MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 4

For an isochoric process the temperature at which the pressure of a gas will be double that of its pressure at 270 $^{0}$ C is

0%
543 $^{0}$ C
0%
813 $^{0}$ C
0%
3 $^{0}$ C
0%
270 $^{0}$ C

Explanation

PV = nRT
Since n and R are constant, we get

$\frac{PV}{T}$ = constant
so,

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$
Here,

$T_{1} = 546K$

$P_{2} = 2P_{1}$

$V_{2} = V_{1}$

$T_{2} = ?$
Putting all the values, we get :

$T_{2}$ = 537

$^{\circ}$ = 813K

A car tyre has air at 1.5 atm at 300 K.If P increases to 1.75 atm with volume same, the temperature will be ____

0%
350 $^{0}$ C
0%
350K
0%
300 $^{0}$ C
0%
300K

Explanation

HINT - Use ideal gas equation PV = nRT for the the cases.

$\textbf{Step 1: Since n and R are constant, we get}$

Use ideal gas equation PV = nRT

here n is number of moles.

P is pressure, V is volume, R is gas constant and T is temperature.

$\dfrac{PV}{T}$ = constant
so,

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$
Here,

$P_{1} = 1.5atm$

$V_{1} = V_{2}$

$T_{1} = 300K$

$P_{2} = 1.75atm$

$T_{2} = ?$

$\textbf{Step 2: Putting all the values, we get : }$

$T_2 = \left( \dfrac{P_2 V_2}{P_1 V_1 } \right) T_1$

$T_2=\dfrac{1.75}{1.5}\times 300$

$=350\ K$

${\textbf{Correct option: B}}$

The volume of a gas is 5 litres at N.T.P. what will be its volume at 273 $^{0}$ C and at a pressure of four atmospheres

0%
5 litres
0%
2 litres
0%
4 litres
0%
2.5 litres

Explanation

from ideal gas equation,

$\frac{P_{1}V_{1}}{T_{1}}= \frac{P_{2}V_{2}}{T_{2}}$
Here,

$P_{1} = 1 atm$

$V_{1} = 5 L$

$T_{1} = 273 K$

$P_{2} = 4 atm$

$V_{2} = ?$

$T_{2} = 546 K$
putting all the values, we get

$V_{2}$ = 2.5 L
So, D is the correct answer.

27 $^{0}$ C and pressure of 30atm is allowed to expand to atmospheric pressure and volume 15 times larger. The final temperature of gas is

0%
123 $^{0}$ C
0%
-123 $^{0}$ C
0%
400K
0%
None

Explanation

from ideal gas equation, PV = nRT
So,

$T = \frac{PV}{nR}$
Since, n is constant, and R is also constant, we can say that T is directly proportional to the product of P and V

$T \alpha PV$
volume is made 15 times and pressure is reduced to

$\frac{1}{30}$ th of it's original value (initial pressure = 30 atm and final pressure = 1 atm)
So, temperature becomes 15 x

$\frac{1}{30}$ times = 0.5 times it's initial value
Initial temperature = 27

$^{\circ}$ C = 300K
So, final temperature = 0.5 x 300 K = 150 K = -123

$^{\circ}$ C
So, B is the correct answer.

A sample of an ideal gas occupies a volume V at pressure P and absolute temperature T. The mass of each molecule is m. The equation for density is

0%
$mKT$
0%
$\dfrac{P}{KT}$
0%
$\dfrac{P}{KTV}$
0%
$\dfrac{Pm}{KT}$

Explanation

let us derive the expression for the density of an ideal gas
density =

$\dfrac{Mass}{Volume} = \dfrac{w}{V}$
Now from ideal gas equation,

$PV = nRT$
And n =

$\dfrac{w}{M}$ where w is the mass of the gas and M is the molar mass of the gas
Putting this in ideal gas equation, we get

$PV = \dfrac{w}{M}RT$
Now substitute the value of V in the density expression to get

$d = \dfrac{PM}{RT}$
So, D is the correct answer.

At 20 $^{o}C$ temperature and 1atmosphere pressure if a gas has a volume of 293 ml .Its volume at NTP is

0%
$546$ cc
0%
$273$ cc
0%
$293$ cc
0%
$124$ cc

Explanation

from ideal gas equation,

$\dfrac{P_{1}V_{1}}{T_{1}}= \dfrac{P_{2}V_{2}}{T_{2}}$

Here,

$P_{1} = 1$ atm

$V_{1} = 293$ ml

$T_{1} = 293$ K

$P_{2} = 1$ atm

$V_{2} = ?$

$T_{2} = 273K$

putting all the values, we get

$V_{2}$ = 273 ml
Hence, option

$B$ is the correct answer.

The pressure of a gas is increased four times and its absolute temperature two times. The ratio of its final volume to its initial volume is

0%
1:2
0%
2:1
0%
1:1
0%
3:1

Explanation

From ideal gas equation, PV = nRT
So, this gives

$V = \frac{nRT}{P}$
so, V is proportional to absolute temperature and inversely proportional to the pressure of the gas
Now temperature is made four times and pressure is made 2 times
So, clearly the volume becomes half of it's original value.
So, A is the correct answer.

A cylinder contains gas at a pressure of 2.5 atm. Due to leakage, the pressure falls to 2 atm, after sometime. The percentage of the gas which is leaked out is

0%
40
0%
15
0%
20
0%
25

Explanation

$\frac { { P }_{ 1 } }{ { d }_{ 1 } } =\frac { { P }_{ 2 } }{ { d }_{ 2 } } \\ \frac { 2.5 }{ { d }_{ 1 } } =\frac { 2 }{ { d }_{ 2 } } \\ { d }_{ 2 }=0.8\times { d }_{ 1 }$
Volume escaped=

$\Delta d=\left( 1-0.8 \right) { d }_{ 1 }=0.2{ d }_{ 1 }\\ Percentage\quad decrease,\\ \frac { \Delta d }{ { d }_{ 1 } } \times 100=20$

The diagram shows the graphs of pressure vs density for a given mass of an ideal gas at two temperatures T $_{1}$ and T $_{2}$ :

0%
T $_{1}$ > T $_{2}$
0%
T $_{1}$ = T $_{2}$
0%
T $_{1}$ < T $_{2}$
0%
Any of the three are possible

Explanation

$PV= nRT$

$P \propto \frac{1}{V}$ is equivalent to

$P$

$\propto$

$d$ .
The slope of P-d curve is proportional to temp
From the diagram, the slopes are equal

$\therefore$ temp T is equal for both.

A given amount of gas is heated until both its pressure and volume are doubled. If initial temperature is 27 $^{0}$ C, its final temperature is

0%
300 K
0%
600 K
0%
1200 K
0%
900K

Explanation

from ideal gas equation, PV = nRT
So,

$T = \frac{PV}{nR}$
Since, n is constant, and R is also constant, we can say that T is directly proportional to the product of P and V

$T \alpha PV$
Since both volume and pressure are doubled, absolute temperature of the gas will become 4 times
Initial temperature = 27

$^{\circ}$ C = 300K
So, final temperature = 4 x 300K = 1200 K
So, C is the correct answer.

One litre of helium under a pressure of 2 atm and at 27 $^{0}$ C is heated until its pressure and volume are doubled. The final temperature attained by the gas is

0%
927 K
0%
927 $^{0}$ C
0%
1200 $^{0}$ C
0%
None

Explanation

from ideal gas equation, PV = nRT
So,

$T = \frac{PV}{nR}$
Since, n is constant, and R is also constant, we can say that T is directly proportional to the product of P and V

$T \alpha PV$
Since both volume and pressure are doubled, absolute temperature of the gas will become 4 times
Initial temperature = 27

$^{\circ}$ C = 300K
So, final temperature = 4 x 300K = 1200 K = 927

$^{\circ}$ C
So, B is the correct answer.

The volume of a gas at $27^o$ C and 2 atmospheric pressure is 2 litres. If the pressure is doubled and absolute temperature is made half, the new volume of gas is

0%
50 ml
0%
500 ml
0%
1000 ml
0%
2000 ml

Explanation

from ideal gas equation, PV = nRT
So, this gives

$V = \frac{nRT}{P}$
so, V is proportional to absolute temperature and inversely proportional to the pressure of the gas
Now temperature is made half and pressure is made 2 times
So, clearly the volume becomes one-fourth of it's original value.
So, B is the correct answer.

A closed vessesl contains 8 gms of oxygen and 7gm of Nitrogen. Total pressure at a certain temperature is 10 atm. When all the oxygen is removed from the system without chage in temperature then the pressure will be

0%
10 x 7/15atm
0%
10 x 8/15atm
0%
10 x 8/16 atm
0%
10 x 8/32 atm

Explanation

$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { { m }_{ 1 }{ M }_{ 2 } }{ { m }_{ 2 }{ M }_{ 1 } } \\ \\ \dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { 8\times 14 }{ 7\times 32 } \\ { P }_{ 1 }=10\times \dfrac { 8 }{ 16 }$

The volume occupied by 8 gm of oxygen at S.T.P. is

0%
11.2 lit
0%
22.4 lit
0%
2.8 lit
0%
5.6 lit

An air bubble rises from bottom to top of a liquid of density 1.5 g/cm $^{3}$ . If its volume is doubled, the depth of liquids is

0%
689 cm
0%
398 cm
0%
760 cm
0%
152 cm

Explanation

In this case,

$(P + \rho g h) V$

$=$

$2V P$
where,

$P$

$=$ atm pressure

$\rho$

$=$ density of the liquid

$h$

$=$ height of the lake

$V$

$=$ volume of the bubble
Solving we get,

$h$

$=$

$\dfrac{P}{\rho g}$
here,

$P$

$=$ 76 X 13.6 X 9.8

$\rho$

$=$ 1.5

$gm/cm^{3}$

$=$ 1500

$kg/m^{3}$
Putting these values we get,

$h$

$=$ 689 cm

If pressure of an ideal gas contains in a closed vessel is increased by 0.5% the increase in temperature is 20k the initial temperature of the gas is

0%
27 $^{0}$ C
0%
127 $^{0}$ C
0%
300 $^{0}$ C
0%
400 $^{0}$ C

Explanation

In this case,

$P_1 = P$ ,

$P_2= 1.05P$

${T_1} = T$ ,

$T_2 = T+20$
Putting these values in

$\dfrac{P_1}{T_1}= \dfrac{P_2} {T_2}$ we get, T = 400 or T =

$127^o$ C

Two closed vessels of the equal volume contain air at 105kPa at 300K and are connected through a narrow tube. If one of the vessels is now maintained at 300K and the other at 400K then the pressure becomes.

0%
120kPa
0%
105kPa
0%
150kPa
0%
300kPa

Explanation

In this case, let the initial pressure = P and final pressure = P'. Since (P/T) = constant, considering the system of two spheres of equal volume, we get,

$\dfrac{P}{300} + \dfrac{P}{300} = \dfrac{P'}{300} + \dfrac{P'}{400}$
here, P = 105k Pa
Solving we get,

$P' = 120 kPa$

At the top of a mountain a thermo meter read 7 $^{0}$ C and barometer reads 70 cm of Hg. At the bottom of the mountain the barometer reads 76cm of Hg and thermometer reads 27 $^{0}$ C. The density of air at the top of mountains is ______ times the density at the bottom.

0%
0.99
0%
0.9
0%
0.89
0%
0.95

Explanation

$\dfrac{P_1}{d_1T_1}=\dfrac{P_2}{d_2T_2}$ (Standard equation of governance)

Dirctly putting up the value,

$\therefore \dfrac{70}{d_1\times ( 7+273)} = \dfrac{76}{d_2\times(27+ 273)}$

$\dfrac{d_1}{d_2} = \cfrac{75}{76}=0.987$

At a given temperature and pressure 64 gm of Oxygen and X gm of H $_{2}$ occupy the same volume. Then x= ......gm

0%
1
0%
2
0%
3
0%
4

Explanation

Since they occupy same volume, the two gases should contain same number of molecules, i.e. they have same number of moles.
Now 1 mole of oxygen = 32gm.
Hence 64 gm = 2 mole of oxygen.

$\therefore$ number of moles of hydrogen is 2.
1 mole of hydrogen= 2gm
Hence X= 4gm

A one litre sphere and a two litre sphere are connected with a capillary tube of negligible volume. They contain an ideal gas at 27 $^{0}$ C at a pressure of 100cm of Hg. Keeping the temperature of one litre sphere constant at 27 $^{0}$ C, if temperature of two litre sphere is increased to 127 $^{0}$ C, then the final pressure is

0%
110 cm of Hg
0%
120 cm of Hg
0%
150 cm of Hg
0%
200 cm of Hg

Explanation

Moles of gas in 1L sphere before increasing temperature

$=\dfrac{P_0 V}{RT}$

Moles of gas in 2L sphere before increasing temperature

$=\dfrac{P_0 (2V)}{RT}$

Moles of gas in 1L sphere after increasing temperature

$=\dfrac{PV}{RT}$

Moles of gas in 2L sphere after increasing temperature

$=\dfrac{P (2V)}{R(T+100)}$

Since the total number of moles remain the same,

$\dfrac{P_0V}{RT}+\dfrac{2P_0 V}{RT}=\dfrac{PV}{RT}+\dfrac{2PV}{R(T+100)}$

Hence,

$P=P_0 (\dfrac{3(T+100)}{3T+100})$

$T=300K$ ,

$P_0=100cm Hg$

$\implies P=120cm Hg$

At the bottom of a lake where the temperature is $${7}^{0}\,$$ bubble of radius $1\, cm$ at the bottom rises to the surface. Where the temperature is $${27}^{0}\, C$$. The radius of the air bubble at the surface is

0%
$3^{1/3}$
0%
$4^{1/3}$
0%
$5^{1/3}$
0%
$6^{1/3}$

Explanation

We know that, by gas law

$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } }$

$\dfrac { 2.8{ \times V }_{ 1 } }{ 280 } =\dfrac { 1\times { V }_{ 2 } }{ 300 }$

${ V }_{ 2 }=3{ V }_{ 1 }$

$R\propto { V }^{ \dfrac { 1 }{ 3 } }$

$\therefore { R }_{ 2 }={ { 3 }^{ \dfrac { 1 }{ 3 } }R }_{ 1 }$

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

0%
T
0%
2T
0%
T $\sqrt{2}$
0%
T/2

Explanation

${ V }^{ 2 }P=Constant\\ { V }^{ 2 }\left( \dfrac { RT }{ V } \right) =Constant\\ VT=Constant\\ \therefore \dfrac { { V }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { V }_{ 2 } }{ { T }_{ 1 } } \\ \\ { T }_{ 2 }=\dfrac { { T }_{ 1 }{ V }_{ 1 } }{ 2{ V }_{ 1 } } \\ \\ { T }_{ 2 }=\dfrac { { T }_{ 1 } }{ 2 }$

Two gases A and B having same pressure P, volume V and temperature T are mixed. if the mixture has volume and temperature as V and T respectively the pressure of mixture is

0%
2P
0%
P
0%
P/2
0%
4P

A vessel is filled with an ideal gas at a pressure of 200 atm and is at a temperature of 27 $^{0}$ C. One half of the mass of the gas is removed from the vessel and the temperature of the remaining gas is increased to 87 $^{0}$ C. At this temperature,the pressure of the gas will be

0%
60 atm
0%
120 atm
0%
360 atm
0%
240 atm

Explanation

${ P }_{ 1 }{ V }_{ 1 }=\dfrac { { m }_{ 1 } }{ M } R{ T }_{ 1 }\\ { P }_{ 2 }{ V }_{ 2 }=\dfrac { { m }_{ 2 } }{ M } R{ T }_{ 2 }\\ { V }_{ 1 }={ V }_{ 2 },{ \quad 2m }_{ 2 }={ m }_{ 1 }\\ \dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { { m }_{ 1 }{ T }_{ 1 } }{ { { m }_{ 2 }T }_{ 2 } } \\ { P }_{ 2 }=\dfrac { 360\times 200 }{ 300\times 2 } \\ \\ { P }_{ 2 }=120atm$

If $\rho$ is the density, m is the mass of 1 molecule and K is the Boltzman constant for a gas then the pressure of the gas is:

0%
$P=\dfrac{\rho KT}{m}$
0%
$P=\dfrac{mKT}{\rho }$
0%
$P=\dfrac{\rho mT}{K}$
0%
$P=\rho KmT$

Explanation

$PV=\dfrac { m }{ M } RT\\ P\dfrac { m }{ \rho } =nkT\\ P=\dfrac { \rho nkT }{ m } ,\quad \\ if\quad n=1,\quad P=\dfrac { \rho kT }{ m }$

Its pressure is increased by $0.4$ %. The initial emperature of the gas is

0%
$250^{0}\ C$
0%
$100^{0}\ C$
0%
$-75^{0}\ C$
0%
$-23^{0}\ C$

Explanation

Since it is a closed system the volume remains constant.

Hence,

$\dfrac { { P }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 } }{ { T }_{ 2 } }$

${ P }_{ 2 }=1.004{ P }_{ 1 }\\ \therefore { T }_{ 2 }=1.004{ T }_{ 1 }\\ but\quad { T }_{ 1 }+1={ T }_{ 2 }\\ \therefore { T }_{ 1 }=250K\\ { T }_{ 1 }=-{ 23 }^{ \circ }$

A given amount of a gas is heated till the volume and pressure both increase by $2\%$ each the percentage change in temperature of the gas is equal to nearly

0%
$2\%$
0%
$3\%$
0%
$4\%$
0%
$1\%$

Explanation

$\dfrac { dP }{ P } +\dfrac { dV }{ V } =\dfrac { dT }{ T } \\ Now\quad dP=0.02P,dV=0.02V\\ 0.02+0.02=\dfrac { dT }{ T } \\ \dfrac { dT }{ T } \times 100=4%=percentage\quad change\quad in\quad temperature$

Two sample of Hydrogen and Oxygen of same mass possess same pressure and volume. The ratio of their temperature is

0%
$1:8$
0%
$1:16$
0%
$8:1$
0%
$16:1$

Explanation

${ P }_{ 1 }{ V }_{ 1 }=\dfrac { { m }_{ 1 } }{ { M }_{ 1 } } R{ T }_{ 1 }\\ { P }_{ 2 }{ V }_{ 2 }=\dfrac { { m }_{ 2 } }{ { M }_{ 2 } } R{ T }_{ 2 }\\ \dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 1 } }{ { M }_{ 2 } } \\ \dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { 1 }{ 16 } \left( { M }_{ 1 }=1gm,{ \ M }_{ 2 }=16gm \right)$

Two gases A&B having same pressure P, volume V and absolute temperature T are mixed. If the mixture has volume and temperature as V & T respectively then the pressure of mixture is

0%
2P
0%
P
0%
P/2
0%
4P

Explanation

Moles of gas A

$=\dfrac{PV}{RT}$

Moles of gas B

$=\dfrac{PV}{RT}$

Moles of the gas mixture

$=\dfrac{P'V}{RT}$

Since the number of moles(mass) remains same before and after mixing,

$\dfrac{PV}{RT}+\dfrac{PV}{RT}=\dfrac{P'V}{RT}$

$\implies P'=2P$

One litre of Helium gas at a pressure of 76 cm - Hg and temperature 27 $^{0}$ C is heated till its pressure and volume are doubled. The final temperature attained by the gas is

0%
900 $^{0}$ C
0%
927 $^{0}$ C
0%
627 $^{0}$ C
0%
327 $^{0}$ C

Explanation

$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \\ \dfrac { PV }{ 300 } =\dfrac { 2P\times 2V }{ { T }_{ 2 } } \\ { T }_{ 2 }=1200K\\ { T }_{ 2 }={ 927 }^{ \circ }C$

The mass of oxygen gas occupy a volume of 11.2 lit at a temperature 27 $^{0}$ C and a pressure of 76 cm of mercury in kilogram is (molecular weight of oxygen =32)

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

Explanation

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

A vessel is filled with an ideal gas at a pressure 27 $^{0}$ C mass of the gas is removed from the vessel & the temp. of the remaining gas is increased to 87 $^{0}$ C . Then the pressure of the gas in the vessel will be

0%
5 atm
0%
6 atm
0%
7 atm
0%
8 atm

Explanation

We have Ideal gas Equation:

$PV=nRT$

First the mass is halved, thus the moles of gas are halved.

$P\propto n$

So,

$P_{1}=\dfrac{P_0}{2}=5atm$

Now the temperature is increased.

$P\propto T$

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}=\dfrac{360}{300}=\dfrac{6}{5}$

$\implies P_2=6atm$

2 grams of monoatomic gas occupies a volume of 2 litres at a pressure of $8.3\times 10^{5}Pa$ and 127 $^{0}$ C . Find the molecular weight of the gas. (R=8.3 joule/mole/K)

0%
2 gram/mole
0%
16 gram/mole
0%
4 gram/mole
0%
32 gram/mole

Explanation

$PV=\dfrac { m }{ M } RT\\ 2\times { 10 }^{ -3 }\times 8.3\times { 10 }^{ 5 }=\dfrac { 2\times { 10 }^{ 3 } }{ M } \times 8.3\times 400\\ M=\dfrac { 2\times { 10 }^{ 3 }\times 8.3\times 400 }{ 2\times { 10 }^{ -3 }\times 8.3\times { 10 }^{ 5 } } \\ M=4{ gram }/{ mole }$

From a disc of radius

$R$ and mass

$M$ , a circular hole of diameter

$R$ , whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

0%
$\dfrac{9MR^{2}}{32}$
0%
$\dfrac{15MR^{2}}{32}$
0%
$\dfrac{13MR^{2}}{32}$
0%
$\dfrac{11MR^{2}}{32}$

Explanation

$PV=\dfrac { m }{ M } RT\\ PV=\dfrac { 8 }{ 32 } RT\\ PV=\dfrac { RT }{ 4 }$

A reciever has a pressure of 144cm of Hg. After two strokes with an exhaust pump, the pressure is 36cm of Hg. After another two strokes the pressure will be.

0%
9cm of Hg
0%
2.4cm of Hg
0%
6cm of Hg
0%
3cm of Hg

Explanation

${ P }_{ n }=P{ x }^{ n }\\ For\quad two\quad strokes,\\ 36=144{ x }^{ 2 }\\ { x }^{ 2 }=\frac { 1 }{ 4 } \\ \therefore x=\frac { 1 }{ 2 } \\ { { P }_{ n } }^{ \backprime }=144{ x }^{ 4 }\\ { { P }_{ n } }^{ \backprime }=144{ \left( \frac { 1 }{ 2 } \right) }^{ 4 }=9cm\quad of\quad Hg$

Three flasks of identical volume are filled separately by

(a) 1 gram of $H_{2}$

(b) 1 gram of $O_{2}$

They are immersed in a tank of water so that all of them attain same temperature. The pressures $P_{1},P_{2}$ and $P_{3}$ have the on.

0%
$P_{1} > P_{2} > P_{3}$
0%
$P_{1} = P_{2} = P_{3}$
0%
$P_{1} < P_{2} < P_{3}$
0%
$P_{3} > P_{1} > P_{2}$

Explanation

We know,

$PV=\dfrac { m }{ M } RT$
Given the flasks have same volume, and mass m=1gm for all the gases, the temperature attained is also same.
Hence the equation can be reduced to,
PM=Constant, where M is the molecular weight.
since

${ M }_{ 1 }{ <M }_{ 2 }{ <M }_{ 3 }\\ { \therefore P }_{ 1 }{ >P }_{ 2 }{ >P }_{ 3 }$

A closed copper vessel contains water equal to half of its volume when the temperature. Of the vessel is raised to 447 $^{0}$ c the pressure of steam in the vessel is (Treat steam as an ideal gas, R=8310 J/k / mole, density of water =1000 kg/m $^{3}$ molecular weight of water =18 )

0%
33.24 x 10 $^{7}$ pa
0%
16.62 x 10 $^{7}$ pa
0%
10.31 x 10 $^{7}$ pa
0%
8.31x 10 $^{7}$ pa

Explanation

Let V be the volume of the copper vessel.

Then considering steam as an ideal gas,

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

Mass of the water in the vessel=

$m=\rho (\dfrac{V}{2})$

$\implies PV=\dfrac{\rho V}{2M}RT$

$\implies P=\dfrac{\rho RT}{2M}=16.62\times 10^7Pa$

A sample of $O_{2}$ gas and a sample of hydrogen gas both have the same no. of moles, the same volume and same pressure. Assuming them to be perfect gases, the ratio of temperature of oxygen gas to the temperature of hydrogen gas is :

0%
1:1
0%
1:2
0%
2:1
0%
1:4

The density of a gas at N.T.P. is 1.5gm/lit. its density at a pressure of 152cm of Hg and temperature 27 $^{0}$ C

0%
$\frac{273}{100}gm/ltr$
0%
$\frac{150}{273}gm/ltr$
0%
$\frac{1}{273}gm/ltr$
0%
1.5 $gm/lit$

Explanation

Since volume is inversely proportional to density,

$\dfrac { { P }_{ 1 } }{ { { d }_{ 1 }T }_{ 1 } } =\dfrac { { P }_{ 2 } }{ { { d }_{ 2 }T }_{ 2 } } \\ \dfrac { 760 }{ 1.5\times 273 } =\dfrac { 152\times { 10 } }{ { d }_{ 2 }\times 300 } \\ { d }_{ 2 }=\dfrac { 273 }{ 100 }gm/lit$

An electric bulb of $250cc$ was sealed off at a pressure $10^{-3}$ mm of Hg and temperature $$27^{0}$$ C. The number of molecules present in the gas is

0%
$8.02 \times 10^{15}$
0%
$6.023 \times 10^{23}$
0%
$8.021 \times 10^{23}$
0%
$6 \times 10^{22}$

Explanation

$PV=nRT\\ n=\dfrac { PV }{ RT } \\ n=\dfrac { 250\times { 10 }^{ -6 }\times 133.33\times { 10 }^{ -3 } }{ 8.314\times 300 } \\ n=1.336\times { 10 }^{ -8 }\\ N=n\times { N }_{ A }\\ N=1.336\times { 10 }^{ -8 }\times 6.022\times { 10 }^{ 23 }\\ N=8.02\times { 10 }^{ 15 }$

Two identical containers connected by a fine capillary tube contain air at N.T.P. if one of those containers is immersed in pure water, boilling under normal pressure then new pressure is

0%
76 cm of Hg
0%
152 cm of Hg
0%
57 cm of Hg
0%
87.76 cm of Hg

Explanation

Since the number of moles of gas in the system remains constant:

Moles in first container earlier+Moles in second container earlier=Moles in first container later+Moles in second container later

$\implies \dfrac{P_0V}{RT_{0}}+\dfrac{P_0V}{RT_0}=\dfrac{P_1V}{RT_{0}}+\dfrac{P_1V}{RT_1}$

$T_0=273\ K,T_1=373\ K,P_0=76\ cm\ of\ Hg$

$\implies P_1=87.76\ cm\ of\ Hg$

A flask is filled with 13 gm of an ideal gas at 27 $^{0}$ C 52 $^{0}$ C mass of the gas that has to be released to maintain the temperature of the gas in the flask 52 $^{0}$ C and the pressure remaining the same is

0%
2.5 gm
0%
2.0 g
0%
1.5 g
0%
1.0 g

Explanation

Since P, V , M and R are constants,

${ m }_{ 1 }{ T }_{ 1 }={ m }_{ 2 }{ T }_{ 2 }\\ 13\times 300={ m }_{ 2 }\times 325\\ { m }_{ 2 }=12gm$
Hence mass to be removed = 13-12= 1gm

At what temperature will the hydrogen molecules escape from Earth's surface?

0%
10 $^{4}$ K
0%
10 $^{3}$ K
0%
10 $^{2}$ K
0%
10 $^{1}$ K

Explanation

We have the escape velocity as

$v_e=11.2 km/s$ , boltzmann constant as

$1.38\times 10^{-23} m^2kgs^{-2}K^{-1}$ and molecular mass of hydrogen as

$2\times 1.66\times 10^{-27} amu$
We have the formula as

$\dfrac{3}{2}kT=\dfrac{1}{2}mv_e^2$
or

$T=\dfrac{mv_e^2}{3k}=\dfrac{2\times 1.66\times 10^{-27}\times 11.2^2\times 10^6}{3\times 1.38\times 10^{-23}}=10^4 K$

During an experiment an ideal gas is found to obey an additional gas law VT =constant. The gas is initially at temperature T and pressure P. When it is heated to the temperature 2T, the resulting pressure is

0%
2P
0%
P/2
0%
4P
0%
P/4

Explanation

The gas follows VT=constant. But it also follows PV=nRT.

Thus

$(\dfrac{nRT}{P})T=$ constant

$\implies \dfrac{T^2}{P}=$ constant

Therefore

$\dfrac{T^2}{P}=\dfrac{(2T)^2}{P'}$

$\implies P'=4P$

A cycle tube has volume

$2000 cm^3$ . Initially the tube is filled to 3/4th of its volume by air at pressure of

$105 N/m^{2}$ . It is to be inflated to a pressure of

$6 \times 10^{5} N/m^{2}$ under isothermal conditions. The number of strokes of pump, which gives

$500 cm^3$ air in each stroke, to inflate the tube is

0%
21
0%
12
0%
42
0%
11

Explanation

In this problem, let us conserve total number of moles before and after,

$(6X{ 10 }^{ 5 })(2000)=(105)(1500)+(101325)(500)n$

Where n is total number of strokes of pump.

Solving this equation given

$n\approx21$ .

If the number of degrees of freedom of a gas is f, then the value of $\gamma$ will be

0%
f
0%
2f
0%
f/2
0%
1+2/f

Explanation

$\gamma= \dfrac{2+f}{f}=1 + \dfrac{2}{f}$ where

$\gamma= \dfrac{C_p}{C_v}$ and f is the no of degrees of freedom.

The total kinetic energy of $8$ litres of helium molecules at $5$ atmosphere pressure will be

0%
$607 Joules$
0%
$6078 Joules$
0%
$607 ergs$
0%
$6078 ergs$

If the pressure of a gas contained in a closed vessel increases by x% when heated by 1 $^{0}$ C , its initial temperature is

0%
$\frac{100}{x}$ Kelvin
0%
$\frac{100}{x}$ Celsius
0%
$\frac{x+100}{x}$ Kelvin
0%
$\frac{100-x}{x}$ Celsius

Explanation

Let the initial pressure and temperature be P and T.

From Ideal Gas Law,

$P\propto T$

$\implies \dfrac{(1+\dfrac{x}{100})P}{P}=\dfrac{T+1}{T}$

$\implies T=\dfrac{100}{x}K$

The average kinetic energy of hydrogen molecule at NTP will be

0%
0.4 x 10 $^{-20}$ Joule /molecule
0%
1.56 x 10 $^{-20}$ Joule / molecule
0%
zero
0%
5.6 x 10 $^{-20}$ Joule /molecule

Explanation

Average kinetic energy:

$KE_{avg}= \frac{3}{2}kT$

$=\dfrac{3}{2}\times 1.38 \times 10^{-23} \times 293$

$=0.40 \times 10^{-20}J$
where k is Boltzman's constant and

$T=(20 + 273)^{\circ}K= 293K$

27 $^{0}$ C is 84cm of Hg. If 25% of the gas is now introduced into the same vessel at the same temperature, the final pressure of the gas will be in cm of Hg

0%
105
0%
100
0%
95
0%
90

Explanation

Pressure=

$h\rho g=\dfrac{nRT}{V}$

$\implies h\propto n\propto m$

Therefore

$\dfrac{h_2}{h_1}=\dfrac{1.25m_{0}}{m_0}$

$\implies h_2=1.25h_1=105cm$

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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