Kinetic Theory Test

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Kinetic Theory Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

An ideal gas follows a process described by $$PV^2=C$$ from $$(P_1, V_1, T_1)$$ to $$(P_2, V_2, T_2)$$ (C is a constant). Then.

A
If $$P_1 > P_2$$ then $$T_2 > T_1$$

B
If $$V_2 > V_1$$ then $$T_2 < T_1$$

C
If $$V_2 > V_1$$ then $$T_2 > T_1$$

D
If $$P_1 > P_2$$ then $$V_2 > V_2$$

Solution

we know $$PV=nRT$$
also $$PV^2=C$$
So $$\dfrac{T^2}{P}=C_1$$
$$\dfrac{T_1^2}{P_1}=\dfrac{T_2^2}{P_2}$$
Also$$T_1V_1=T_2V_2$$
So answer B is correct.
Question 2
1 / -0

At what temperature hydrogen molecules will escape from the earth's surface?
(Take mass of hydrogen molecule $$ = 0.34 \times 10^{-26} $$ kg, Boltzman constant $$ = 1.38 \times 10^{-23} $$ J/K, radius of the earth $$ = 6.4 \times 10^{6} $$ and acceleration due to gravity $$ = 9.8 m/s^2 $$

A
10 K

B
$$ 10^2 K $$

C
$$ 10^3 K $$

D
$$ 10^4 K $$

Solution

The root mean square velocity of gas
$$ v_{mns} = \sqrt {\dfrac {3KT}{m}} .. (i) $$
Escape velocity of gas molecules
$$ v_{es} = \sqrt {2gR_e} ...(ii) $$
At the root mean square velocity of gas, molecule must be equal to the escape velocity.
From Eqs. (i) and (ii), we get
$$ \sqrt { \dfrac {3KT}{m} } = \sqrt{2gR_e} $$
$$ T = \dfrac {2gR_e m }{3K} $$
$$\approx 10^{4}K$$
Question 3
1 / -0

If at the same temperature and pressure, the densities of two diatomic gases are $${ d }_{ 1 }$$ and $${ d }_{ 2 }$$ respectively, the ratio of mean kinetic energy per molecule of gases will be :

A
$$1 : 1$$

B
$${ d }_{ 1 }:{ d }_{ 2 }$$

C
$$\sqrt { { d }_{ 1 } } :\sqrt { { d }_{ 2 } } $$

D
$$\sqrt { { d }_{ 2 } } :\sqrt { { d }_{ 1 } } $$

Solution

Mean kinetic energy of gas depends only on the temperature. Here temperature is given same, so ratio of kinetic energies will be $$1 : 1$$.
Question 4
1 / -0

$$4.48L$$ of an ideal gas at S.T.P requires $$12$$ calories to raise its temperature by $${15}^{o}C$$ at constant volume. The $${C}_{p}$$ of the gas is:

A
$$3cal$$

B
$$4cal$$

C
$$7cal$$

D
$$6cal$$

Solution

At STP, 22.4L of ideal gas = 1 mole

Heat required $$=n{ C }_{ V }\Delta T$$

$$\therefore \quad 12=\dfrac{4.48}{22.4}\times { C }_{ V }\times 15$$

$$\therefore \quad { C }_{ V }=0.4cal/mol.K$$

Now, $${ C }_{ P }-{ C }_{ V }=R$$

$$\therefore \quad { C }_{ P }={ C }_{ V }+R$$

$$\therefore \quad { C }_{ P }=0.4+5.6$$ (R=5.6 cal/mol.K)

$$\therefore \quad \boxed { { C }_{ P }=6Cal } $$

Hence, the correct option is $$D$$
Question 5
1 / -0

In Haber's process of ammonia manufacture:
$${ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\longrightarrow 2{ NH }_{ 3 }(g)$$, $$\Delta { H }_{ { 25 }^{ o }C }^{ o }=-92.2kJ$$
Molecules $${N}_{2}(g)$$ $${H}_{2}(g)$$ $${NH}_{3}(g)$$
$${C}_{P}J{ K }^{ -1 }{ mol }^{ -1 }$$ $$29.1$$ $$28.8$$ $$35.1$$
If $${C}_{P}$$ is independent of temperature,then reaction at $${100}^{o}C$$ compared to that of $${25}^{o}C$$ will be:

A
more endothermic

B
less endothermic

C
more exothermic

D
less exothermic

Solution
Question 6
1 / -0

The mean energy of a molecule of an ideal gas is

A
$$KT$$

B
$$\dfrac { 1 }{ 2 } KT$$

C
$$\dfrac { 3 }{ 2 } KT$$

D
$$2 KT$$

Solution

Mean energy of a molecule of an ideal gas $$E =\dfrac{1}{2}mv^2_{rms}$$
Rms velocity of the molecules $$v_{rms} = \sqrt{\dfrac{3nRT }{m}}$$
where $$nR=K$$
$$\therefore$$ $$E =\dfrac{1}{2}m\bigg(\dfrac{3KT}{m}\bigg)$$
$$\implies$$ $$E= \dfrac{3}{2}KT$$
Question 7
1 / -0

Two identical containers P and Q with frictionless piston contain the same ideal gas at the same temperature and the same volume. The mass of gas in P is $$m_1$$ and that in Q is $$m_2$$. The gas in each cylinder is now allowed to expand isothermally to the same final volume $$1.5$$V. The changes pressure in P and Q are found to be $$\Delta$$p and $$2\Delta$$p respectively then.

A
$$\displaystyle m_2=2m_1$$

B
$$\displaystyle m_1 =2m_2$$

C
$$\displaystyle m_1=\frac{m_2}{4}$$

D
$$\displaystyle m_1m_2=1$$

Solution

From ideal gas equation, we have
$$pV=mRT$$
So $$\displaystyle\frac{p}{m}=\frac{RT}{V}$$
As both gases expand isothermally to the same volume
So, $$RT/V=$$ constant
$$\therefore \displaystyle\frac{p}{m}=$$ constant
$$\therefore \displaystyle\frac{p_1}{p_2}=\frac{m_1}{m_2}\Rightarrow \frac{\Delta p}{2\Delta p}=\frac{m_1}{m_2}$$
$$\therefore m_2=2m_1$$.
Question 8
1 / -0

A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at $$300\ K$$. The ratio of the average rotational kinetic energy per $$O_{2}$$ molecule to per $$N_{2}$$ molecule is

A
$$1 : 1$$

B
$$1 : 2$$

C
$$2 : 1$$

D
Depends on the moments of inertia of the two molecules

Solution

A diatomic gas has 2 degrees of freedom associated with rotational motion. Law of equipartition of energy states that the rotational kinetic energy per degree of freedom is $$\dfrac{1}{2} KT$$. Since two gases are at same temperature their rotational kinetic energies will be equal.
Hence Option A is correct answer.
Question 9
1 / -0

Equal volumes of monoatomic and diatomic gases at the same temperature are given equal quantities of heat. Then,

A
the temperature of diatomic gas will be more

B
the temperature of monoatomic gas will be more

C
the temperature of both will be zero

D
nothing can be said

Solution

Energy required to increase the temperature of one mole of monoatomic gas by $$\Delta T$$=$$3/2\times R\times \Delta T$$
Energy required to increase the temperature of one mole of a diatomic gas by $$\Delta T$$=$$5/2\times R\times \Delta T$$
So, for equal volume i.e. equal moles equal amount of heat will increase the temperature of a monoatomic gas more than the diatomic gas
Question 10
1 / -0

The temperature at which the mean KE of the molecules of gas is one-third of the mean KE of its molecules at $$180^o C$$ is

A
$$-122^o C$$

B
$$-90^o C$$

C
$$60^o C$$

D
$$151^o C$$

Solution

Kinetic energy of a gas is directly proportional to its temperature
$$\therefore \dfrac {K_1}{K_2}=\dfrac{T_1}{T_2}$$
$$\dfrac {K}{K/3}=\dfrac{273+180}{T_2}\Rightarrow 3=\dfrac{453}{T_2}$$
$$T_2=151^oC$$

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Molecules	$${N}_{2}(g)$$	$${H}_{2}(g)$$	$${NH}_{3}(g)$$
$${C}_{P}J{ K }^{ -1 }{ mol }^{ -1 }$$	$$29.1$$	$$28.8$$	$$35.1$$

Kinetic Theory Test - 47

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Kinetic Theory Test - 47

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Question 1

If $$P_1 > P_2$$ then $$T_2 > T_1$$

If $$V_2 > V_1$$ then $$T_2 < T_1$$

If $$V_2 > V_1$$ then $$T_2 > T_1$$

If $$P_1 > P_2$$ then $$V_2 > V_2$$

Solution

Question 2

10 K

$$ 10^2 K $$

$$ 10^3 K $$

$$ 10^4 K $$

Solution

Question 3

$$1 : 1$$

$${ d }_{ 1 }:{ d }_{ 2 }$$

$$\sqrt { { d }_{ 1 } } :\sqrt { { d }_{ 2 } } $$

$$\sqrt { { d }_{ 2 } } :\sqrt { { d }_{ 1 } } $$

Solution

Question 4

$$3cal$$

$$4cal$$

$$7cal$$

$$6cal$$

Solution

Question 5

more endothermic

less endothermic

more exothermic

less exothermic

Solution

Question 6

$$KT$$

$$\dfrac { 1 }{ 2 } KT$$

$$\dfrac { 3 }{ 2 } KT$$

$$2 KT$$

Solution

Question 7

$$\displaystyle m_2=2m_1$$

$$\displaystyle m_1 =2m_2$$

$$\displaystyle m_1=\frac{m_2}{4}$$

$$\displaystyle m_1m_2=1$$

Solution

Question 8

$$1 : 1$$

$$1 : 2$$

$$2 : 1$$

Depends on the moments of inertia of the two molecules

Solution

Question 9

the temperature of diatomic gas will be more

the temperature of monoatomic gas will be more

the temperature of both will be zero

nothing can be said

Solution

Question 10

$$-122^o C$$

$$-90^o C$$

$$60^o C$$

$$151^o C$$

Solution

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