Kinetic Theory Test

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Kinetic Theory Test - 39

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Question 1
1 / -0

Maxwell's laws of distribution of velocities shows that

A
the number of molecules with most probable velocity is very large

B
the number of molecules with most probable velocity is small

C
the number of molecules with most probable velocity is zero

D
the number of molecules with most probable velocity is exactly equal to 1

Solution

The form of Maxwell's velocity distribution function is gaussian type. So the maximum of this function represents the speed at which most of the molecules travel. This speed is known as most probable speed.
Question 2
1 / -0

A vessel contains air at a temperature of $$15^{0}C$$ and 60% R.H. What will be the R.H if it is heated to $$20^{0}C$$? (S.V.P at $$15^{0}C$$ is 12.67 & at $$20^{0}C$$ is 17.36mm of Hg respectively)

A
262%

B
26.2%

C
44.5%

D
46.2%

Solution

By definition,
$$\displaystyle\ \frac{V.P }{S.V.P }\times100$$ =$$\displaystyle\ \frac{60}{100}$$ (both VP and SVP are in room temperature)
$$= \displaystyle\ \frac{(V.P)_{15}}{12.67}$$
$$(V.P)_{15} = 7.6 mm$$ $$of$$ $$Hg$$
Now since unsaturated vapour obeys gas equation & mass of water remains constant so
$$P$$ = $$\left (\displaystyle\ \frac{nRT}{V} \right)$$
$$\Rightarrow$$ $$P\ \alpha \ T$$
$$\displaystyle\ \frac{(V.P)_{15}}{(V.P)_20}$$
$$= \displaystyle\ \frac{273+15}{273+20}$$
$$\Rightarrow$$ $$(RH)_{20}$$= 7.73 mm of Hg
So $$(R.H)_{20} =\displaystyle\ \frac{(V.P)_{20}}{(S.V.P)_{20}}\times 100$$
$$= 44.5$$%
Question 3
1 / -0

'N' molecules each of mass 'm', of gas A and '2N' molecules, each of mass '2m', of gas B are contained in the same vessel. Which is maintained at a temperature T. The mean square of the velocity of molecules of B type is denoted by $$V^{2}$$ and the mean square of the X component of the velocity of A type is denoted by $$\omega^{2}$$, $$\displaystyle\ \frac{\omega^{2}}{V^{2}}=$$

A
2

B
1

C
1/3

D
2/3

Solution

Total KE of A type of molecules =
$$\displaystyle\ \frac{3}{2} m\omega^{2}$$
Total KE of A type of molecule is
$$K.E_{A}$$ = $$\displaystyle\ \frac{1}{2}[(V_{r.ms})^{2}_{x}+(V_{rms})^{2}_{y}+(V_{rms})^{2}_{z}]$$
but $$(V_{rms})_{x}$$ = $$\omega$$
So $$(V_{rms})_{y}$$ = $$(V_{rms})_{z}$$ = $$\omega$$
Total KE of B type molecules
$$= \displaystyle\ \frac{1}{2}\times2mv^{2}$$ = $$m.v^{2}$$
Now, $$\displaystyle\ \frac{3}{2}\times m\omega^{2}$$ = $$m v^{2}$$
or $$(\omega^{2}/v^{2})$$ = $$\displaystyle\ \frac{2}{3}$$
Question 4
1 / -0

A vessel has 6g of hydrogen at pressure P and temperature 500K. A small hole is made in it so that hydrogen leaks out. How much hydrogen leaks out if the final pressure is $$P/2$$ and temperature falls to 300K?

A
2g

B
3g

C
4g

D
1g

Solution

$$PV$$ = $$\displaystyle\ \frac{m}{M}RT$$
Initally, $$PV$$ = $$\displaystyle\ \frac{6}{M}R\times500$$
Finally, $$\displaystyle\ \frac{P}{2}V$$ = $$\displaystyle\ \frac{(6-x)}{M}R\times300$$ ( if x g gas leaks out)
Hence, $$2 = \displaystyle\ \frac{6}{6-x}\times\displaystyle\ \frac{5}{3}$$
$$\therefore x = 1$$gram
Question 5
1 / -0

Graph of specific heat at constant volume for a monatomic gas is:

A

B

C

D

Solution

For an ideal gas specific heat $$C_v$$ is independent of temperature and for diatomic molecules $$C_v = \dfrac{3}{2}R$$ for all temperature conditions.

So the graph between $$C_v$$ and T will be a straight line parallel to the x-axis.
Question 6
1 / -0

A graph is plotted with PV/T on y-axis and mass of the gas along x-axis for different gases. The graph is

A
a straight line parallel to x-axis for all the gases

B
a straight line passing through origin with a slope having a constant value for all the gases

C
a straight line passing through origin with a slope having different values for different gases

D
a straight line parallel to y-axis for all the gases

Solution

$$\displaystyle\ \frac{PV}{T}$$ = nR = $$\left( \displaystyle\ \frac{m}{M}\right)R$$
Or, $$\displaystyle\ \frac{PV}{T} = \left (\displaystyle\ \frac{R}{M}\right)m$$
i.e., $$\displaystyle\ \frac{PV}{T}$$ versus on graph is straight line passing through origin with slope R/M. i.e., the slope depends on molecular mass of the gas M and is different for different gases
Question 7
1 / -0

A gas mixture consists of 2 moles of oxygen and 4 moles of Argon at temperature T. Neglecting all vibrational moles, the total internal energy of the system is

A
4RT

B
15RT

C
9RT

D
11RT

Solution

Internal energy of 2 moles of oxygen
$$U_{2}$$ = $$\mu\left(\displaystyle\ \frac{5}{2} RT\right)$$
$$=2. \displaystyle\ \frac{5}{2}RT$$ = $$5RT$$
Internal energy of 4 moles of Argon.
$$U_{AR} = \mu \left(\displaystyle\ \frac{3}{2}RT\right)$$
$$=4. \displaystyle\ \frac{3}{2}RT$$ = $$6RT$$
$$therefore$$ Total internal energy
$$U = U_{O_{2}}+ U_{AR}$$
$$=11RT$$
Question 8
1 / -0

If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by $$1^{0}C$$, the initial temperature must be

A
250K

B
$$250^{0}C$$

C
2500K

D
$$25^{0}C$$

Solution

We know that $$\displaystyle\ \frac{P_{1}}{T_{1}}$$ = constant for constant volume.
Let the initial pressure be $$P_1$$ and initial temperature be $$T_1$$.
Thus final pressure $$P_2 =1.004 P_1 $$ and final temperature $$T_2 = T_1 + 1$$.
Using $$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$$
So, $$\displaystyle\ \frac{P_{1}}{T_{1}}$$ = $$\displaystyle\ \frac{1.004 P_{1}}{T_{1}+1}$$
Or $$T_1 + 1 = 1.004 T_1$$
Or $$0.004 T_1 = 1$$
$$\Rightarrow T_1 =250K$$
Question 9
1 / -0

An ideal gas is found to obey an additional law $$VP^{2}$$ = constant. The gas is initially at temperature T and volume V. When it expands to a volume 2V, the temperature becomes :

A
$$\displaystyle\ \frac{T}{\sqrt{2}}$$

B
$$2T$$

C
$$\displaystyle\ \frac{2T}{\sqrt{2}}$$

D
$$4T$$

Solution

We are given an ideal gas which is following an additional law,
$$PV^2=$$ a constant ....................(1)

As the gas is ideal, the gas law is
$$PV=nRT$$

Taking the value of $$P$$ from the above equation, we get
$$P=\dfrac{nRT}{V}$$

Putting this value in equation (1), we get
$$\dfrac{nRT(V)^2}{V} =$$ constant
As $$n$$ and $$R$$ are constant, above equation becomes,
$$TV=$$ a constant .....................(2)

Now, the gas is expanding from initial volume $$V_1$$ to final volume $$V_2$$ by $$2$$ units, it becomes
$$V_2=2V_1$$

From equation (2), we get
$$T_1 V_1 = T_2 V_2$$
$$T_1 V_1 = T_2 (2V_1)$$
$$T_2=\dfrac{T_1}{2}$$
So the correct answer is $$\dfrac{T}{2}$$
Question 10
1 / -0

The heat capacity at constant volume of a sample of a monoatomic gas is $$35\ J/K$$. Find the number of moles.

A
$$12.81 \ \ mol $$

B
$$21.81 \ \ mol $$

C
$$4.81 \ \ mol $$

D
$$2.81 \ \ mol $$

Solution

For monoatomic gas, degrees of freedom is 3.

Since $${ C }_{ V }=\dfrac { f }{ 2 } nR$$

Hence, $$35=\dfrac { 3 }{ 2 } n(8.314)$$

$$n=\dfrac { 70 }{ 3\times 8.314 } =2.81mol$$

Answer is $$2.81mol.$$

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Re-Attempt Weekly Quiz Competition

Kinetic Theory Test - 39

Result

Kinetic Theory Test - 39

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SHARING IS CARING

Question 1

the number of molecules with most probable velocity is very large

the number of molecules with most probable velocity is small

the number of molecules with most probable velocity is zero

the number of molecules with most probable velocity is exactly equal to 1

Solution

Question 2

262%

26.2%

44.5%

46.2%

Solution

Question 3

2

1

1/3

2/3

Solution

Question 4

2g

3g

4g

1g

Solution

Question 5

Solution

Question 6

a straight line parallel to x-axis for all the gases

a straight line passing through origin with a slope having a constant value for all the gases

a straight line passing through origin with a slope having different values for different gases

a straight line parallel to y-axis for all the gases

Solution

Question 7

4RT

15RT

9RT

11RT

Solution

Question 8

250K

$$250^{0}C$$

2500K

$$25^{0}C$$

Solution

Question 9

$$\displaystyle\ \frac{T}{\sqrt{2}}$$

$$2T$$

$$\displaystyle\ \frac{2T}{\sqrt{2}}$$

$$4T$$

Solution

Question 10

$$12.81 \ \ mol $$

$$21.81 \ \ mol $$

$$4.81 \ \ mol $$

$$2.81 \ \ mol $$

Solution

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