Kinetic Theory Test

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Kinetic Theory Test - 49

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Question 1
1 / -0

Two identical glass bulbs are interconnected by a thin glass tube at $$0^{\circ}C$$. A gas is filled at N.T.P. in these bulb is placed in ice and another bulb is placed in hot bath, then the pressure of the gas becomes $$1.5\ times$$. The temperature of hot bath will be

A
$$100^{\circ}$$

B
$$182^{\circ}C$$

C
$$256^{\circ}C$$

D
$$546^{\circ}C$$

Solution

Quantity of gas in these bulbs is constant i.e,
Initial no. of moles in both bulb $$=$$ final no. of moles
$$\Rightarrow n_1+n_2={ n }_{ 1 }^{ ' }+{ n }_{ 2 }^{ ' }$$
$$\Rightarrow \dfrac{PV}{R(273)}+\dfrac{PV}{R(273)}=\dfrac{1.5PV}{R(273)}+\dfrac{1.5PV}{R(T)}$$
$$\Rightarrow \dfrac{2}{273}=\dfrac{1.5}{273}+\dfrac{1.5}{T}$$
$$\Rightarrow T=819K$$
$$=546°C$$
Hence, the answer is $$546°C.$$
Question 2
1 / -0

A vessel of volume $$0.3 \ { { m }^{ 3 } }$$ contains Helium at $$20.0$$. The average kinetic energy per molecule for the gas is:

A
$$6.07\times { 10 }^{ -21 }J$$

B
$$7.3\times { 10 }^{ 3 }J$$

C
$$14.6\times { 10 }^{ 3 }J$$

D
$$12.14\times { 10 }^{ -21 }J$$

Solution

Given temperature of gas $$=20°C$$
                                            $$=293K$$
$$\Rightarrow$$ Average translational kinetic energy $$=\dfrac{3}{2}KT$$
                                                                 $$=\dfrac{3}{2}\times1.38\times10^{-23}\times293$$
                                                                 $$=6.07\times10^{-21}J$$
Hence, the answer is $$6.07\times10^{-21}J.$$
Question 3
1 / -0

A monoatomic ideal gas undergoes a process in which the ratio of $$P$$ to $$V$$ at any instant is constant and equal to unity. The molar heat capacity of gas is:

A
$$1.5\ R$$

B
$$2.0\ R$$

C
$$2.5\ R$$

D
$$0$$

Solution

A monoatomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equals 1.
$$\dfrac{P}{V}=$$ constant
$$PV^{−1}=$$ constant
$$PV^{\gamma}= $$ constant
$$\gamma =−1$$

Monatomic gas:
$$C_V=\dfrac{3}{2}R$$

Molar heat capacity $$=C_V+\dfrac{R}{1−\gamma}$$

$$=\dfrac{3}{2}R+\dfrac{R}{1−(−1)}$$

$$=\dfrac{3}{2}R+\dfrac{R}{2}=\dfrac{4R}{2}=2R$$
Question 4
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One Kg of a diatomic gas is at a pressure of $$8\times 10^4$$N/ $$m^{2}$$. The density of the gas is $$4$$kg/ $$m^{3}$$. The energy of the gas due to its thermal motion will be

A
$$3\times 10^4$$J

B
$$5\times 10^4$$J

C
$$6\times 10^4$$J

D
$$7\times 10^4$$J

Solution

For a diatomic molecular gas, internal energy per molecule is
$$U=\dfrac{5}{2}NK_BT$$
$$PV=NK_BT$$
From equation is (i) and (ii)
$$U=\dfrac{5}{2}$$PV
Here, $$P=8\times 10^4N$$ $$m^{-2}, \rho =4kg$$ $$m^{-3}$$, $$m=1$$kg
$$\therefore U=\dfrac{5}{2}\times 8\times 10^4\times \dfrac{1}{4}=5\times 10^4$$J
Question 5
1 / -0

The internal energy of one gram of helium at $$100$$K and one atmospheric pressure is?

A
$$100$$J

B
$$1200$$J

C
$$300$$J

D
$$500$$J

Solution

The helium molecule is monoatomic and hence its internal energy per molecule is $$\dfrac{3}{2}K_B$$ T(where $$k_B$$ is the Boltzmann, constant). The internal energy per mole is therefore is $$\dfrac{3}{2}RT$$. One gram of helium is one fourth mole and hence its internal energy is $$\dfrac{1}{4}\times \dfrac{3}{2}R\times 100=300$$J,( By taking the value of R to be approximately $$8$$J $$mol^{-1}$$ $$K^{-1}$$)
Question 6
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Which one of the following is not an assumption of kinetic theory of gases?

A
The volume occupied by the molecules of the gas is negligible

B
The force of attraction between the molecules is negligible

C
The collision between the molecules are elastic

D
All molecules have same speed
Solution
The correct answer is option (A).

Explanation: The assumptions of the kinetic theory of gases are:
Gases consist of a large number of molecules.
Molecules of gases move rapidly and randomly with different speeds
The volume of each molecule of gas is negligible compared to the volume occupied by the gas itself.
Elastic collisions occur when one molecule collides with another or when they collide with the walls of the container.
There is no force of attraction or repulsion between gas molecules.
The average kinetic energy of gases depends on the Kelvin temperature.
Hence, we see that all molecules do not have same speed according to the assumption of kinetic theory of gases.
The correct answer is option (D).
Question 7
1 / -0

If the pressure and the volume of certain quantity of ideal gas are halved, then its temperature

A
Is doubled

B
Becomes one-fourth

C
Remains constant.

D
Become four times

Solution

According to ideal gas equation
$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$ or $$T_2=\dfrac{P_2V_2}{P_1V_1}$$
Here, $$P_1=P, V_1=V, T_1=T, P_2=\dfrac{P}{2}, V_2=\dfrac{V}{2}, T_2=?$$
$$\therefore T_2=\dfrac{T\left(\dfrac{P}{2}\right)\left(\dfrac{V}{2}\right)}{PV}$$, $$T_2=\dfrac{T}{4}$$.
Question 8
1 / -0

When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increase the internal energy of gas is?

A
$$\frac{2}{5}$$

B
$$\frac{3}{5}$$

C
$$\frac{3}{7}$$

D
$$\frac{5}{7}$$

Solution

$$f$$= $$∆U$$/$$∆Q$$
=$$nC_{v}∆T$$/$$nC_{p}∆T$$
=$$C_{v}$$/$$C_{p}$$
=$$1$$/$$Y$$
$$Y$$=$$7$$/$$5$$ for ideal gas
$$f$$=$$5$$/$$7$$
Question 9
1 / -0

A monoatomic ideal gas, initially at temperature $$T_1$$. is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to temperature $$T_2$$ by releasing the piston suddenly. If $$L_1$$ and $$L_2$$ are the lengths of the gas column before and after expansion respectively, then $$\frac{T_1}{T_2}$$ is given by

A
$$(\dfrac{L_1}{L_2})^{2/3}$$

B
$$\dfrac{L_1}{L_2}$$

C
$$\dfrac{L_2}{L_1}$$

D
$$(\dfrac{L_2}{L_1})^{2/3}$$

Solution
Question 10
1 / -0

The average kinetic energy of $$O_2$$ at a particular temperatures is $$0.768$$ eV. The average kinetic energy of $$N_2$$ molecules in eV at the same temperature is?

A
$$0.0015$$

B
$$0.0030$$

C
$$0.048$$

D
$$0.768$$

Solution

Ideal gas law, says that

$$PV = N K T$$.

This means that:

Average Kinetic Energy, $$K_{avg} = \dfrac{3}{2} k T$$

This is a very important fact, as it tells the fundamentals about the temperature. The absolute temperature of the ideal gas is directly proportional to the average kinetic energy. So, changes in pressure and/or volume will result in changes in temperature too. Also, we can see that the average kinetic energy of the ideal gas

molecule will depend on temperature only and not on the gas nature/mass. Therefore, if the temperature is the same, then the kinetic energy of $$O_{2} $$ and $$N_{2}$$ both will be equal.

Hence average K.E. for $$N_{2}$$ will also be 0.768 eV.

Option D is correct.

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Kinetic Theory Test - 49

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Kinetic Theory Test - 49

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Question 1

$$100^{\circ}$$

$$182^{\circ}C$$

$$256^{\circ}C$$

$$546^{\circ}C$$

Solution

Question 2

$$6.07\times { 10 }^{ -21 }J$$

$$7.3\times { 10 }^{ 3 }J$$

$$14.6\times { 10 }^{ 3 }J$$

$$12.14\times { 10 }^{ -21 }J$$

Solution

Question 3

$$1.5\ R$$

$$2.0\ R$$

$$2.5\ R$$

$$0$$

Solution

Question 4

$$3\times 10^4$$J

$$5\times 10^4$$J

$$6\times 10^4$$J

$$7\times 10^4$$J

Solution

Question 5

$$100$$J

$$1200$$J

$$300$$J

$$500$$J

Solution

Question 6

The volume occupied by the molecules of the gas is negligible

The force of attraction between the molecules is negligible

The collision between the molecules are elastic

All molecules have same speed

Solution

Question 7

Is doubled

Becomes one-fourth

Remains constant.

Become four times

Solution

Question 8

$$\frac{2}{5}$$

$$\frac{3}{5}$$

$$\frac{3}{7}$$

$$\frac{5}{7}$$

Solution

Question 9

$$(\dfrac{L_1}{L_2})^{2/3}$$

$$\dfrac{L_1}{L_2}$$

$$\dfrac{L_2}{L_1}$$

$$(\dfrac{L_2}{L_1})^{2/3}$$

Solution

Question 10

$$0.0015$$

$$0.0030$$

$$0.048$$

$$0.768$$

Solution

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