MCQ Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 1

An ideal gas goes through a reversible cycle

$a\rightarrow b\rightarrow c\rightarrow d$ has the V- T diagram shown below. Process

$d\rightarrow a$ and

$b\rightarrow c$ are adiabatic.
The corresponding P-V diagram for the process is ( all figure are schematic and not drawn to scale).

Explanation

In process

$a \rightarrow b$ and

$c \rightarrow d$ the volume is directly proportional to temperature of gas, hence the process is isobaric, however, the slope of curve is different in each case. using ideal gas equation

$PV=nRT \Rightarrow V=\dfrac{nR}{P} T$

slope

$m=\dfrac{nR}{P}$ for a high pressure the slope of the V-T curve will be less. In given diagram the slope of

$c \rightarrow d$ process is lower than

$a \rightarrow b$ process, hence process

$a \rightarrow b$ is at higher pressure, from which option B and D can be ruled out.

Now initial rate of change of volume w.r.t. temperature is very less from

$b \rightarrow c$ and

$d \rightarrow a$ as shown in V-T diagram.

hence slope of P-V curve should tend to

$\lim_{\Delta V\rightarrow 0}\dfrac{\partial P }{\partial v}=\infty$ as initial change in volume is nearly zero, which can be verified in option A

Hence correct answer is option A.

A leak proof cylinder of length

$1m$ , made of a metal which has low coefficient of expansion is floating vertically in water at

$0^oC$ such that its height above the water surface is

$20cm$ . When the temperature of water is increased to

$4^oC$ , the height of the cylinder above the water surface becomes

$21cm$ . The density of water at

$T = 4^oC$ , relative to the density at

$T = 0^oC$ is close to:

0%
$1.26$
0%
$1.03$
0%
$1.04$
0%
$1.01$

Explanation

Let

$A$ be the area of cross section of the cylinder.

$p_0$ be the density of water at

$0^oC$

$p_4$ be the density of water at

$4^oC$ .

Total length of the cylinder is

$100cm$

Given that the length of the cylinder above the surface of water is

$20cm$

$0^oC$ , the weight of the cylinder is balanced by the buoyant force.

$p_0A \times 0.8 \times g = w$ [Buoyant force

$= pA lg$ ] ...(i)

At 4^oC$$, thef weight of the cylinder is balanced by the buoyant force.

$p_4A\times 0.79\times g = w$ ...(ii)

Dividing (i) by (ii)

$\dfrac{p_0\times 0.8}{p_4 \times 0.79} = 1$ ,

$p_4 = \dfrac{p_0\times 0.8}{0.79} = 0.012p_0$

$\dfrac{p_4}{p_0} = 1.012 \simeq 1.01$

1477496_1697194_ans_4c304cb2f30a4ae2a62121d343119eff.png

Which of the following shows the correct relationship between the pressure

$'P'$ and density

$\rho$ of an ideal gas at constant temperature ?

Explanation

From ideal gas law:

$PV=nRT$

$P=\cfrac{MnRT}{MV}, \quad M:$ Molecular mass number

But

$\rho=Mn/V$

$P=\rho RT/M$

At constant temperature,

$P \propto \rho$

Hence, the correct graph is (d).

Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2:The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4:The ratio of their densities is

0%
1 : 4
0%
1 : 2
0%
6 : 9
0%
8 : 9

Explanation

$Sol$ : (D)

$PV =nRT =\dfrac{m }{M }RT$

$\Rightarrow PM =\rho RT$

$\dfrac{\rho _{1}}{\rho _{2}}=\dfrac{P _{1}M _{1}}{P _{2}M _{2}}=(\dfrac{P _{1}}{P _{2}})\times(\dfrac{M _{1}}{M _{2}})=\dfrac{4}{3}\times\dfrac{2}{3}=\dfrac{8}{9}$
Here

$\rho _{1}$ and

$\rho _{2}$ are the densities gases in the vessel containing the mixture.

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?

0%
mKT
0%
P/(kT)
0%
Pm/(kT)
0%
P/(kTV)

Explanation

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=

$\dfrac {\text{Total mass} }{\text volume}$

$\rho= \dfrac{Nm}{V}=\dfrac{Pm}{kT}$

$\textbf{Hence option C correct}$

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

0%
$1 : 2$
0%
$2 : 3$
0%
$2 : 1$
0%
$2 : 5$

Explanation

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,

$\dfrac {m_{H}}{m_{He}} = \dfrac {0.74\times 2}{0.88\times 4} = \dfrac {1.48}{3.52} = \dfrac {2}{5}$ .

A hot and a cold body are kept in vacuum separated from each other. Which of the following causes decrease in temperature of the hot body?

0%
Radiation
0%
Convection
0%
Conduction
0%
Temperature remains unchanged

$0.5$ moles of an ideal gas at constant temperature

$27^{\circ}C$ kept inside a cylinder of length

$L$ and cross-section area

$A$ closed by a massless piston.
The cylinder is attached with a conducting rod of length

$L$ , cross-section area

$(\dfrac{1}{9})m^{2}$ and thermal conductivity

$k$ , whose other end is maintained at

$0^{\circ}C$ . If piston is moved such that rate of heat flow through the conducing rod is constant then velocity of piston when it is at height

$\dfrac{L}{2}$ from the bottom of cylinder is:
[Neglect any kind of heat loss from system]

0%
$\left (\dfrac {k}{R}\right )m/sec$
0%
$\left (\dfrac {k}{10R}\right )m/sec$
0%
$\left (\dfrac {k}{100R}\right )m/sec$
0%
$\left (\dfrac {k}{1000R}\right )m/sec$

Explanation

$\dfrac {\triangle Q}{\triangle t} = \dfrac {\triangle W}{\triangle t} =$ work done per unit time

$= \dfrac {ka\theta}{L}$

$\dfrac {dW}{dt} = P \dfrac {dv}{dt} = k \dfrac {a\theta}{L}, P = \dfrac {nRT}{V}$

$\Rightarrow \dfrac {0.5R(300}{V} A. \dfrac {dl}{dt} = \dfrac {ka\theta}{L}$

$\Rightarrow \dfrac {0.5R(300)}{A.\dfrac {L}{2}}A.v = \dfrac {ka\theta}{L}$

$\Rightarrow v = \dfrac {ka}{R}\left (\dfrac {27}{300}\right ) = \dfrac {k}{100R}$ .

The difference between volume and pressure coefficients of an ideal gas is :

0%
$\dfrac{1}{273}$
0%
$273$
0%
$\dfrac{2}{273}$
0%
Zero

Explanation

Volume coefficient:

$\alpha$
Pressure coefficient:

$\beta$

$\alpha=\beta=\frac{1}{273}$

A sealed glass jar is full of water. When it is kept in a freezing mixture, it is broken because :

0%
water expands from $4^{o}C$ to $0^{o}C$
0%
ice expands while melting
0%
water expands due to freezing
0%
ice expands since its temperature falls below $0^{0}C$

Explanation

Anomalous expansion of water occurs when cooled from

$4^0$ C to

$0^0$ C.
Hence, the sealed glass jar will be broken.

The gas law

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

0%
isothermal change only
0%
adiabatic change only
0%
both isothermal & adiabatic
0%
neither isothermal nor adiabatic

Explanation

The given relation is always valid for ideal gas whether the process is adiabatic or isothermal because its the fundamental equation of ideal gas.

$P \times V=n \times R \times T$

List - I List - II

a. Temperature of water e.273 K

at the bottom of a lake

b. $\gamma _{r}$ of water is - ve f. 277 K

between

c. $\gamma _{r}$ of water is positive g. above 277 K

d. Temperature of water h. 273K and

just below ice layer in 277 K

a Lake

0%
a- g, b-h, c-e, d-f
0%
a-h, b-e, c-f, d-g
0%
a-e, b-f, c-g, d-h
0%
a-f, b-h, c-g, d-e

Explanation

Water has maximum density at

$4^{\circ}C=277K$ . Hence water at this temperature is settled at the bottom.

Below this temperature, liquid water has negative coefficient of expansion. Hence as ice is melted, the volume of water initially decreases.

Water at $4^{0}$ C is filled to the brim of 2 beakers. A boy performed an experiment A of heating and B of cooling. The observation was :

0%
water level in B went down and in A water flowed out
0%
water flowed out in both the cases
0%
water flowed out in B and water level went down in A
0%
water level went down in both cases

Explanation

Water has anomalous expansion from

$4^0$ C to

$0^0$ C.
Water expands when heated from

$4^0$ C onwards.
Since water expands in both the cases, it will overflow from both experiments A and B.

The P-T graph for the given mass of an ideal gas is shown in figure. Then the volume

0%
increases
0%
decreases
0%
remains constant
0%
data insufficient

A gas in an airtight container is heated from 25 $^{o}$ C to 90 $^{o}$ C. The density of gas will :

0%
increase slightly
0%
increase considerably
0%
remain the same
0%
decrease slightly

A sealed glass jar is full of water. When its temperature is decreased to $0^{o}C$ :

0%
The glass jar remains as it is with ice
0%
The glass jar remains as it is with water
0%
The glass jar contains half the amount of ice mixed with water
0%
The glass jar breaks due to the formation of ice

Explanation

From

$4^0$ C to

$0^0$ C water will expand. while ice is getting formed.
Hence, the glass jar will break.

The molar gas constant is the same for all gases because at the same temperature and pressure, equal volumes of gases have the same :

0%
number of molecules
0%
average potential energy
0%
ratio of specific heats
0%
density

Explanation

We know,

$R=\frac { PV }{ nT }$
assuming P,V and T to be constant.
Also we know, that equal volume of gases contain equal number of molecules, hence the value of R is also constant.

If the volume of the gas is to be increased by 4 times :

0%
temperature and pressure must be doubled
0%
at constant P, the temperature must be increased by 4 times
0%
at constant T, the pressure must be increased by 4 times
0%
it cannot be increased

Explanation

$PV=RT$
At constant pressure temp has to be increased by 4 times to increase volume by 4 times.

A gas at temperature 27 $^{0}$ C and pressure 30 atmospheres is allowed to expand to one atmospheric pressure. If the volume becomes 10 times its initial volumes, the final temperature becomes :

0%
100 $^{o}$ C
0%
373 K
0%
373 $^{o}$ C
0%
-173 $^{o}$ C

Explanation

$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \\ \dfrac { 30V }{ 300 } =\dfrac { 1\times 10V }{ { T }_{ 2 } } \\ { T }_{ 2 }=100K\\ { T }_{ 2 }=-173^0C$

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 :

0%
200K
0%
100K
0%
200 $^{0}$ C
0%
100 $^{0}$ C

The Universal gas constant may be expressed as :

a) 8.31 J/mole-K c) 2.00 J/mole-K

b) 8.31 cal/mole-K d) 2.00 cal/mole-K

0%
a,c
0%
a,d
0%
b,c
0%
b,d

The parameter that determine the physical state of gas are :

a) Pressure b) Volume

c) Number of moles d) Temperature

0%
a & b
0%
a,b & c
0%
a,b & d
0%
a,c & d

16 gm of $O_{2}$ gas and x gm of $H_{2}$ gas occupy the same volume at the same temperature and pressure. Then x is :

0%
1/2gm
0%
1gm
0%
8gm
0%
16 gm

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)

0%
0.001456
0%
0.01456
0%
0.1456
0%
1.1456

Explanation

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

Select the correct formula :

(where k=Boltzmann's constant, R= gas constant, n= moles, r = density, M= molecular weight, p= pressure, T= kelvin temperature, V= volume)

a) k=RN $_{av}$

b) $r=\dfrac{nM}{V}$

c) $\dfrac{p}{r}=\dfrac{RT}{M}$

d) R=kN $_{av}$

0%
a,b,c
0%
a,b,d
0%
b,c,d
0%
a,c,d

Explanation

(a)

$k= R/N_{av}$
(b) r is density it is clearly satisfying the relation. nM gives toatal mass and when divided by volume it gives density of the substance
(c)We know that , PV=nRT
PV=wRT/M
PV/w=RT/M
P/r=RT/M
(d)

$k=R/N_{av}$ therefore,

$R=kN_{av}$
hence (b) (c) (d) are correct
Hence option(C)

In the equation PV=constant, the numerical value of constant depends upon

a) temperature b) mass of the gas

c) system of units used d) nature of the gas

0%
a & b
0%
b & c
0%
c & d
0%
all

How many times is the weight of the air filling a room in winter ( $7^{0}C$ ) greater than weight in summer $(37^{0}C)$ ? Assume that pressure is constant.

0%
1.11
0%
0.90
0%
11.1
0%
9.0

Explanation

Ideal gas follows: PV=nRT

$\implies PM_{0}=dRT$ where

$d$ is the density of the gas

$\implies d\propto \dfrac{1}{T}$

$\implies$ weight of the gas

$\propto \dfrac{1}{T}$

$\implies \dfrac{W_{winter}}{W_{summer}}=\dfrac{273+37}{273+7}=1.11$

Two containers of equal volume containing the same gas at pressure $P_{1}$ and $P_{2}$ and absolute temperature $T_{1}$ and $T_{2}$ respectively were connected with narrow capillary tube. The gas reaches a common pressure P and a common temperature T. The ratio P/T is equal to :

0%
$\dfrac{P_{1}}{T_{1}} +\dfrac{P_{2}}{T_{2}}$
0%
$\dfrac{1}{2}\left ( \dfrac{P_{1}}{T_{1}} +\dfrac{P_{2}}{T_{2}}\right )$
0%
$\dfrac{P_{1}T_{2}+P_{2}T_{1}}{T_{1}+T_{2}}$
0%
$\dfrac{P_{1}T_{2}-P_{2}T_{1}}{T_{1}-T_{2}}$

Explanation

Since the mass remains constant before and after opening the tube, moles of the gases remain same.

Thus

$\dfrac{P_1V}{RT_1}+\dfrac{P_2V}{RT_2}=\dfrac{P(2V)}{RT}$

$\implies \dfrac{P}{T}=\dfrac{1}{2}(\dfrac{P_1}{T_1}+\dfrac{P_2}{T_2})$

The fastest mode of transmission of heat is Radiation.

0%
True
0%
False

When a body has the same temperature as that of its surroundings :

0%
It does not radiate heat
0%
It radiates same quantity of heat as it receives from the surroundings
0%
It radiates less quantity of heat as it receives from the surroundings
0%
It radiates more quantity of heat as it receives from the surroundings.

Which of the following methods of flow of heat is (are) based on gravitational attraction?

0%
Conduction
0%
Convection
0%
Radiation
0%
All of these

In a room containing air, heat can go from one place to another by :

0%
conduction
0%
convection
0%
radiation
0%
all of these

Ventilators are provided at the top of a room:

0%
To bring oxygen for breathing.
0%
So that sunlight may enter the room.
0%
To maintain convectional currents to keep the air fresh in the room.
0%
To provide an outlet for carbon dioxide.

A) Heat transfer by conduction and convection require a material medium.

B) Heat transfer by radiation doesn't require a material medium.

0%
both A and B are true
0%
A is true but B is false
0%
B is true but A is false
0%
both A and B are false

In the following, which statement is correct :
a) A hot body emits hot radiations only.
b) A cold body absorbs the radiations only.
c) A cold body emits cold radiations only.
d) All the bodies emit and absorb radiations simultaneously.

0%
a
0%
b
0%
c
0%
d

Newton's law of cooling is applied in laboratory for the determination of the

0%
Specific heat of gases
0%
Latent heat of gases
0%
Specific heat of liquids
0%
Latent heat of liquids

Explanation

When a body with mass 'm' and specific heat 's' cools, the rate of loss of heat is given by

$\dfrac{dQ}{dt}$ =

$-ms$

$\dfrac{d\theta}{dt}$

where,

$\theta$ is the temperature of the body

$t$ is time

Sun's heat reaches the earth by the process of:

0%
Conduction
0%
Convection
0%
Radiation
0%
Scattering

Radiation is passing through a transparent medium, then :

0%
the temperature of medium increases
0%
the temperature of medium decreases
0%
the temperature of medium does not alter
0%
the temperature of medium first increases and then becomes steady

Density of a material varies with temperature.

0%
True
0%
False

When the pressure of a gas is changed, then:

0%
The density of the gas also changes.
0%
The ratio of the pressure to the density remains unaffected.
0%
The velocity of the sound remains unaffected.
0%
The value of $\gamma$ changes.

Explanation

When the pressure of a gas is changed, then the ratio of the pressure to the density remains unaffected as

$\displaystyle = \frac{P}{\rho} = \frac{RT}{M}$

Keeping the number of moles, volume and temperature the same, which of the following are the same for all ideal gases?

0%
rms speed of molecules
0%
Density
0%
Pressure
0%
Average magnitude of momentum

Explanation

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

by keeping the number of moles, volume and temperature same, pressure will also remains same (

$\because$ R is constant)

Thermal radiations are similar to

0%
$\alpha$ rays
0%
X-rays
0%
cathode rays
0%
none of these

How is the velocity of thermal radiations (

$v$ ) related to the velocity of light (

$c$ )?

0%
$v < c$
0%
$v > c$
0%
$v = c$
0%
the relation depends upon the wavelength of the radiations

A hot and a cold body are kept in vacuum separated from each other. Which of the following causes decrease in temperature of the hot body?

0%
conduction
0%
radiations
0%
convection
0%
the temperature of the bodies remain unchanged

The density of water at 4

$^o$ C is

0%
minimum
0%
maximum
0%
$10^{3}Kg/m^{3}$
0%
$1Kg/m^{3}$

Explanation

Water has its maximum density of

$1g/cm^3$ or

$1000 kg/m^3$ at

$4^{\circ} C$ . When the temperature changes from either greater or less than 4 degrees, the density will become less than

$1 g/cm^3$ . Water has the maximum density of

$1 g/cm^3$ only when it is pure water.

In which part of electromagentic spectrum do the heat radiations lie?

0%
visible
0%
violet
0%
ultraviolet
0%
infrared

The material of prism used for obtaining spectrum of heat radiations is

0%
rock salt
0%
quartz
0%
flint glass
0%
crown glass

Due to anomalous behaviour of water, for the same mass, the volume of ice is more than that of water.

0%
True
0%
False

Explanation

Generally, the volume of substances increases on heating and decreases on cooling. But when water is heated from

$0^oC$ to

$4^oC$ , its volume decreases and increases only after

$4^oC$ . This unusual behaviour of water from

$0^oC$ to

$4^oC$ is known as anomalous expansion of water.

When water is heated from

$0^{\circ}C$ to

$10^{\circ}C$ , its volume:

0%
increases
0%
decreases
0%
does not changes
0%
first decreases and then increases

Explanation

Water exhibits an anomalous behaviour. It contracts on heating between

$0 ^{ o }{ C }$ and

$4 ^{ o }{ C }$ , so the volume of the water decreases. Above

$4 ^{ o }{ C }$ volume increases upon heating. Thus, it has highest density at

$4 ^{ o }{ C }$

Heat is transmitted from higher to lower temperature through actual mass motion of the molecules in

0%
conduction
0%
convection
0%
radiation
0%
None of these

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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