day when the temperature is $$5^{0}C$$, then the true pressure is:

A
77.38 cm of Hg

B
75.8 cm of Hg

C
74 cm of Hg

D
80 cm of Hg

Solution

Pressure = Actual pressure - Observed pressure
Volume = A(Barometer length - observed length)
$$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \\ \Rightarrow \dfrac { (76-74.5)(90-74.5) }{ 288 } =\dfrac { (P-75.8)(90-75.8) }{ 278 } \\ \Rightarrow P=77.38 \ \ cm \ \ of \ Hg$$
Question 7
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Internal energy of $$n_{1}$$ moles of $$H_{2}$$ at temperature $$T$$ is equal to the internal energy of $$n_{2}$$ moles of He at temperature $$2T$$. Then the ratio $$\dfrac{n_{1}}{n_{2}}$$ is

A
3/5

B
2/3

C
6/5

D
3/7

Solution

Internal energy per degree of freedom: $$E=\frac{1}{2}kT$$ where k is the Boltzman's constant and T is the temp in Kelvin.
For n_1 moles of H_2 $$E_1= n_1 \times f_1\times \frac{1}{2}kT_1$$
For n_2 moles of He $$E_2= n_2 \times f_2 \times \frac{1}{2}kT_2$$
Given $$E_1=E_2$$ and $$T_2=2T_1$$
$$f_1$$=5 since $$H_2$$ is a diatomic molecule and has 5 degrees of freedom where as He is mono atomic and has 3 degrees of freedom.
$$ \therefore n_1 \times f_1\times \dfrac{1}{2}kT_1= n_2 \times f_2 \times \dfrac{1}{2}kT_2$$
$$ \therefore \dfrac{n_1}{n_2}=\dfrac{f_2T_2}{f_1T_1}=\dfrac {3 \times 2T_1}{5T_1}=\dfrac {6}{5}$$
Question 8
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A current passes through a resistor. Let $$K_{1}$$ and $$K_{2}$$ represent the average kinetic energy of the conduction electrons and the metal ions respectively, then:

A
$$K_{1}$$ < $$K_{2}$$

B
$$K_{1} = K_{2}$$

C
$$K_{1}$$ > $$K_{2}$$

D
Any of these 3 may be correct

Solution
Question 9
1 / -0

A smooth vertical tube with two different cross-sections is open at both ends. They are fitted with pistons of different areas of cross- section and each piston moves within a particular section. One mole of a gas enclosed between the pistons which are tied with non-stretchable threads. The difference in cross-sectional area of pistons $$10\mathrm{c}\mathrm{m}^{2}$$ The mass of gas confined between pistons is 5kg. The outside pressure is 1 atmosphere$$=10^{5}N/m^{2}$$. By how many degrees must the gas between pistons be heated to shift the piston by 5 cm?
Given $$R=8.3$$.

A
$$0.9^{\mathrm{o}}\mathrm{C}$$

B
$$1.2^{\mathrm{o}}\mathrm{C}$$

C
$$1.5^{\mathrm{o}}\mathrm{C}$$

D
$$0.5^{\mathrm{o}}\mathrm{C}$$

Solution

Let us consider $$s_1$$ and $$ s_2$$ as the cross-section area of the lower and upper piston respectively.
Let $$P_o$$ be the atmospheric pressure and P be the gas pressure.At equilibrium, the downward force and upward forces balances,

i.e., $$P_o + Ps_1 + mg = Ps_2 + P_{o}s_{1} $$ ($$Force = Pressure \times Area$$}

$$ P = P_{o} + \dfrac{mg}{s_1 - s_2}$$ -- (1)

Let $$l_1$$ and $$l_2$$ be the length of string in lower and upper section.
Initial volume of the gas = $$ l_1 s_1 + l_2 s_2$$

When the pistons shifts by a distance $$x$$,

Final volume = $$(l_1 - x)s_1 + (l_2 + x)s_2$$

We have $$ PV = RT$$. Applying this before heating and after heating the gas,

$$P(l_1 s_1 + l_2 s_2) = RT$$ -- (2)

$$P[(l_1 -x)s_1 + (l_2 + x)s_2] = R (T + \Delta T)$$

$$P[(l_1 s_1 + l_2 s_2)+ (s_2 - s_1)x]= R (T + \Delta T)$$

$$[P(l_1 s_1 + l_2 s_2)+ P(s_2 - s_1)x]= R (T + \Delta T)$$

$$P(l_1 s_1 + l_2 s_2)+ P \Delta S x= R (T + \Delta T)$$

Substituting for $$P(l_1 + l_2)$$ from equation 2,

$$RT+ P \Delta S x= R (T + \Delta T)$$

Substituting for $$P$$ from equation 1,

$$RT+ P_{o} + \dfrac{mg}{s_1 - s_2} \Delta S x= R (T + \Delta T)$$

$$\implies \Delta T = \dfrac{1}{R} [P_o + mgx)$$

$$ = \dfrac {1}{8.3} [10^5 \times 5 \times 10^{-2} \times 10 \times 0^{-4} + 5 \times 9.81 \times 5 \times 10^{-2} ]$$

$$ = 0.9$$
Question 10
1 / -0

A horizontal uniform glass tube of 100cm length is sealed at both ends contains 10 cm mercury column in the middle the temperature and pressure of air on either side of mercury column are respectively 31$$^{0}$$C and 76cm of mercury if the air column at one end is kept at 0$$^{0}$$C and the other end at 273$$^{0}$$C the pressure of air which is at 0$$^{0}$$C is (in cm of Hg )

A
$$76$$

B
$$88.2$$

C
$$102.4$$

D
$$122$$

Solution

Let $$x$$ be the displacement of the mercury column on the side whose temperature is changed to $$0^{\circ}C$$. New pressure on both sides must be same so that the mercury column remains in equilibrium.
$$\dfrac{PV}{T}$$ is constant for air column.
$$\implies \dfrac{P_{0}(45)}{273+31}=\dfrac{P_1 (45-x)}{273+0}$$

and $$\dfrac{P_0 (45)}{273+31}=\dfrac{P_1(45+x)}{273+273}$$

Therefore, $$x=15\ cm$$ and thus $$P=102.4\ cm\ of\ Hg$$