Kinetic Theory Test

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Kinetic Theory Test - 77

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Weekly Quiz Competition

Question 1
1 / -0

The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be

A
$$PV=\Big(\dfrac{5}{32}\Big)RT$$

B
$$PV = 5 RT$$

C
$$PV = \Big(\dfrac{5}{2}\Big)RT$$

D
$$PV = \Big(\dfrac{5}{16}\Big)RT$$

Solution

Given,
$$m=5g$$
$$M=32g$$ (molecular mass of the oxygen molecule)
Number of moles, $$n=\dfrac{m}{M}$$
$$n=\dfrac{5}{32}$$
From the equation of gas,
$$PV=nRT$$
$$PV=(\dfrac{5}{32})RT$$
Question 2
1 / -0

The maximum high temperature molar heat capacity at constant volume to be expected for acetylene which is linear molecule is:

A
9 cal/deg-mol

B
12 cal/deg-mol

C
19 cal/deg-mol

D
14 cal/deg-mol

Solution
Question 3
1 / -0

Volume versus temperature graphs for a given mass of an ideal gas are shown in figure at two different values of constant pressure. What can be inferred about relation between $$P_1$$ and $$P_2$$?

A
$$P_1 > P_2$$

B
$$P_1 = P_2$$

C
$$P_1 < P_2$$

D
Data is insufficient

Solution

As pressure and quantity of gas in system are contant. So, by ideal gas equation $$PV=nRT$$
$$V\propto T$$ as $$n,R$$ and $$P$$ are constant
$$\cfrac{{V}_{1}}{{T}_{1}}=$$ costant or slope of graph is constant
$$V=\cfrac{nRT}{P}$$
$$\cfrac{dV}{dT}=\cfrac{nR}{P}$$ so $$\cfrac{dv}{dT}$$ increase when $$P$$ decreases
$$\cfrac{dV}{dT}\propto \cfrac{1}{P}$$
as slope of $${P}_{1}$$ is smaller than $${P}_{2}$$. Hence $${P}_{1}> {P}_{2}$$ verifies option (a)
Question 4
1 / -0

In the given figure conainer $$A$$ holds an ideal gas at a pressure of $$5.0 \times 10^{5}\ Pa$$ and a temperature of $$300\ K.$$ It is connected by the tin tube (and a closed valve) to container $$B$$, with four times the volume of $$A$$. Container $$B$$ hold same ideal gas at a pressure of $$1.0\ \times 10^{5}\ Pa$$ and a temperature of $$400\ K$$. The valve is opened to allow the pressure to equalize, but the temperature of each container kept constant at its initial value. The final pressure in the two containers will be close to :

A
$$2.0 \times 10^{5}\ Pa$$

B
$$2.5 \times 10^{5}\ Pa$$

C
$$3.0 \times 10^{5}\ Pa$$

D
$$3.5 \times 10^{5}\ Pa$$

Solution
Question 5
1 / -0

Liquid is filled in a vessel which is kept in a room with temperature $$20^\circ C$$. When the temperature of the liquid is $$80^\circ C$$, then it losses heat at the rate of 60 cal/sec.What will be the rate of loss of heat when the temperature of the liquid is $$40^\circ C$$.

A
180 cal/sec

B
40 cal/sec

C
30 cal/sec

D
20 cal/sec

Solution

We know rate of heat transfer
$$ q= h A\Delta T$$
or
$$q= C\Delta T$$
where C is constant

$$60= C.(80-20)$$
$$60= C\times 60$$
$$c=1$$

for $$q= 40^\circ$$
$$q=c (40-20)$$
$$=1\times 20$$
$$q=20 cal/sec$$
Question 6
1 / -0

The rms speed of an ideal diatomic gas at temperature T is v . when gas dissociates into atoms then its new rms speed becomes double The temperature at which the gas dissociated into atoms are:-

A
$$T$$

B
$$\sqrt{2}T$$

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

D
$$2T$$

Solution
Question 7
1 / -0

a container is filled with 20 moles of an ideal diatomic gas at an absolute temperature t when heat is supplied to gas temperature remains constant but 8 moles dissociate into the atom. heat energy given to gas is?

A
$$4\ RT$$

B
$$6\ RT$$

C
$$3\ RT$$

D
$$RT$$

Solution
Question 8
1 / -0

Root mean square speed of the molecules of ideal gas is $$v$$. If pressure is increased two times at constant temperature, the $$rms$$ speed will become:

A
$$2v$$

B
$$\dfrac{v}{2}$$

C
$$4v$$

D
$$v$$

Solution

$$v_{rms}=\sqrt{\dfrac{3RT}{M}}$$
As T is constant
Though pressure is increased $$v_{rms} $$ remains unchanged
Question 9
1 / -0

Two perfect gases having masses $$m_{1}$$ and $$m_{2}$$ at temperature $$T_{1}$$ and $$T_{2}$$ are mixed without any loses of internal kinetic energy of the molecules. The molecular weights of the gases are $$M_{1}$$ and $$M_{2}$$. What is the final temperature of the mixture?

A
$$\dfrac {m_{1}T_{1} + m_{2}T_{2}}{m_{1} + m_{2}}$$

B
$$\dfrac {M_{1}T_{1} + M_{2}T_{2}}{M_{1} + M_{2}}$$

C
$$\dfrac {\dfrac {m_{1}}{M_{1}} T_{1} + \dfrac {m_{2}}{M_{2}}T_{2}}{\left (\dfrac {m_{1}}{M_{1}} + \dfrac {m_{2}}{M_{2}}\right )}$$

D
$$\dfrac {\dfrac {M_{1}}{m_{1}} T_{1} + \dfrac {M_{2}}{m_{2}}T_{2}}{\dfrac {M_{1}}{m_{1}} + \dfrac {M_{2}}{m_{2}}}$$

Solution

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Question 10
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A particle of mass m is moving along the x-axis with speed v when it collides with a particle of mass $$2$$ M initially at rest. After the collision, the first particle has come to rest and the second particle has split into two equal mass pieces that are shown in the figure. Which of the following statements correctly describes the speeds of the piece? $$(\theta > 0)$$.

A
Each piece moves with speed v

B
Each piece moves with speed $$v/2$$

C
One of the pieces moves with speed $$v/2$$, the other with speed greater than $$v/2$$

D
Each piece moves with speed greater than $$v/2$$

Solution

Let $${ V }_{ A }^{ / }$$ and $${ V }_{ B }^{ / }$$ be the velocity of A and B after collision.
From the conservation of momentum,
$$\quad { M }_{ A }{ V }_{ A }+{ M }_{ B }{ V }_{ B }={ M }_{ A }{ V }_{ A }^{ / }+\quad { M }_{ B }^{ / }{ V }_{ B }^{ / }$$-------- (1)
But, $${ M }_{ B }=2m\quad $$ and $${ M }_{ A }=m$$
From (1), => mv+ 0= 0+ 2($$\frac { 1 }{ 2 } \times 2m$$)v'Cos$$\theta $$
=> v=2v'Cos$$\theta $$
v'= $$\frac { v }{ 2Cos\theta } =\frac { 1 }{ Cos\theta } (\frac { v }{ 2 } )$$
Hence each piece moves with speed greater than $$\dfrac{v}{2}$$