Kinetic Theory Test

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Kinetic Theory Test - 40

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Question 1
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Directions For Questions

Consider air to be a diatomic gas with average mole moasses 29g/mole. A mass 1.45kg of air is contained in a cylinder with a piston at $$27^{0}C$$ and pressure $$1.5\times10^{5} N/m^{2}$$. Energy is given to the system as heat and the system is allowed to expand till the final pressure $$3.5\times10^{5} N/m^{2}$$. As the gas expands, pressure and volume follow the relation
$$ V= KP^{2}$$ where K is constant (R =8.3J/(mole-K)

...view full instructions

Final volume of the system will be nearly -

A
$$6.2m^{3}$$

B
$$2.8m^{3}$$

C
$$4.5m^{3}$$

D
$$1.4m^{3}$$

Solution

$$PV = nRT$$
Initial volume
$$= \displaystyle\ \frac{ 1.45\times10^{3}}{29} \times \displaystyle\ \frac{8.3\times300}{1.5\times10^{5}}m^{3}$$
$$\displaystyle\ \frac{V_{i}}{P_{i^{2}}}$$ =$$\displaystyle\ \frac{V_{f}}{P_{f}^{2}}$$ $$\Rightarrow$$ $$V_{f}$$ = $$\left (\displaystyle\ \frac{3.5}{1.5}\right)^{2} V_{i}$$ = $$4.5 m^{3}$$
$$T = \displaystyle\ \frac{P_{f}V_{f}}{nR} = 3800K$$
$$\triangle U = nC_{V}\triangle T = n \displaystyle\ \frac{5R}{2} \triangle T = 3.6\times10^{6}J$$
Question 2
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If the kinetic energy of the molecules in $$5$$ litres of helium at $$2$$ $$atm$$ is $$E$$. What is the kinetic energy of molecules in $$15$$ litres of oxygen at $$3$$ $$atm$$ in terms of $$E$$?

A
$$7.5E$$

B
$$7E$$

C
$$8.5E$$

D
$$8E$$

Solution

Kinetic energy of n moles is given by,

$$K=\dfrac { f }{ 2 } nRT=\dfrac { f }{ 2 } PV$$

Thus, $$E=\dfrac { 3 }{ 2 } (5)(2)=15$$

$$E'=\dfrac { 5 }{ 2 } (15)(3)=112.5=7.5E$$

Answer is $$7.5E$$
Question 3
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The average kinetic energy of the molecules of an ideal gas at $$10^{\circ}C$$ has the value E. The temperature at which the kinetic energy of the same gas becomes 2E is

A
$$5^{\circ}C$$

B
$$10^{\circ}C$$

C
$$40^{\circ}C$$

D
None of these

Solution

The average kinetic energy of gas molecules $$T$$
Thus $$\dfrac{E_2}{E_1}=\dfrac{T_2}{T_1}=2$$
$$\implies T_2=2\times T_1$$
$$=2\times 283K=566K=293^{\circ}C$$
Question 4
1 / -0

$$\Delta C_p$$ for change $$N_2(g)+3H_2 (g)= 2N\! H_3(g)$$ is:

A
$$C_{pNH_3}-(C_{pN_2})$$

B
$$2C_{pNH_3}-(C_{pN_2}+3C_{pH_2})$$

C
$$2C_{pNH_3}-(C_{pH_2})$$

D
$$2C_{pNH_3}+(C_{pN_2}+3C_{pH_2})$$

Solution

$$\Delta C_p$$ represents the change in the heat capacity at constant pressure. It is equal to the difference in the heat capacities of the products and the reactants.
$$\Delta C_p = C_p \: of \: products - C_p \: of \: reactants$$.
$$\Delta C_p$$ for change $$N_2(g)+3H_2 (g)= 2N\! H_3(g)$$ is given by the following expression.
$$\Delta C_p=2C_{pNH_3}-(C_{pN_2}+3C_{pH_2})$$
Question 5
1 / -0

When x amount of heat is given to a gas at constant pressure, it performs $$\displaystyle \frac{x}{3}$$ amount of work. The average number of degrees of freedom per molecule of the gas is-

A
3

B
4

C
5

D
6

Solution

$$\displaystyle \frac{W}{Q}=\frac{P\Delta V}{nC_{P}\Delta T}=\frac{nR\Delta T}{nC_{P}\Delta T}=\frac{x/3}{x}$$ (standard result)

$$\displaystyle \Rightarrow C_{P}=3R=\left ( \frac{f}{2}+1 \right )R\Rightarrow f=4$$
Question 6
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At ordinary temperatures, the molecules of a diatomic gas have only translational and rotational kinetic energies. At high temperatures, they may also have vibrational energy. As a result of this compared to lower temperatures, a diatomic gas at higher temperatures will have-

A
lower molar heat capacity

B
higher molar heat capacity

C
lower isothermal compressibility

D
higher isothermal compressibility

Solution

Vibrational energy involves additional degrees of freedom. Thus the degrees of freedom for a diatomic gas increases at higher temperatures.
Molar heat capacity is proportional to the number of degrees of freedom of the gas.
Thus the molar heat capacity also increases for a diatomic gas at higher temperatures.
Question 7
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Which of the following expands most on the rise of an equal amount of temperature?

A
Liquid

B
Solid

C
Gas

D
Cannot say

Solution

Gases have low intermolecular forces of attraction among solid, liquids and gases and hence they expand the most when appliedthe same amount of heat as others.
Question 8
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STATEMENT-1 : According to kinetic theory of gases the internal energy of a given sample of an ideal gas is only kinetic.
STATEMENT-2 : The ideal gas molecules exert force on each other only when they collide.

A
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

C
STATEMENT-1 is True, STATEMENT-2 is False

D
STATEMENT-1 is False, STATEMENT-2 is True

Solution

The ideal gas molecules exert force on each other only when they collide. Thus the only phenomenon occurring is the collision of the molecules due to the kinetic energy of the molecules.
Hence the internal energy of a given sample of ideal gas is only kinetic.
Question 9
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When the temperature is increased:

A
speed with which the molecules vibrate increases

B
speed with which the molecules vibrate decreases

C
there is no change in the speed of the molecules

D
the speed of the molecules remains constant

Solution

$$\textbf{Explanation}$$
$$\bullet$$Kinetic energy of particle can be given by the formula:
$$K = \dfrac{3}{2} K_b\ T$$, where $$K_b$$ is Boltzmann’s constant and T is temperature.

$$\bullet$$As temperature increases, it will results into increase of kinetic energy. Electrons or atoms in gaseous states are bound with either energy states or covalent bonds which results into increment of vibrational energy.
So, speed with which the molecules vibrate increases.

Hence, option A is the correct answer.
Question 10
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One mole of a diatomic gas undergoes a process $$P=\frac {P_0}{1+\left (\frac {V}{V_0}\right )^3}$$, where $$P_0, V_0$$ are constants. The translational kinetic energy of the gas when $$V = V_0$$ is given by

A
$$\frac {5P_0V_0}{4}$$

B
$$\frac {3P_0V_0}{4}$$

C
$$\frac {3P_0V_0}{2}$$

D
$$\frac {5P_0V_0}{2}$$

Solution

$$\displaystyle P=\dfrac {P_0}{1+\left (\dfrac {V}{V_0}\right )^3}=\dfrac {P_0}{2}\Rightarrow T=\dfrac {P_0V_0}{2R}$$

$$\therefore$$ Translational kinetic energy is equal to $$\dfrac {3}{2}RT=\dfrac {3R}{2}\dfrac {P_0V_0}{2R}=\dfrac {3P_0V_0}{4}$$

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Kinetic Theory Test - 40

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Kinetic Theory Test - 40

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SHARING IS CARING

Question 1

$$6.2m^{3}$$

$$2.8m^{3}$$

$$4.5m^{3}$$

$$1.4m^{3}$$

Solution

Question 2

$$7.5E$$

$$7E$$

$$8.5E$$

$$8E$$

Solution

Question 3

$$5^{\circ}C$$

$$10^{\circ}C$$

$$40^{\circ}C$$

None of these

Solution

Question 4

$$C_{pNH_3}-(C_{pN_2})$$

$$2C_{pNH_3}-(C_{pN_2}+3C_{pH_2})$$

$$2C_{pNH_3}-(C_{pH_2})$$

$$2C_{pNH_3}+(C_{pN_2}+3C_{pH_2})$$

Solution

Question 5

3

4

5

6

Solution

Question 6

lower molar heat capacity

higher molar heat capacity

lower isothermal compressibility

higher isothermal compressibility

Solution

Question 7

Liquid

Solid

Gas

Cannot say

Solution

Question 8

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is True

Solution

Question 9

speed with which the molecules vibrate increases

speed with which the molecules vibrate decreases

there is no change in the speed of the molecules

the speed of the molecules remains constant

Solution

Question 10

$$\frac {5P_0V_0}{4}$$

$$\frac {3P_0V_0}{4}$$

$$\frac {3P_0V_0}{2}$$

$$\frac {5P_0V_0}{2}$$

Solution

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