Kinetic Theory Test

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Kinetic Theory Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

The kinetic energy of $$ 1g $$ molecule of a gas at normal temperature and pressure is :

A
$$1.3$$ x $$10^2 J $$

B
$$2.7$$ x $$10^2 J $$

C
$$0.56$$ x $$10^4 J $$

D
$$3.4$$ x $$10^3 J $$

Solution

There is no information about gas so the translational kinetic energy is given as,

$$K = \dfrac{3}{2}RT$$

$$ = \dfrac{3}{2} \times 8.31 \times \left( {273 + 0} \right)$$

$$ = 3.4 \times {10^3}\;{\rm{J}}$$

Thus, the kinetic energy of a gas molecule is $$3.4 \times {10^3}\;{\rm{J}}$$.
Question 2
1 / -0

In an experiment, 1.35 mol of oxygen (O2) are heated at constant pressure starting at 11.0ºC. How much heat must be added to the gas to double its volume?

A

$$1.12\times 10^4 J$$

B

$$1.40\times 10^4$$

C
$$2.12\times 10^4$$J

D
$$3.12×10^4 J$$

Solution

Since oxygen $$(O_2)$$ is a diatomic gas
$$\therefore C_v=5/2R$$
So, $$C_p=C_v+R\\ \quad=5/2R+R\\ \quad =7/2R$$
At constant pressure
If we double the volume the temperature will be doubled.
$$T_i=11.0°C=(11.0+273)K=284K$$
So, $$T_p=2(284)K=568K$$
To obtain heat, $$Q=nC_p(T_p-T_i)$$
$$Q=(1.35mol)(7/2R)(508K-284K)\\Q=1.12\times10^4J$$
Question 3
1 / -0

$$m$$ grams of a gas of molecular weight $$M$$ is flowing in an insulated tube with velocity $$v$$. If the system is suddenly stopped then the rise in its temperature will be (γ= ratio of specific heats, R=universal gas constant, J=Mechanical equivalent of heat)

A
$$\dfrac {Mv^{2}(\gamma - 1)}{2RJ}$$

B
$$\dfrac {Mv^{2}\gamma}{2RJ}$$

C
$$\dfrac {v^{2}}{2Js}$$

D
$$\dfrac {Mv^{2}(\gamma - 1)}{2MRJ}$$

Solution
Question 4
1 / -0

Two cylinders contain the same amount of ideal monatomic gas. The same amount of heat is given to two cylinders. If the temperature rise in cylinder A is $$T_0$$ then temperature rise in cylinder B will be :

A
$$\dfrac{4}{3}T_0$$

B
$$2T_0$$

C
$$\dfrac{T_0}{2}$$

D
$$\dfrac{5}{3}T_0$$

Solution

$$Q=nC_vT_0$$
$$Q_A=nC_p\triangle T_A$$
$$Q_B=nC_v\triangle T_B$$
$$Q_A=Q_B$$
$$nC_pT_0=nC_v\triangle T$$
$$\triangle T=T_0\times \dfrac{C_p}{C_v}$$
$$=T_0\times \dfrac{5}{3}.$$
Hence, the answer is $$\dfrac{5}{3}T_0.$$
Question 5
1 / -0

By the ideal gas law, the pressure of $$0.60$$ moles $${NH}_{3}$$ gas in a $$3.00\ L$$ vessel at $${25}^{o}C$$ is, given that $$R=0.082\ L$$ atm $${mol}^{-1}{k}^{-1}$$:

A
$$48.9\ atm$$

B
$$4.89\ atm$$

C
$$0.489\ atm$$

D
$$489\ atm$$

Solution

Pressure is given as $$P=\dfrac{nRT}{V}=\dfrac{0.6\times .082\times (25+273)}{3}=4.89atm$$
Option B is correct.
Question 6
1 / -0

Mean kinetic energy (or average energy) per gm. of a molecule of a monoatomic gas is given by :

A
$$\dfrac{3RT}{2}$$

B
$$\dfrac{kT}{2}$$

C
$$\dfrac{RT}{3}$$

D
$$\dfrac{3kT}{2}$$

Solution
Question 7
1 / -0

Valency of carbon is 4. From this, we understand that there are ____chemical bond/bonds between the carbon atom and one oxygen atom in the compound-carbon dioxide.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

there are 2 chemical bonds between the carbon atom and each oxygen atom
in this way as the 2 oxygen atoms form 2 bonds each , the valency of oxygen is satisfied
as the carbon atom forms 4 bonds in total with the 2 oxygen atoms , its valency is also satisfied
Question 8
1 / -0

If temperature of body increases by 10%, then increase in radiated energy of the body is :

A
10 %

B
40 %

C
46 %

D
1000 %

Solution
Question 9
1 / -0

RMS velocity of an ideal gas at $$27^o C$$ is $$500{m/s}$$, Temperature is increased four times, RMS velocity will become.

A
$$1000{m/s}$$

B
$$560{m/s}$$

C
$$2000m/s$$

D
None of these

Solution

$$V_{rms}=\sqrt{\cfrac{3RT}{M}}=500$$ $${m/s}$$
$$V_{rms}^{\prime}=\sqrt{\cfrac{3RT^{\prime}}{M}} =\sqrt{\cfrac{3R(4T)}{M}}$$
$$V_{rms}^{\prime}=2\sqrt{\cfrac{3RT}{M}}=2V_{rms}=1000$$ $${m/s}$$
Question 10
1 / -0

The amount of heat energy required to raise the temperature of $$1\ g$$ of helium in a container of volume $$10L$$, from $$T_{1}\ K$$ to $$T_{2}\ K$$ is ($$N_{a} =$$ Avogadros number, $$k_{B}=$$ Boltzmann constant)

A
$$\dfrac { 3 }{ 2 } { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right) $$

B
$$\dfrac { 3 }{ 7} { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right) $$

C
$$\dfrac { 3 }{ 4 } { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right) $$

D
$$\dfrac { 3 }{ 8 } { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right) $$

Solution

The process is being completed at a constant volume because the volume of the container is fixed.
so the heat energy required will be $$Q=nC_v \Delta T$$
Where $$C_v$$ is specific heat at constant volume and $$n$$ is number of moles given by $$n=\dfrac{1gm}{4gm/mole}=\dfrac{1}{4}mole$$
and the specific heat is given as $$C_v=\dfrac{fN_ak_B}{2}$$
The degree of freedom, $$f$$ has value $$3$$ for a mono-atomic gas.
$$N_a$$ is Avogadro Number and $$k_B$$ is Boltzman's constant
putting all above values in the expression for $$Q$$ we get $$Q=\dfrac{3}{8}N_ak_B \Delta T=\dfrac{3}{8}N_ak_B(T_2 -T_1)$$
Option D is correct.

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Kinetic Theory Test - 52

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Kinetic Theory Test - 52

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Question 1

$$1.3$$ x $$10^2 J $$

$$2.7$$ x $$10^2 J $$

$$0.56$$ x $$10^4 J $$

$$3.4$$ x $$10^3 J $$

Solution

Question 2

$$1.12\times 10^4 J$$

$$1.40\times 10^4$$

$$2.12\times 10^4$$J

$$3.12×10^4 J$$

Solution

Question 3

$$\dfrac {Mv^{2}(\gamma - 1)}{2RJ}$$

$$\dfrac {Mv^{2}\gamma}{2RJ}$$

$$\dfrac {v^{2}}{2Js}$$

$$\dfrac {Mv^{2}(\gamma - 1)}{2MRJ}$$

Solution

Question 4

$$\dfrac{4}{3}T_0$$

$$2T_0$$

$$\dfrac{T_0}{2}$$

$$\dfrac{5}{3}T_0$$

Solution

Question 5

$$48.9\ atm$$

$$4.89\ atm$$

$$0.489\ atm$$

$$489\ atm$$

Solution

Question 6

$$\dfrac{3RT}{2}$$

$$\dfrac{kT}{2}$$

$$\dfrac{RT}{3}$$

$$\dfrac{3kT}{2}$$

Solution

Question 7

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 8

10 %

40 %

46 %

1000 %

Solution

Question 9

$$1000{m/s}$$

$$560{m/s}$$

$$2000m/s$$

None of these

Solution

Question 10

$$\dfrac { 3 }{ 2 } { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right) $$

$$\dfrac { 3 }{ 7} { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right) $$

$$\dfrac { 3 }{ 4 } { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right) $$

$$\dfrac { 3 }{ 8 } { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right) $$

Solution

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