Kinetic Theory Test

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Kinetic Theory Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from $$T_{1}K$$ to $$T_{2}K$$ is

A
$$\displaystyle \frac{3}{8}N_{a}k_{B}(T_{2}-T_{1})$$

B
$$\displaystyle \frac{3}{2}N_{a}k_{B}(T_{2}-T_{1})$$

C
$$\displaystyle \frac{3}{4}N_{a}k_{B}(T_{2}-T_{1})$$

D
$$\displaystyle \frac{3}{4}N_{a}k_{B}\left ( \frac{T_{2}}{T_{1}} \right )$$

Solution

$$\displaystyle Q=\frac{f}{2}nR\Delta T$$
$$\displaystyle \frac{3}{2}\times \frac{1}{4}\times k_{B}N_{a}\Delta T$$
$$=\displaystyle \frac{3}{8}N_{a}k_{B}(T_{2}-T_{1})=\frac{3}{8}N_{a}k_{B}(T_{2}-T_{1})$$
Question 2
1 / -0

The ratio of the specific heats $$\dfrac {C_p}{C_v}=\gamma$$ in terms of degree of freedom (n) is given by:

A
$$\left (1+\dfrac {2}{n}\right )$$

B
$$\left (1+\dfrac {n}{2}\right )$$

C
$$\left (1+\dfrac {1}{n}\right )$$

D
$$\left (1+\dfrac {n}{3}\right )$$

Solution

The internal energy for 1 mole of gas is given as $$U=\dfrac{n}{2}RdT=C_vdT$$
where n is degrees of freedom
$$C_p-C_v= R$$
$$Cp=(1+\dfrac{n}{2}) R$$
and $$\dfrac{C_p}{C_v}=\gamma=\dfrac{(1+\dfrac{n}{2}) R}{\dfrac{n}{2}} =1+\dfrac{2}{n}$$
Question 3
1 / -0

A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. which of the following gives the density of the gas?

A
mKT

B
P/(kT)

C
Pm/(kT)

D
P/(kTV)

Solution

Hint: Use ideal gas equation.

Step 1: Note the given values
Volume =V
Pressure =P
Temperature =T
Mass of each molecule =m

Step 2: Calculate density of gas using ideal gas equation .
From the ideal gas law,
$$PV=NkT$$
$$\implies \dfrac{N}{V}=\dfrac{P}{kT}$$

Density of gas= $$\dfrac {\text{Total mass} }{\text volume}$$
$$\rho= \dfrac{Nm}{V}=\dfrac{Pm}{kT}$$
$$\textbf{Hence option C correct}$$
Question 4
1 / -0

Degree of freedom for polyatomic gas is

A
$$\ge 4$$

B
$$\ge 5$$

C
$$\ge 6$$

D
$$\ge 7$$

Solution

A diatomic molecule posses $$3$$ translational, $$2$$ rotational and $$1$$ vibrational degree of freedom which gives a total of $$6$$ degrees of freedom. A polyatomic gas has more than $$2$$ atoms in one molecule of it so it must have more than or equal to $$6$$ degrees of freedom.
Question 5
1 / -0

A vessel of volume $$20\ L$$ contains a mixture of hydrogen and helium at temperature of $$27^{\circ}C$$ and pressure $$2\ atm$$. The mass of mixture is $$5\ g$$. Assuming the gases to be ideal, the ratio of mass of hydrogen to that of the helium in the given mixture will be

A
$$1 : 2$$

B
$$2 : 3$$

C
$$2 : 1$$

D
$$2 : 5$$

Solution

Let there are $$n_{1}$$ moles of hydrogen and $$n_{2}$$ moles of helium in the given mixture. As $$Pv = nRT$$
Then the pressure of the mixture
$$P = \dfrac {n_{1}RT}{V} + \dfrac {n_{2}RT}{V} = (n_{1} + n_{2}) \dfrac {RT}{V}$$
$$\Rightarrow 2\times 101.3\times 10^{3} = (n_{1} + n_{2}) \times \dfrac {(8.3\times 300)}{20\times 10^{-3}}$$
or, $$(n_{1} + n_{2}) = \dfrac {2\times 101.30 \times 10^{3} \times 20\times 10^{-3}}{(8.3)(300)}$$
or, $$n_{1} + n_{2} = 1.62 .... (1)$$
The mass of the mixture is (in grams)
$$n_{1}\times 2 + n_{2} \times 4 = 5$$
$$\Rightarrow (n_{1} + 2n_{2}) = 2.5 ..... (2)$$
Solving the eqns. $$(1)$$ and $$(2)$$, we get
$$n_{1} = 0.74$$ and $$n_{2} = 0.88$$
Hence, $$\dfrac {m_{H}}{m_{He}} = \dfrac {0.74\times 2}{0.88\times 4} = \dfrac {1.48}{3.52} = \dfrac {2}{5}$$.
Question 6
1 / -0

The gas law $$\left [ \dfrac{PV}{T} \right ]=$$ constant is true for

A
isothermal change only

B
adiabatic change only

C
both isothermal & adiabatic

D
neither isothermal nor adiabatic

Solution

The given relation is always valid for ideal gas whether the process is adiabatic or isothermal because its the fundamental equation of ideal gas.
$$P \times V=n \times R \times T$$
Question 7
1 / -0

The temperature of a gas contained in a closed vessel is increased by 2 K when the pressure is increased by 2%. The initial temperature of the gas is :

A
200K

B
100K

C
200$$^{0}$$C

D
100$$^{0}$$C

Solution

In this case, T1 = T, T2 = T +2; P1 = P P2 = 1.02P
Putting these values in P1/T1 = P2/T2 , we get, T = 100K
Question 8
1 / -0

The mass of oxygen gas (in Kilo grams) occupying a volume of 11.2 litre at a temperature 27$$^{0}$$C and a pressure of 76cm of mercury is :
(Molecular weight of oxygen = 32)

A
0.001456

B
0.01456

C
0.1456

D
1.1456

Solution

$$PV=\dfrac { m }{ M } RT\\ 11.2\times { 10 }^{ -3 }\times 1.01\times { 10 }^{ 5 }=\dfrac { m }{ 32\times { 10 }^{ -3 } } \times 8.314\times 300\\ m=0.01456$$
Question 9
1 / -0

From what minimum height, a block of ice has to be dropped in order that it may melt completely on hitting the ground :

A
$$mgh$$

B
$$\dfrac{mgh}{l}$$

C
$$\dfrac{l}{g}$$

D
$$\dfrac{h}{lg}$$

Solution

Let the mass of the ice be m then
$$m \times g \times h = m \times l$$ , where l is the latent heat of melting.
$$h = \dfrac{l}{g}$$
Question 10
1 / -0

If for a gas $$\dfrac{R}{C_{v}}=0.67$$, then the gas is made up of molecules which are :

A
Diatomic

B
Monoatomic

C
Polyatomic

D
Mixture of Diatomic & Polyatomic

Solution

Given $$\dfrac {R}{C_v}=0.67$$
or
$$\dfrac {R}{f\dfrac{R}{2}}=0.67$$
or
$$f=\dfrac{2}{0.67}=3$$
The degree of freedom is 3. So the given gas is monoatomic.

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Kinetic Theory Test - 11

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Kinetic Theory Test - 11

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Question 1

$$\displaystyle \frac{3}{8}N_{a}k_{B}(T_{2}-T_{1})$$

$$\displaystyle \frac{3}{2}N_{a}k_{B}(T_{2}-T_{1})$$

$$\displaystyle \frac{3}{4}N_{a}k_{B}(T_{2}-T_{1})$$

$$\displaystyle \frac{3}{4}N_{a}k_{B}\left ( \frac{T_{2}}{T_{1}} \right )$$

Solution

Question 2

$$\left (1+\dfrac {2}{n}\right )$$

$$\left (1+\dfrac {n}{2}\right )$$

$$\left (1+\dfrac {1}{n}\right )$$

$$\left (1+\dfrac {n}{3}\right )$$

Solution

Question 3

mKT

P/(kT)

Pm/(kT)

P/(kTV)

Solution

Question 4

$$\ge 4$$

$$\ge 5$$

$$\ge 6$$

$$\ge 7$$

Solution

Question 5

$$1 : 2$$

$$2 : 3$$

$$2 : 1$$

$$2 : 5$$

Solution

Question 6

isothermal change only

adiabatic change only

both isothermal & adiabatic

neither isothermal nor adiabatic

Solution

Question 7

200K

100K

200$$^{0}$$C

100$$^{0}$$C

Solution

Question 8

0.001456

0.01456

0.1456

1.1456

Solution

Question 9

$$mgh$$

$$\dfrac{mgh}{l}$$

$$\dfrac{l}{g}$$

$$\dfrac{h}{lg}$$

Solution

Question 10

Diatomic

Monoatomic

Polyatomic

Mixture of Diatomic & Polyatomic

Solution

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