Kinetic Theory Test

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Kinetic Theory Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

A flask is filled with 13 gm of an ideal gas at 27$$^{0}$$C its temperature is raised to 52$$^{0}$$C . The mass of the gas that has to be released to maintain the temperature of the gas in the flask at 52$$^{0}$$C and the pressure remaining the same is

A
2.5 gm

B
2.0 g

C
1.5 g

D
1.0 g

Solution

Since P, V , M and R are constants,
$${ m }_{ 1 }{ T }_{ 1 }={ m }_{ 2 }{ T }_{ 2 }\\ 13\times 300={ m }_{ 2 }\times 325\\ { m }_{ 2 }=12gm$$
Hence mass to be removed = 13-12= 1gm
Question 2
1 / -0

At what temperature will the hydrogen molecules escape from Earth's surface?

A
10$$^{4}$$ K

B
10$$^{3}$$ K

C
10$$^{2}$$ K

D
10$$^{1}$$ K

Solution

We have the escape velocity as $$v_e=11.2 km/s$$, boltzmann constant as $$1.38\times 10^{-23} m^2kgs^{-2}K^{-1}$$and molecular mass of hydrogen as $$2\times 1.66\times 10^{-27} amu $$
We have the formula as
$$\dfrac{3}{2}kT=\dfrac{1}{2}mv_e^2$$
or
$$T=\dfrac{mv_e^2}{3k}=\dfrac{2\times 1.66\times 10^{-27}\times 11.2^2\times 10^6}{3\times 1.38\times 10^{-23}}=10^4 K$$
Question 3
1 / -0

During an experiment an ideal gas is found to obey an additional gas law VT =constant. The gas is initially at temperature T and pressure P. When it is heated to the temperature 2T, the resulting pressure is

A
2P

B
P/2

C
4P

D
P/4

Solution

The gas follows VT=constant. But it also follows PV=nRT.
Thus $$(\dfrac{nRT}{P})T=$$constant
$$\implies \dfrac{T^2}{P}=$$constant
Therefore $$\dfrac{T^2}{P}=\dfrac{(2T)^2}{P'}$$
$$\implies P'=4P$$
Question 4
1 / -0

If the pressure of a gas contained in a closed vessel increases by x% when heated by 1$$^{0}$$C , its initial temperature is

A
$$\frac{100}{x}$$ Kelvin

B
$$\frac{100}{x}$$ Celsius

C
$$\frac{x+100}{x}$$ Kelvin

D
$$\frac{100-x}{x}$$ Celsius
Solution
Let the initial pressure and temperature be P and T.
From Ideal Gas Law,
$$P\propto T$$
$$\implies \dfrac{(1+\dfrac{x}{100})P}{P}=\dfrac{T+1}{T}$$
$$\implies T=\dfrac{100}{x}K$$
Question 5
1 / -0

The average kinetic energy of hydrogen molecule at NTP will be

A
0.4 x 10$$^{-20}$$ Joule /molecule

B
1.56 x 10$$^{-20}$$ Joule / molecule

C
zero

D
5.6 x 10$$^{-20}$$ Joule /molecule

Solution

Average kinetic energy:
$$KE_{avg}= \frac{3}{2}kT$$
$$=\dfrac{3}{2}\times 1.38 \times 10^{-23} \times 293$$
$$=0.40 \times 10^{-20}J$$
where k is Boltzman's constant and $$T=(20 + 273)^{\circ}K= 293K$$
Question 6
1 / -0

The pressure of a certain mass of gas at 27$$^{0}$$C is 84cm of Hg. If 25% of the gas is now introduced into the same vessel at the same temperature, the final pressure of the gas will be in cm of Hg

A
105

B
100

C
95

D
90

Solution

Pressure=$$h\rho g=\dfrac{nRT}{V}$$
$$\implies h\propto n\propto m$$
Therefore $$\dfrac{h_2}{h_1}=\dfrac{1.25m_{0}}{m_0}$$
$$\implies h_2=1.25h_1=105cm$$
Question 7
1 / -0

A cycle tube has volume $$2000 cm^3$$. Initially the tube is filled to 3/4th of its volume by air at pressure of $$105 N/m^{2}$$. It is to be inflated to a pressure of $$6 \times 10^{5} N/m^{2}$$ under isothermal conditions. The number of strokes of pump, which gives $$500 cm^3$$ air in each stroke, to inflate the tube is

A
21

B
12

C
42

D
11

Solution

In this problem, let us conserve total number of moles before and after,

$$(6X{ 10 }^{ 5 })(2000)=(105)(1500)+(101325)(500)n$$

Where n is total number of strokes of pump.
Solving this equation given $$n\approx21$$.
Question 8
1 / -0

If the number of degrees of freedom of a gas is f, then the value of $$\gamma $$ will be

A
f

B
2f

C
f/2

D
1+2/f

Solution

$$\gamma= \dfrac{2+f}{f}=1 + \dfrac{2}{f}$$ where $$\gamma= \dfrac{C_p}{C_v}$$ and f is the no of degrees of freedom.
Question 9
1 / -0

The total kinetic energy of $$8$$ litres of helium molecules at $$5 $$ atmosphere pressure will be

A
$$607 Joules$$

B
$$6078 Joules$$

C
$$607 ergs$$

D
$$6078 ergs$$

Solution
Question 10
1 / -0

A uniform tube with piston in the middle and containing a gas at 0$$^{0}$$C is heated to 100$$^{0}$$C at one side . If the piston moves 5 cm, find the length of the tube containing the gas at 100$$^{0}$$C.

A
16.75 cm

B
37.5 cm

C
28.25 cm

D
12.5 cm

Solution

As $$\dfrac{PV}{T}$$ is constant for gas in both sections of the cylinder.
Let the length and area of the tube be $$L$$ and $$A$$, respectively.
For section with constant temperature,
$$P_{0}V_{0}=P_1V_1$$
$$\implies P_ (A\dfrac{L}{2})=P_1(A(\dfrac{L}{2}-5))$$
Initial temperature $$T_0 = 0^oC = 273 \textrm{K}$$
Final temperature $$T_1 = 100^oC = 100+273 = 373 \textrm{K}$$
Similarly for section whose temperature increased,
$$\dfrac{P_0V_0}{T_0}=\dfrac{P_1V_1}{T_1}$$
$$\implies \dfrac{P_0(A\dfrac{L}{2})}{273}=\dfrac{P_1(A(\dfrac{L}{2}+5))}{373}$$
$$\implies L=65cm$$
So the length of tube containing gas at $$100^{\circ}$$ is $$\dfrac{L}{2}+5=37.5cm$$

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Kinetic Theory Test - 29

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Kinetic Theory Test - 29

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Question 1

2.5 gm

2.0 g

1.5 g

1.0 g

Solution

Question 2

10$$^{4}$$ K

10$$^{3}$$ K

10$$^{2}$$ K

10$$^{1}$$ K

Solution

Question 3

2P

P/2

4P

P/4

Solution

Question 4

$$\frac{100}{x}$$ Kelvin

$$\frac{100}{x}$$ Celsius

$$\frac{x+100}{x}$$ Kelvin

$$\frac{100-x}{x}$$ Celsius

Solution

Question 5

0.4 x 10$$^{-20}$$ Joule /molecule

1.56 x 10$$^{-20}$$ Joule / molecule

zero

5.6 x 10$$^{-20}$$ Joule /molecule

Solution

Question 6

105

100

95

90

Solution

Question 7

21

12

42

11

Solution

Question 8

f

2f

f/2

1+2/f

Solution

Question 9

$$607 Joules$$

$$6078 Joules$$

$$607 ergs$$

$$6078 ergs$$

Solution

Question 10

16.75 cm

37.5 cm

28.25 cm

12.5 cm

Solution

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