Kinetic Theory Test

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Kinetic Theory Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

n moles of an ideal monoatomic gas undergoes an isothermal expansion at temperature T during which its volume becomes 4 times. The work done on the gas and change in internal energy of the gas respectively is

A
n RT Ln 4,0

B
- n RT Ln 4,0

C
n RT Ln 4 $$\frac { 3 n R T } { 2 }$$

D
- n RT Ln 4, $$\frac { 3 n R T } { 2 }$$

Solution

$$\begin{array}{l} w=nRT\, \, \, \ln { \left( { \frac { { 4v } }{ v } } \right) } \\ =nRT\, \, \ln { 4 } \, \, \, \, \, \, \& \, \, \, \Delta u=0 \end{array}$$
$$\therefore$$ Option $$A$$ is correct.
Question 2
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Which of the following is incorrect relation fot a perfect gas?

A
$${ \quad \left( \frac { \eth U }{ \eth V } \right) }_{ T }=0$$

B
$${ \quad \left( \frac { \eth H }{ \eth P } \right) }_{ T }=0$$

C
$${ \quad \left( \frac { \eth T }{ \eth P } \right) }_{ H }=0$$

D
$${ \quad \left( \frac { \eth U }{ \eth T } \right) }_{ V }=0$$

Solution

$$(A)\left(\dfrac{\partial U}{\partial V}\right)_T=0$$
because perfect gas equation is $$PV=nRT.$$
$$\left(\dfrac{\partial H}{\partial P}\right)_T=\left(\dfrac{\partial T}{\partial P}\right)_H=\left(\dfrac{\partial U}{\partial b}\right)_V$$
Question 3
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An ideal gas is expanding such that $$PT = constant$$. The coefficient of volume expansion of the gas is

A
$$\dfrac {1}{T}$$

B
$$\dfrac {2}{T}$$

C
$$\dfrac {3}{T}$$

D
$$\dfrac {4}{T}$$

Solution
Question 4
1 / -0

The Kinetic energy of translation of $$20$$ gram of oxygen at $${47^0}C$$ is [ Molecular weight of $${O_2} = 32$$ gram/ mol and $$R = 8.3J/mol/K$$]

A
2490 ergs

B
2490 J

C
249 J

D
960 J

Solution
Question 5
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One $$kg$$ of diatomic gas is at a pressure of $$8\times 10^{4}\ N/m^{2}$$. The density of the gas is $$4\ kg/m^{3}$$. What is the energy of the gas due to its thermal motion?

A
$$5\times 10^{4} J$$

B
$$6\times 10^{4} J$$

C
$$7\times 10^{4} J$$

D
$$3\times 10^{4} J$$

Solution

Thermal energy corresponds to internal energy.
Mass$$=1$$kg
Density$$=8$$kgper $${m}^{3}$$
Volume $$=\dfrac{Mass}{Density}=\dfrac{1}{8}{m}^{3}$$
Pressure$$=8\times{10}^{4}$$Nper sq.m
$$\therefore$$ Internal Energy$$=\dfrac{5}{2}PV=5\times {10}^{4}$$J
Question 6
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The kinetic energy of 1 gram mole of a gas at normal temperature and pressure is ( R = 8.31 J/mole- K)

A
0.56 $$\times { 10 }^{ 4 }J$$

B
1.3 $$\times { 10 }^{ 2 }J$$

C
2.7 $$\times { 10 }^{ 2 }J$$

D
3.4 $$\times { 10 }^{ 3 }J$$

Solution

Given,
$$R=8.31J/mol K$$
$$T=273K$$
Kinetic energy,
$$K=\dfrac{3}{2}RT$$
$$K=\dfrac{3}{2}\times 8.31\times 273=3.4\times 10^3 J$$
The correct option is D.
Question 7
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A mixture Of $$n _ { 2 }$$ moles of mono atomic gas and $$n _ { 2 }$$ moles of diatomic gas has $$\frac { C _ { p } } { C _ { V } } = y = 1.5$$

A
$$n _ { 1 } = n _ { 2 }$$

B
$$2 n _ { 1 } = n _ { 2 }$$

C
$$n _ { 1 } = 2 n _ { 2 }$$

D
$$2 n _ { 1 } = 3 n _ { 2 }$$

Solution

$$\begin{array}{l} As\, { y_{ mix } }=\dfrac { { { C_{ { p_{ mix } } } } } }{ { { C_{ { v_{ mix } } } } } } where\, { C_{ { p_{ mix } } } }=\dfrac { { { n_{ 1 } }{ C_{ { p_{ 1 } } } }+{ n_{ 2 } }{ C_{ { p_{ 2 } } } } } }{ { { n_{ 1 } }+{ n_{ 2 } } } } \\ and\, \, { C_{ { v_{ mix } } } }=\dfrac { { { n_{ 1 } }{ C_{ { v_{ 1 } } } }+{ n_{ 2 } }{ C_{ { v_{ 2 } } } } } }{ { { n_{ 1 } }+{ n_{ 2 } } } } \\ So,\, \, { y_{ mix } }=\dfrac { { { n_{ 1 } }{ C_{ { p_{ 1 } } } }+{ n_{ 2 } }{ C_{ { p_{ 2 } } } } } }{ { { n_{ 1 } }+{ n_{ 2 } } } } \\ Given\, ,\, for\, monoatomic\, { C_{ p } },\dfrac { 5 }{ 2 } R\, and\, { C_{ { v_{ 1 } } } }=\dfrac { 3 }{ 2 } R \\ For\, diatomic\, { C_{ { p_{ 2 } } } }=\dfrac { { 7R } }{ 2 } \, and\, { C_{ { v_{ 2 } } } }=\dfrac { 5 }{ 2 } R \\ { y_{ mix } }=\dfrac { { { n_{ 1 } }\times \dfrac { 5 }{ 2 } R+{ n_{ 2 } }\times \dfrac { 7 }{ 2 } R } }{ { { n_{ 1 } }\times \dfrac { 3 }{ 2 } R+{ n_{ 2 } }\times \dfrac { 5 }{ 2 } R } } =\dfrac { { 5{ n_{ 1 } }+7{ n_{ 2 } } } }{ { 3{ n_{ 1 } }+5{ n_{ 2 } } } } =\dfrac { 3 }{ 2 } \\ 10{ n_{ 1 } }+14{ n_{ 2 } }=9{ n_{ 1 } }+15{ n_{ 2 } } \\ { n_{ 1 } }={ n_{ 2 } } \\ Hence, \\ option\, \, A\, \, is\, correct\, \, answer. \end{array}$$
Question 8
1 / -0

An ideal gas at pressure P is adiabatically compressed so that its density becomes n times the initial value. The final pressure of the gas will be $$ ( \gamma= C_p/C_v) $$

A
$$ n^\gamma P $$

B
$$ n^{-\gamma} P $$

C
$$ n^{\gamma-1} P $$

D
$$ n^{1-\gamma} P $$

Solution
Question 9
1 / -0

When a monoatomic gas expands at consatnt pressure, the percentage of heat supplied that increases temperature of the gas and in doing external work in expansion at constant pressure is

A
$$100%,0$$

B
$$60%,40%$$

C
$$40%,60%$$

D
$$75%,25%$$
Question 10
1 / -0

If the law of equipartition of energy is applied to solid elements like silver, the heat capacity of one gram atom of the solid is

A
$$\dfrac5 2 R$$

B
$$\dfrac32 R$$

C
$$R$$

D
$$3 R$$

Solution

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Re-Attempt Weekly Quiz Competition

Kinetic Theory Test - 60

Result

Kinetic Theory Test - 60

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SHARING IS CARING

Question 1

n RT Ln 4,0

- n RT Ln 4,0

n RT Ln 4 $$\frac { 3 n R T } { 2 }$$

- n RT Ln 4, $$\frac { 3 n R T } { 2 }$$

Solution

Question 2

$${ \quad \left( \frac { \eth U }{ \eth V } \right) }_{ T }=0$$

$${ \quad \left( \frac { \eth H }{ \eth P } \right) }_{ T }=0$$

$${ \quad \left( \frac { \eth T }{ \eth P } \right) }_{ H }=0$$

$${ \quad \left( \frac { \eth U }{ \eth T } \right) }_{ V }=0$$

Solution

Question 3

$$\dfrac {1}{T}$$

$$\dfrac {2}{T}$$

$$\dfrac {3}{T}$$

$$\dfrac {4}{T}$$

Solution

Question 4

2490 ergs

2490 J

249 J

960 J

Solution

Question 5

$$5\times 10^{4} J$$

$$6\times 10^{4} J$$

$$7\times 10^{4} J$$

$$3\times 10^{4} J$$

Solution

Question 6

0.56 $$\times { 10 }^{ 4 }J$$

1.3 $$\times { 10 }^{ 2 }J$$

2.7 $$\times { 10 }^{ 2 }J$$

3.4 $$\times { 10 }^{ 3 }J$$

Solution

Question 7

$$n _ { 1 } = n _ { 2 }$$

$$2 n _ { 1 } = n _ { 2 }$$

$$n _ { 1 } = 2 n _ { 2 }$$

$$2 n _ { 1 } = 3 n _ { 2 }$$

Solution

Question 8

$$ n^\gamma P $$

$$ n^{-\gamma} P $$

$$ n^{\gamma-1} P $$

$$ n^{1-\gamma} P $$

Solution

Question 9

$$100%,0$$

$$60%,40%$$

$$40%,60%$$

$$75%,25%$$

Question 10

$$\dfrac5 2 R$$

$$\dfrac32 R$$

$$R$$

$$3 R$$

Solution

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