Kinetic Theory Test

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Kinetic Theory Test - 73

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Weekly Quiz Competition

Question 1
1 / -0

A closed container of volume 0.02 m$$^3$$ contains a mixture of neon and argon gases at a temperature of 27$$^{0}$$ C and at a pressure of $$1\times 10^{5}N/m^{2}$$. The total mass of the mixture is 28 g. If the gram molecular weights of neon and argon are 20 and 40 respectively, the masses of the individual gases in the container are respectively(assuming them to be ideal) [R = 8.314 J/mol K]

A
16 gm, 12 gm

B
4 gm, 24 gm

C
6 gm, 22 gm

D
12 gm, 16 gm

Solution

$$n=\frac { PV }{ RT } \\ n=\frac { 1\times { 10 }^{ 5 }\times 0.02 }{ 8.314\times 300 } \\ n=0.802\\ { n }_{ 1 }{ +n }_{ 2 }=0.802\\ { 20n }_{ 1 }{ +20n }_{ 2 }=16\quad \left( Multiplying\quad by\quad 20 \right) \\ { w }_{ 1 }{ +w }_{ 2 }=28\\ { n }_{ 1 }{ M }_{ neon }{ +n }_{ 2 }{ M }_{ argon }=28\\ 20{ n }_{ 1 }{ +40n }_{ 2 }=28\\ Solving\quad to\quad two\quad equations\quad we\quad get,\\ 20{ n }_{ 2 }=12\\ { n }_{ 2 }=0.6\\ \therefore { w }_{ 2 }=0.6\times 40=24gms\\ { w }_{ 1 }=28-24=4gms$$
Question 2
1 / -0

An ideal monoatomic gas $$(C_{v}=1.5\ R)$$ initially at $$298$$ K and $$1.013$$ atm expands adiabatically irreversibly until it is in equilibrium with a constant external pressure of $$0.1013$$ atm. The final temperature (in Kelvin) of the gas is :

A
$$190.7$$

B
$$119.7$$

C
$$278$$

D
$$273$$

Solution

We know the relation, $$C_{v}=\dfrac{R}{\gamma-1}$$, where $$\gamma =\dfrac{C_p}{C_v}$$ and $$C_p - C_v = R$$

$$PV^{\gamma}=$$ constant for adiabatic process (reversible)

$$W=C_{v}(T_{2}-T_{1})=-RP_{ext}\left ( \dfrac{T_{2}}{P_{2}}-\dfrac{T_{1}}{P_{1}} \right )$$

Substituting the values, we get-

$$1.5\times R\times (-298+T_{2})=-R \times (0.1013)\left ( \dfrac{T_{2}}{0.1013}-\dfrac{298}{1.013} \right )$$

$$\implies$$ $$1.5\times R\times (T_{2}-298)=-R \times(T_{2}-29.8)$$

$$\implies$$$$1.5\times T_{2}-1.5\times 298=-T_{2}+29.8$$

$$\implies$$ $$2.5\times T_{2}=1.6\times 298$$

$$\implies$$ $$T_{2}=\dfrac{1.6\times 298}{2.5}$$ $$=190.72\ K$$

Option A is correct.
Question 3
1 / -0

A closed hollow insulated cylinder is filled with gas at 0$$^{0}$$C and also contains an insulated piston of negligible weight and negligible thickness at the the middle point. The gas at one side of the piston is heated to 100$$^{0}$$C . If the piston moves 5cm, the length of the hollow cylinder is

A
13.65 cm

B
27.3 cm

C
64.6 cm

D
54.6 cm

Solution

$$\dfrac{PV}{T}$$ is constant for gas in both sections of the cylinder.

For section with constant temperature,
$$P_{0}V_{0}=P_1V_1$$
$$\implies P_ (A\dfrac{L}{2})=P_1(A(\dfrac{L}{2}-5))$$

Similarly for section whose temperature increased,
$$\dfrac{P_0V_0}{T_0}=\dfrac{P_1V_1}{T_1}$$
$$\implies \dfrac{P_0(A\dfrac{L}{2})}{273}=\dfrac{P_1(A(\dfrac{L}{2}+5))}{373}$$
$$\implies L=64.6cm$$
Question 4
1 / -0

Two identical containers each of volume V$$_{0}$$ are joined by a small pipe. The containers contain identical gases at temperature T$$_{0}$$ and pressure P$$_{0}$$. One container is heated to temperature 2T$$_{0}$$ while maintaining the other at the same temperature. The common pressure of the gas is P and n is the number of moles of gas in container at temperature 2T$$_{0}$$.

A
$$P=2P_{0}$$

B
$$P=\dfrac{4}{3}P_{0}$$

C
$$n=\dfrac{2P_{0}V_{0}}{3RT_{0}}$$

D
$$n=\dfrac{3P_{0}V_{0}}{2RT_{0}}$$

Solution

Since the total number of moles in the system remains the same before and after heating,

$$\dfrac{P_0V_0}{T_0}+\dfrac{P_0V_0}{T_0}=\dfrac{PV_0}{T_0}+\dfrac{PV_0}{2T_0}$$

$$\implies P=\dfrac{4}{3}P_0$$
Question 5
1 / -0

How many degrees of freedom have the gas molecules if at S.T.P the gas density is 1.3 kg/m$$^{3}$$ and the velocity of sound in the gas is 330 m/sec?

A
$$5$$

B
$$6$$

C
$$4$$

D
$$3$$

Solution

Velocity of sound in the gas is given by
$$v=\sqrt \frac {\gamma P}{\rho}$$ where
P is pressure, $$\rho$$ is the density of the gas and $$\gamma=\frac {C_p}{C_v}$$ is the ratio of specific heats at constant pressure to constant volume.
$$ \therefore \gamma = \dfrac {330^2 \times 1.3}{1.013 \times 10^5} =1.4$$

But, $$\gamma = \dfrac{2+f}{f}$$ where f is the no of degrees of freedom.
Solving $$ 1.4= \dfrac{2}{f} +1$$
$$ \therefore \dfrac{2}{f} = 0.4$$
$$ \therefore f=5$$
Question 6
1 / -0

A motor draws some gas from an adiabatic container of volume 2 liters having a monatomic gas at a pressure 4 atm. After drawing the gas the pressure in the container reduces to 1 atm. If the motor converts 10% of the energy contained in the drawn gas and the output of the motor is 10p joule, then p is
( Use $$1atm=10^5 pascal)$$

A
9

B
4

C
8

D
6

Solution

We have internal energy of a gas in a container,
$$U = \dfrac{f}{2} n R T $$
$$ = \dfrac {f}{2} PV$$ (as $$ PV = nRT$$)
$$ = \dfrac {3}{2} PV$$ (as $$f = 3$$ for a monoatomic gas)
The initial internal energy is given by
$$U_i = \dfrac {3}{2} \times 4 \times 2 = 12 atm-litre$$
The final internal energy is given by
$$U_f = \dfrac {3}{2} \times 1 \times 2 = 3 atm-litre$$
Total energy of the gas drawn $$ = 9 atm-litre = 911.7 J$$ ($$1 atm-litre = 101.3J$$)
Output of the motor = 10 % of 911.7 = 91.2 J ~ 90 J =10 p
Hence $$p = 9$$
Question 7
1 / -0

Which of the following will have maximum total kinetic energy at temperature 300 K ?

A
$$1 kg, H_{2}$$

B
$$1 kg , He$$

C
$$\frac{1}{2}kg H_{2}+\frac{1}{2}kg He$$

D
$$\frac{1}{4}kg H_{2}+\frac{3}{4}kg He$$

Solution

We have total Kinetic energy,
$$U = \frac {f}{2} nRT$$
The total kinetic energy in the case of 1 kg of $$H_2$$ is the greatest as the degree of freedom is 5 and number of moles is the highest as it contains 1 kg mass.
Question 8
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
$$5 RT$$

C
$$15 RT$$

D
$$11 RT$$

Solution

The internal energy of $$n$$ moles of a gas is $$u = \dfrac {1}{2} nFRT$$
where $$F =$$ number of degrees of freedom.
The internal energy of $$2\ moles$$ of oxygen at temperature $$T$$ is
$$u_{1} = \dfrac {1}{2}\times 2\times 5RT = 5RT (F = 5$$ for oxygen molecule)
Total internal energy of $$4$$ moles of argon at temperature $$T$$ is
$$= u_{2} = \dfrac {1}{2}\times 4\times 3RT = 6RT$$
Total internal energy $$= u_{1} + u_{2} = 11\ RT$$.
Question 9
1 / -0

Maxwell"s velocity distribution curve is given for the same quantity for two different temperatures. For the given curves then which of the folowing relation is true?

A
$$T_{1}> T_{2}$$

B
$$T_{1}< T_{2}$$

C
$$T_{1}\leq T_{2}$$

D
$$T_{1}= T_{2}$$

Solution

Most probable velocity $$v_{mp} = \sqrt{\dfrac{2kT}{m}}$$
As T increases $$v_{mp}$$ increases.
$$(v_{mp})_{T_2} \gt (v_{mp})_{T_1} $$
$$\Rightarrow T_2 \gt T_1$$
$$\Rightarrow T_1 \lt T_2$$
Question 10
1 / -0

One mole of an ideal gas $$ [{C}_{v,m} = \frac {5}{2}R ] $$ at $$300$$ K and $$5$$ atm is expanded adiabatically to a final pressure of $$2$$ atm against a constant pressure of $$2$$ atm. The final temperature (in Kelvin) of the gas is:

A
$$270.2$$

B
$$273.0$$

C
$$248.5$$

D
$$200.5$$

Solution

Since the process is an adiabatic processs, $$q = 0$$ and hence, $$ \Delta U = w $$

$$ n{C}_{v,m}({T}_{2} - {T}_{1}) = - {P}_{ext} \left [\dfrac{nR{T}_{2}}{{P}_{2}} - \dfrac{nR{T}_{1}}{{P}_{1}}\right ]$$

For one mole, we have $$ {C}_{v,m}({T}_{2} - {T}_{1}) = {P}_{ext} R \left [\dfrac {{T}_{1}}{{P}_{1}} - \dfrac {{T}_{2}}{{P}_{2}} \right ] $$

$$\dfrac{5}{2} R ({T}_{2} - 300) = 2 \times R \left [\dfrac {300}{5} - \dfrac {{T}_{2}}{2} \right ] $$ $$ \implies {T}_{2} \approx 248.5\ K$$

Hence, the final temperature of the gas is $$248.5\ K$$.

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Kinetic Theory Test - 73

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Kinetic Theory Test - 73

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Question 1

16 gm, 12 gm

4 gm, 24 gm

6 gm, 22 gm

12 gm, 16 gm

Solution

Question 2

$$190.7$$

$$119.7$$

$$278$$

$$273$$

Solution

Question 3

13.65 cm

27.3 cm

64.6 cm

54.6 cm

Solution

Question 4

$$P=2P_{0}$$

$$P=\dfrac{4}{3}P_{0}$$

$$n=\dfrac{2P_{0}V_{0}}{3RT_{0}}$$

$$n=\dfrac{3P_{0}V_{0}}{2RT_{0}}$$

Solution

Question 5

$$5$$

$$6$$

$$4$$

$$3$$

Solution

Question 6

9

4

8

6

Solution

Question 7

$$1 kg, H_{2}$$

$$1 kg , He$$

$$\frac{1}{2}kg H_{2}+\frac{1}{2}kg He$$

$$\frac{1}{4}kg H_{2}+\frac{3}{4}kg He$$

Solution

Question 8

$$4 RT$$

$$5 RT$$

$$15 RT$$

$$11 RT$$

Solution

Question 9

$$T_{1}> T_{2}$$

$$T_{1}< T_{2}$$

$$T_{1}\leq T_{2}$$

$$T_{1}= T_{2}$$

Solution

Question 10

$$270.2$$

$$273.0$$

$$248.5$$

$$200.5$$

Solution

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