Kinetic Theory Test

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Kinetic Theory Test - 75

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Weekly Quiz Competition

Question 1
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One box containing $$1$$ mole of $$He$$ at $$7/3 T_0$$ and other box containing $$1$$ mole of a polyatomic gas $$(\gamma=1.33)$$ at $$T_0$$ are placed together to attain thermal equilibrium. The final temperature becomes $$T_f$$. Then :

A
$$T_f=\dfrac {9}{13}T_0$$

B
$$T_f=\dfrac {13}{9}T_0$$

C
$$T_f=\dfrac {5}{2}T_0$$

D
$$T_f=\dfrac {3}{2}T_0$$

Solution

At thermal equilibrium, both have the same temperature and heat given by one gas equal to the heat absorbed by other gas.

Heat released by $$He=$$ Heat taken up by polyatomic gas

$$1\times C_v\times \Delta T=1\times C_v\times \Delta T$$

$$1\times \dfrac {3R}{2}(7/3T_0-T_f)=1\times 3R\times (T_f-T_0)$$

$$\left [C_v=\dfrac {3}{2}\ R\ \text {for He and}\ C_v=\frac {3}{2}R+X=\underset {\text {for polyatomic gas}}{\dfrac {3}{2}R+\dfrac {3}{2}R=3R}\right ]$$

$$\therefore \dfrac {7}{2}T_0-\dfrac {3}{2}T_f=3T_f-3T_0$$

$$\dfrac {13}{2}T_0=\dfrac {9}{2}T_f$$

$$\therefore T_f=\dfrac {13}{9}T_0$$
Question 2
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Directions For Questions

Two closed identical conducting containers are found in the laboratory of an old scientist. For the verification of the gas some experiments are performed on the two boxes and the results are noted

Experiment 1: When the two containers are weight
$$W_{A}$$ = 225g
$$W_{B}$$ = 160g and mass of evacuated container in both the case
$$W_{C}$$ = 100g.

Experiment 2:
When two containers are given same amount of heat same temperature rise is recorded. The pressure changes are found
$$\triangle P_{A}$$ = 2.5 atm, $$\triangle P_{B}$$ = 1.5 atm

Required data for unknown gas: Mono (molar gas)
$$He=4g, Ne=20g, Ar=40g, Kr=84g, Xe=131g, Rn=222g$$

Dia (molar gas)
$$H_{2} = 2g, F_{2}= 19g, N_{2} = 28g, O_{2}= 32, Cl_{2} = 71g$$

...view full instructions

Identify the type of gas filled in container A and B respectively

A
Mono, mono

B
Dia, dia

C
Mono, dia

D
Dia, Mono

Solution

Since the temperature rise and heat supplied to both the container is same.

$$Q=n_AC_{VA}\Delta T=n_BC_{VB}\Delta T$$ $$\Rightarrow n_AC_{VA}=n_BC_{VB}\ -(1)$$

also, $$\Delta P_{A}V=n_A R\Delta T\ -(2)$$

and, $$\Delta P_{B}V=n_B R\Delta T\ -(3)$$

Dividing $$(2)$$ by $$(3)$$,

$$\dfrac{n_A}{n_B}=\dfrac{\Delta P_A}{\Delta P_B}=2.5/1.5=5/3$$

$$\Rightarrow n_A=5n_B/3$$

From equation $$(1)$$

$$C_{VB}/C_{VA}=5/3=(5R/2)/(3R/2)$$

$$C_{VB}=5R/2$$, so $$B$$ is diatomic.

$$C_{VA}=3R/2$$, so $$A$$ is Mono-atomic.
Question 3
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In a certain gas $$\displaystyle \frac{2}{5}$$th of the energy of molecules is associated with the rotation of molecules and the rest of it is associated with the motion of the centre of mass. How much energy must be supplied to one mole of this gas at constant volume to raise the temperature by $$1^{\circ}C$$?

A
$$20.8 J$$

B
$$28.8 J$$

C
$$26.8 J$$

D
$$25.8 J$$

Solution

Since two fifth energy is associated with the rotation, thus there are five degrees of freedom out of which two are rotational and three translational.

Thus $${ e }_{ total }=\dfrac { 5 }{ 2 } RT$$

$$\Delta E=\dfrac { 5 }{ 2 } (8.314)1=20.8J$$
Question 4
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A monoatomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equal to unity. The molar heat capacity of gas is:

A
$$\frac {4R}{2}$$

B
$$\frac {3R}{2}$$

C
$$\frac {5R}{2}$$

D
zero

Solution

$$\textbf{Given:-}$$ Mono-atomic ideal gas
$$\dfrac{p}{V} = 1$$

$$\textbf{Solution:-} $$

The molar heat capacity for any process is given by a following expression

$$C = C_{V} + \dfrac{R}{1 - \gamma } when pV^{\gamma } = constant$$

$$\dfrac{C_{P}}{C_{V}} = \gamma $$

$$\dfrac{p}{V} = 1 ie, pV^{-1} = constant$$

$$C = \dfrac{3}{2}R + \dfrac{R}{1 - (-1)} = \dfrac{3}{2}R + \dfrac{R}{2} = \dfrac{4}{2}R$$

$$\textbf{Hence the correct option is A}$$
Question 5
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Directions For Questions

$$200$$ g of a gas $$X$$ $$(C_P=0.125 \ cal/g$$ and $$C_v \ =0.075 \ cal/g$$) is placed in a closed container of $$5$$ litre volume at pressure $$P$$ and temperature $$27^oC$$. The gas is heated from $$27^oC$$ to $$327^oC$$. It shows positive deviation, i.e., Z > 1 at high pressure.

...view full instructions

Which of the following statements is wrong about the gas?

A
The gas is monoatomic

B
The gas is $$He$$

C
The gas is a rare gas or inert gas

D
The number of molecules of gas is $$3.01 \times 10^{24}$$

Solution

Given,
$$C_p=0.125 \cal/g, C_v = 0.075\ cal/g$$

Also as we know,
$$C_p-C_v = \dfrac {R}{M}$$

$$M=\dfrac {2}{0.125-0.075}=40$$ so molar mass of gas is 40.

Also, $$\dfrac {C_p}{C_v}=\dfrac {0.125}{0.075}=\dfrac {5}{3}=1.66$$.

Therefore, gas is monoatomic and also the molar mass of gas is 40 $$g/mol$$.

Moles of gas $$X$$ = $$5$$ moles $$= 5 \times 6.023 \times 10^{23} = 3.01 \times 10^{24}$$ molecules.

Thus, gas is $$Ar$$ which is an inert gas.
Question 6
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Figure shows the variation of the internal energy $$U$$ with density $$\rho$$ of one mole of an ideal monatomic gas for thermodynamic cycle ABCA. Here process AB is a part of rectangular hyperbola

A
process AB is isothermal & net work in cycle is done on gas

B
process AB is isothermal & net work in cycle is done by the gas

C
process AB is isobaric & net work in cycle is done on the gas

D
process AB is adiabatic & net work in cycle is done by gas

Solution

The internal energy is given by $$U = C_V T$$
A rectangular hyperbola in $$x$$ and $$y$$ axes is given by $$xy = \textrm{constant}$$.
Thus, the leg $$AB$$ in the cycle corresponds to $$\rho U = \textrm{constant}$$, i.e., $$\rho C_VT = \textrm{const}$$ ....(1)
We also know $$P =\rho RT$$ .....(2)
From (1) and (2), we get $$ P = \textrm{constant}$$
Thus, the process AB is "Isobaric".

In the graph, from A to B, the internal energy decreases. Thus, the temperature decreases.
Hence, the density increases, i.e., volume decreases.
Work done by the gas is $$W = P\Delta V$$
As, $$\Delta V$$ is negative, work done by the gas is negative. i.e., Work is done on the gas.
Hence correct answer is option C.
Question 7
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Keeping the number of moles, volume and pressure the same, which of the following are the same for all ideal gas?

A
rms speed of a molecule

B
density

C
temperature

D
average of magnitude of momentum

Solution

Ideal gas equation $$\rightarrow PV=nRT$$
Given that $$n_1V_1P$$ are same.
So, $$T$$ i.e, temperature is same for all ideal gas.
Hence, the answer is Temperature.
Question 8
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One mole of an ideal gas $$\left (C_{v,m}\, =\, \displaystyle \frac {5}{2} R \right )$$ at $$300$$ K and $$5$$ atm is expanded adiabatically to a final pressure of $$2$$ atm against a constant pressure of $$2$$ atm. Final temperature of the gas is:

A
$$270$$ K

B
$$273$$ K

C
$$248.5$$ K

D
$$200$$ K

Solution

For adiabatic expansion against a constant external pressure:

$$W = ‐P_{ex}\Delta V = C_v\Delta T$$,

$$P_{ex}(V_f – V_i) = C_v(T_f – T_i)$$

$$C_v T_i + P_{ex} V_i = C_v T_f + P_{ex} Vf$$

Also,

$$V_f= d\dfrac {nRT_f}{P_{ex}} , \quad \quad \quad V_i= \dfrac {nRT_i}{P_i}$$

$$C_v T_i + nR T_i \left ( \dfrac{P_{ex}}{P_i } \right) = C_v T_f + P_{ex} \left (\dfrac{nR T_f}{P_{ex}} \right ) = T_f [C_v + nR]$$

$$T_f = T_i \times \dfrac{ \left [ \dfrac{5}{2}n R + nR \left ( \dfrac{P_{ex} }{P_i } \right ) \right ]}{[\frac{5}{2}nR + nR]} = T_i \times \dfrac {\left [\frac{5}{2} + \left ( \dfrac{P_{ex}}{P_i } \right ) \right ]}{\left (\dfrac{5}{2} + 1 \right )}$$

$$T_f= 300 \times \dfrac {\left [ \dfrac{5}{2}+ \dfrac{2}{5} \right ]} {\left [\dfrac{7}{2} \right ]} = 300\times 0.828 = 248.5K$$

Hence, the correct option is $$C$$
Question 9
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A sample of a fluorocarbon was allowed to expand reversibly and adiabatically to twice its volume. In the expansion the temperature dropped from 298.15 to 248.44 K. Assume the gas behaves perfectly. Estimate the value of $$C_{V_1 m}$$.

A
$$C_{V_1m} = 31.6 Jk^{1} mol^{1}$$

B
$$C_{V_1m} = 3.16 Jk^{1} mol^{1}$$

C
$$C_{V_1m} = 316 Jk^{1} mol^{1}$$

D
None of these

Solution

Process reversibly adiabatic
$$T_1 = 298.15 K {\;} V_2 = 2V_1$$

$$T_2 = 248.44 K {\;} Pv^{\gamma}= K {\;} PV = nRT$$

$$\dfrac {T}{V}\cdot V^{\gamma} K T_1V_1^{\gamma1} = T_2V_2^{\gamma1}$$

$$\left (\dfrac {T_1}{T_2}\right )=\left (\dfrac {V_2}{V_1}\right )^{\gamma-1} {\;} \left (\dfrac {298.15}{248.44}\right )=2^{\gamma-1}$$

$$1.2=2^{\gamma-1} {;} log 1.2=log 2. (\gamma-1)$$

$$\gamma-1=\dfrac {log 1.2}{10g 2} {\;} \gamma-1=0.263$$

Now $$nC_v(T_2-T_1)=\dfrac {P_2V_2-P_1V_1}{\gamma-1}$$

$$C_{V_1m}=\left (\dfrac {R}{\gamma-1}\right )=\dfrac {nR(T_2-T_1)}{(\gamma-1)}$$

$$C_{V_1m}=\dfrac {8.314}{0.263} {\;} C_{V_1m}=31.61Jk^{1} mol^{1}$$
Question 10
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The quantity $$\dfrac {2U}{fkT}$$ represents (where $$U=$$ internal energy of gas)

A
mass of the gas

B
kinetic energy of the gas

C
number of moles of the gas

D
number of molecules in the gas

Solution

According to the law of Equipartition of energy to the molecule of a gas states that for each degree of freedom(f) for a molecule the associated energy is $$\dfrac{1}{2}KT$$
So, let $$U$$ be the total energy of all the molecules $$n$$ each having degree of freedom equal to $$1.$$
So, $$U=nf\dfrac{1}{2}KT$$
$$\Rightarrow \dfrac{2U}{fKT}=n$$
Hence, the answer is number of molecules in the gas.

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Kinetic Theory Test - 75

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Kinetic Theory Test - 75

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Question 1

$$T_f=\dfrac {9}{13}T_0$$

$$T_f=\dfrac {13}{9}T_0$$

$$T_f=\dfrac {5}{2}T_0$$

$$T_f=\dfrac {3}{2}T_0$$

Solution

Question 2

Mono, mono

Dia, dia

Mono, dia

Dia, Mono

Solution

Question 3

$$20.8 J$$

$$28.8 J$$

$$26.8 J$$

$$25.8 J$$

Solution

Question 4

$$\frac {4R}{2}$$

$$\frac {3R}{2}$$

$$\frac {5R}{2}$$

zero

Solution

Question 5

The gas is monoatomic

The gas is $$He$$

The gas is a rare gas or inert gas

The number of molecules of gas is $$3.01 \times 10^{24}$$

Solution

Question 6

process AB is isothermal & net work in cycle is done on gas

process AB is isothermal & net work in cycle is done by the gas

process AB is isobaric & net work in cycle is done on the gas

process AB is adiabatic & net work in cycle is done by gas

Solution

Question 7

rms speed of a molecule

density

temperature

average of magnitude of momentum

Solution

Question 8

$$270$$ K

$$273$$ K

$$248.5$$ K

$$200$$ K

Solution

Question 9

$$C_{V_1m} = 31.6 Jk^{1} mol^{1}$$

$$C_{V_1m} = 3.16 Jk^{1} mol^{1}$$

$$C_{V_1m} = 316 Jk^{1} mol^{1}$$

None of these

Solution

Question 10

mass of the gas

kinetic energy of the gas

number of moles of the gas

number of molecules in the gas

Solution

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