Thermal Properties of Matter Test

Result

Thermal Properties of Matter Test - 36

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The pressure of a gas in a container is 10$$^{-11}$$ pascal at 27$$^o$$C. The number of molecules per unit volume of vessel will be

A
$$6 \times 10^{23} cm^{-3}$$

B
$$2.68 \times 10^{19} cm^{-3}$$

C
$$2.5 \times 10^{6} cm^{-3}$$

D
$$2400 cm^{-3}$$

Solution

$$10^5 Pa \equiv 6.023 \times 10^{23}$$ (equivalence equation according to information given in question)
$$10^{-11} Pa \equiv 6.023 \times 10^7$$
$$22400 cc \Rightarrow 6.023 \times 10^7$$
$$1 cc \Rightarrow \displaystyle \frac{6.023 \times 10^7}{2400} =2400 cm^{-3} $$
Question 2
1 / -0

Equal masses of N$$_2$$ and O$$_2$$ gases are filled in vessels A and B. The volume of vessel B is double of A. The ratio of pressure in vessel A and B will be

A
16 : 7

B
16 : 14

C
32 : 7

D
32 : 28

Solution

Moles of $$N_2=n_A\dfrac{m}{28}$$

Moles of $$O_2=n_B\dfrac{m}{32}$$
Given that $$V_B=2V_A$$
$$P\propto \dfrac{n}{V}$$
$$\implies \dfrac{P_A}{P_B}=\dfrac{n_A}{n_B}\dfrac{V_B}{V_A}=\dfrac{16}{7}$$
Question 3
1 / -0

The dimension of universal gas constant R are

A
$$M^2 L^2 T^{-2}$$

B
$$ML^2 T^{-2} \theta^{-1}$$

C
$$M^2 L^2 T^{-2} \theta^{-2}$$

D
$$MLT^{-2} \theta^{-2}$$

Solution

using ideal gas equation
$$R=\dfrac { PV }{ nT } $$
dimensions of pressure $$M{ L }^{ -1 }{ T }^{ -2 }$$
dimensions of Volume $${L}^{3}$$
number of moles is dimensionless
dimensions of Temperature $$\theta$$
so, dimensions of R$$ \dfrac { (M{ L }^{ -1 }{ T }^{ -2 })({ L }^{ 3 }) }{ \theta } =M{ L }^{ 2 }{ T }^{ -2 }{ \theta }^{ -1 }$$
Hence,Option B is correct
Question 4
1 / -0

In outer space there are 10 molecules per cm$$^3$$ on an average and the temperature there is 3K. The average pressure of this light gas is

A
$$10^5 Nm^{-2}$$

B
$$5 \times 10^{-14} Nm^{-2}$$

C
$$0.4 \times 10^{-16} Nm^{-2}$$

D
$$4.14 \times 10^{-16} Nm^{-2}$$

Solution

$$P= \displaystyle \frac{nKT}{V} = \frac{10 \times 1.38 \times 10^{-23} \times 3}{10^{-6}}$$
$$= 4.14 \times 10^{-16} Nm^{-2}$$
Question 5
1 / -0

A cylinder contains 2kg of air at a pressure of 10$$^5$$Pa. If 2 kg more air is pumped into it, keeping the temperature constant, the pressure will be

A
$$10^{10} Pa$$

B
$$2 \times 10^5 Pa$$

C
$$10^5 Pa$$

D
$$0.5 \times 10^5 Pa$$

Solution

Since both volume and Temperature is constant,
$$\dfrac { P }{ n } =constant=\dfrac { P }{ m } \\ n=no.of\quad moles=\dfrac { mass\quad of\quad air(m) }{ Molecular\quad weight\quad of\quad water(M) } $$
$$\dfrac { { 10 }^{ 5 } }{ 2 } =\dfrac { P }{ (2+2) } \\ \Rightarrow P=2\times{ 10 }^{ 5 }Pa$$
Hence, Option B is correct
Question 6
1 / -0

One mole of a gas at a pressure 2 Pa and temperature 27$$^o$$C is heated till both pressure and volume are doubled. What is the temperature of the gas?

A
1200 K

B
900 K

C
600 K

D
300 K

Solution

We have the relation PV=nRT (ideal gas equation)

T=$$\dfrac{PV}{nR}$$

Given, that pressure and volume are doubled i.e, $$P_1=2P$$,$$V_1=2V$$.

$$T_1=\dfrac{P_1V_1}{nR}$$

$$=\dfrac{\left( 2P \times 2V\right)}{nR}$$

$$=4T$$

Given that T=27$$^o$$C i.e, 300K

$$T_1=4\times 300$$

$$=1200K$$
Question 7
1 / -0

Which of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium?

A
Kinetic energy

B
Momentum

C
Density

D
Speed

Solution

In case of ideal gases the average velocity is always zero. Hence the average momentum is zero.
Whereas average speed is non- zero so the kinetic energy also non-zero, as these two are scalar quantities.
Question 8
1 / -0

The pressure of a gas filled in a closed vessel increases by 0.4%. When temperature is increases by 1$$^o$$C the initial temperature of the gas is

A
250$$^o$$C

B
250 K

C
250$$^o$$F

D
2500$$^o$$C

Solution

Since volume is constant i.e number of moles is constant.
$$P_1$$ and $$T_1$$ are the initial pressure and temperature.
According to the question,
$${ P }_{ 2 }=1.004{ P }_{ 1 }\\ { T }_{ 2 }={ T }_{ 1 }+1$$
$$\therefore \dfrac { { P }_{ 1 } }{ { T }_{ 1 } } =\dfrac{P_2}{T_2}=\dfrac { (1+0.004){ P }_{ 1 } }{ { T }_{ 2 } } \\ \Rightarrow { T }_{ 2 }=1.004{ T }_{ 1 }=({ T }_{ 1 }+1)$$
$$0.004T_1 = 1$$
$$\\ \Rightarrow { T }_{ 1 }=250K$$
Hence, Option B is correct
Question 9
1 / -0

The temperature of water at the surface of a deep lake is $$2^{\circ}C$$. The temperature expected at the bottom is

A
$$0^{\circ}$$

B
$$2^{\circ}$$

C
$$4^{\circ}$$

D
$$-6^{\circ}$$

Solution

Water exhibits an anomalous behaviour; it contracts on heating between $$0 ^{ o }{ C }$$ and $$4 ^{ o }{ C }$$. When the surface of the lake cools down above $$4 ^{ o
}{ C }$$ the cool water at the surface flows to the bottom because of its greater density. But when the surface temperature reaches to $$2 ^{ o }{ C }$$, the water near the surface is less dense than the warmer water below. Hence the downword flow ceases, and the water near the surface remains colder than that at the bottom.
So the temperature expected at the bottom is $$4 ^{ o }{ C }$$
option (C) is the correct answer.
Question 10
1 / -0

A closed cubicle box made of perfectly insulating material has walls of thickness $$8$$ cm and the only way for heat to enter and leave the box is through the metal plugs $$A$$ and $$B$$ each of cross section $$12cm^2$$ and length $$8$$ cm fixed in the opposite walls of the box as shown in figure. Outer surface of $$A$$ is kept at $$100^oC$$ while outer surface of $$B$$ is at $$4^oC$$. The thermal conductivity of the material of the plugs is $$0.5cal/s/cm/\space^oC$$. A source of energy generating $$36$$cal/s is enclosed inside the box. The equillibrium temperature of the inner surface of the box (assuming that it is same at all points on the inner surface) is

A
$$38^oC$$

B
$$57^oC$$

C
$$76^oC$$

D
$$85^oC$$

Solution

The principle of conservation of energy is used to solve this question.
Heat from $$A$$ to source + $$36$$ = Heat from source to $$B$$.
$$\dfrac{KA}{L}(100-T)+36=\dfrac{KA}{L}(T-4)$$
$$T$$=Temp of source
$$K$$=0.5 cal/s/cm/degC
$$L=8 cm$$
$$A=12 cm^2$$
Substituting the above values, we get $$T= 76\ ^oC$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Thermal Properties of Matter Test - 36

Result

Thermal Properties of Matter Test - 36

Score

-

Rank

-

SHARING IS CARING

Question 1

$$6 \times 10^{23} cm^{-3}$$

$$2.68 \times 10^{19} cm^{-3}$$

$$2.5 \times 10^{6} cm^{-3}$$

$$2400 cm^{-3}$$

Solution

Question 2

16 : 7

16 : 14

32 : 7

32 : 28

Solution

Question 3

$$M^2 L^2 T^{-2}$$

$$ML^2 T^{-2} \theta^{-1}$$

$$M^2 L^2 T^{-2} \theta^{-2}$$

$$MLT^{-2} \theta^{-2}$$

Solution

Question 4

$$10^5 Nm^{-2}$$

$$5 \times 10^{-14} Nm^{-2}$$

$$0.4 \times 10^{-16} Nm^{-2}$$

$$4.14 \times 10^{-16} Nm^{-2}$$

Solution

Question 5

$$10^{10} Pa$$

$$2 \times 10^5 Pa$$

$$10^5 Pa$$

$$0.5 \times 10^5 Pa$$

Solution

Question 6

1200 K

900 K

600 K

300 K

Solution

Question 7

Kinetic energy

Momentum

Density

Speed

Solution

Question 8

250$$^o$$C

250 K

250$$^o$$F

2500$$^o$$C

Solution

Question 9

$$0^{\circ}$$

$$2^{\circ}$$

$$4^{\circ}$$

$$-6^{\circ}$$

Solution

Question 10

$$38^oC$$

$$57^oC$$

$$76^oC$$

$$85^oC$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.