An ideal gas has volume $$V$$ and pressure $$P$$. The total translational kinetic energy of all the molecule of the gas is.

greater than PV if the gas is diatomic

PV only if the gas is monoatomic

PV in all cases antainar is evacuated to leave just one

MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 12

2 moles of an ideal monoatomic gas at temperature

T_{0}

$T_0$ is mixed wth 4 moles of another ideal monoatomic gas at temperature

2 T_{0}

$2T_0$ then the temperature of the mixture is:

0%
$\frac{5}{3} T_0$
0%
$\frac{3}{2} T_0$
0%
$\frac{4}{3} T_0$
0%
$\frac{5}{4} T_0$

A sample of gas is at

0^{\circ} C

$0 ^ { \circ } \mathrm { C }$ .The temperature at which its rms speed of the molecules will be doubled is

0%
$103 ^ { \circ } \mathrm { C }$
0%
$273 ^ { \circ } \mathrm { C }$
0%
$723 ^ { \circ } \mathrm { C }$
0%
$819 ^ { \circ } \mathrm { C }$

The translational kinetic energy of molecules of one mole of a mono atomic gas is

U = \frac{3 N K T}{2}

$U=\dfrac{3NKT}2$ . The value of atomic specific heat of gas under constant pressure will be

0%

$\dfrac32 R$
0%
$\dfrac52 R$
0%
$\dfrac72 R$
0%
$\dfrac92 R$

'n' moles of an ideal gas undergoes a process A

\to

$\rightarrow$ B as shown in figure. The maximum temperature of the gas during the process will be :

0%
$\dfrac{9P_0V_0}{nR}$
0%
$\dfrac{9P_0V_0}{4nR}$
0%
$\dfrac{3P_0V_0}{2nR}$
0%
$\dfrac{9P_0V_0}{2nR}$

Explanation

\begin{array}{l} F r o m t h e f i g u r e \\ E q u a t i o n o f l i n e A B \\ y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \\ P - P_{0} = \frac{2 P_{0} - p_{0}}{V_{0} 2 V_{0}} (V - 2 V_{0}) \\ = \frac{- P_{0}}{V_{0}} (V - 2 V_{0}) \\ P = \frac{- P_{0}}{V_{0}} V + 3 P_{0} \\ P V = - \frac{P_{0}}{V_{0}} V + 3 P_{0} V \\ n R T = - \frac{P_{0}}{V_{0}} V + 3 P_{0} V \\ T = \frac{1}{n R} (- \frac{P_{0}}{V_{0}} V + 3 P_{0} V) \\ \frac{d T}{d V} = 0 (f o r max i m u m t e m p e r a t u r e) \\ - \frac{P_{0}}{V_{0}} 2 V + 3 P_{0} = 0 \\ - \frac{P_{0}}{V_{0}} 2 V = - 3 P_{0} \\ V = \frac{3}{2} V_{0} (c o n d i t i o n f o r max i m u m t e m p e r a t u r e) \\ T_{max} = \frac{1}{n R} (\frac{P_{0}}{V_{0}} \times \frac{9}{4} V_{0}^{2} + 3 P_{0} \times \frac{3}{2} V_{0}) \\ = \frac{1}{n R} (- \frac{9}{4} P_{0} V_{0} + \frac{9}{2} P_{0} V_{0}) \\ = \frac{9}{4} \frac{P_{0} V_{0}}{n R} \end{array}

$\begin{array}{l} From\, the\, figure \\ Equation\, of\, line\, AB \\ y-{ y_{ 1 } }=\dfrac { { { y_{ 2 } }-{ y_{ 1 } } } }{ { { x_{ 2 } }-{ x_{ 1 } } } } \left( { x-{ x_{ 1 } } } \right) \\ P-{ P_{ 0 } }=\dfrac { { 2{ P_{ 0 } }-{ p_{ 0 } } } }{ { { V_{ 0 } }2{ V_{ 0 } } } } \left( { V-2{ V_{ 0 } } } \right) \\ =\dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } \left( { V-2{ V_{ 0 } } } \right) \\ P=\dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } } \\ PV=-\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } }V \\ nRT=-\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } }V \\ T=\dfrac { 1 }{ { nR } } \left( { -\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } }V } \right) \\ \dfrac { { dT } }{ { dV } } =0\, \, \left( { for\, \max imum\, temperature } \right) \\ -\frac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } 2V+3{ P_{ 0 } }=0 \\ -\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } 2V=-3{ P_{ 0 } } \\ V=\dfrac { 3 }{ 2 } { V_{ 0 } }\, \, \left( { condition\, for\, \max imum\, \, temperature } \right) \\ { T_{ \max } }=\dfrac { 1 }{ { nR } } \left( { \dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } \times \dfrac { 9 }{ 4 } V_{ 0 }^{ 2 }+3{ P_{ 0 } }\times \frac { 3 }{ 2 } { V_{ 0 } } } \right) \\ =\frac { 1 }{ { nR } } \left( { -\dfrac { 9 }{ 4 } { P_{ 0 } }{ V_{ 0 } }+\dfrac { 9 }{ 2 } { P_{ 0 } }{ V_{ 0 } } } \right) \\ =\dfrac { 9 }{ 4 } \dfrac { { { P_{ 0 } }{ V_{ 0 } } } }{ { nR } } \end{array}$

Hence, the option

B

$B$ is the correct answer.

A circular disc of mass

m

$m$ and radius

r

$r$ is rolling about its axis with a constant speed

v

$v$ . Its kinetic energy is

0%
$\cfrac{1}{4}mv^2$
0%
$\cfrac{1}{2}mv^2$
0%
$\cfrac{3}{4}mv^2$
0%
$mv^2$

An ideal gas is initially at

P_{l}, V_{1}

$P_{l}, V_{1}$ is expanded to

P_{2}, V_{2}

$P_{2}, V_{2}$ and then compressed adiabatically to the same volume

V_{l}

$V_{l}$ and pressure

P_{3}

$P_{3}$ .If W is the net work done by the gas in complete process which of the

0%
$W>0 ; P_{3}>P_{1}$
0%
$W<0 ; P_{3}>P_{1}$
0%
$W<0;{P_3}$
0%
$P_{2}>P_{1}$ ; $W<0$

Explanation

Hence, option

(D)

$(D)$ is correct answer.
1380775_1500553_ans_e7eac230606548cabd9bb4a5018acc59.jpg

When work done by a system was 10 J, the increase in the internal energy of the system was 30 J. The heat 'q' supplied to the system was :

0%
10 J
0%
+20 J
0%
40 J
0%
-20 J

The kinetic energy of molecular translation per unit volume of a gas at 2 atmosphere pressure will be nearly

0%
$3\times { 10 }^{ 5 }J$
0%
$3\times { 10 }^{ 6 }J$
0%
$3\times { 10 }^{ 7 }J$
0%
$3\times { 10 }^{ 8 }J$

When a certain amount of ethylene was burnt

6226

$6226$ kJ heat was evolved. If heat of combustion of ethylene is

1411

$1411$ kJ/mol the volume of

O_{2}

$O_2$ ( at NTP) that entered into the reaction is

0%
$296.5$ mL
0%
$296.5$ litre
0%
$6226 \times 22.4$ litre
0%
$22.4$ litre

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping form the Earth's atmosphere? (Given : mass of oxygen molecule

(m) = 2.76 \times 10^{- 26}

$\left( m \right) =2.76\times { 10 }^{ -26 }$ kg, Boltzmann constant

(K_{B} = 2.76 \times 10^{- 23} {J K}^{- 1})

$\left( { K }_{ B }=2.76\times { 10 }^{ -23 }{ JK }^{ -1 } \right)$

0%
$2.508\times { 10 }^{ 4 }K$
0%
$8.360\times { 10 }^{ 4 }K$
0%
$5.016\times { 10 }^{ 4 }K$
0%
$1.254\times { 10 }^{ 4 }K$

Explanation

At the minimum velocity with which the body must be projected vertically upwards so that it could escape from the earth's atmosphere is its escape velocity

V_{e}

$V_{e}$

v_{l} = \sqrt{2 g R}

$v_{l}=\sqrt{2gR}$

Substituting the value of g

(9.8 m s^{- 2})

$(9.8 ms^{-2})$ and radius of earth

(R = 6.4 \times 10^{6} m)

$(R=6.4\times 10^{6}m)$ We get

v_{e} = \sqrt{2 \times 9.8 \times 64 \times 10^{6}}

$v_{e}=\sqrt{2\times9.8\times 64\times 10^{6}}$

≅ 11.2 k m s^{- 1} = 11200 m s^{- 1}

$\cong 11.2kms^{-1}=11200ms^{-1}$

Let the temperature of molecule be

T

$T$ when it attains

v_{e}

$v_{e}$ .

According to the question,

v_{r m s} = v_{e}

$v_{rms}=v_{e}$

where

v_{r m s}

$v_{rms}$ is the

r m s

$rms$ speed of the oxygen molecule,

\Rightarrow \sqrt{\frac{3 k_{B} T}{m o_{2}}} = 11.2 \times 10^{3}

$\Rightarrow \sqrt{\dfrac{3k_{B}T}{mo_{2}}}=11.2\times 10^{3}$

T = \frac{(11.2 \times 10^{3})^{2} (m o_{2})}{(3 k_{B})}

$T=\dfrac{(11.2\times 10^{3})^{2}(mo_{2})}{(3k_B)}$

Substituting the given values i.e

R_{B} = 1.38 \times 10^{- 23} J K^{- 1}

$R_{B}=1.38\times 10^{-23} JK^{-1}$ and

m o_{2} = m = 2.76 \times 10^{- 26} k g

$mo_2=m=2.76\times 10^{-26}kg$

We get,

T = \frac{(11.2 \times 10^{3})^{2} (2.76 \times 10^{- 26})}{(3 \times 1.38 \times 10^{- 23})}

$T=\dfrac{(11.2\times 10^{3})^{2}(2.76\times 10^{-26})}{(3\times 1.38\times 10^{-23})}$

= 8.3626 \times 10^{4} k

$=8.3626\times 10^{4}k$

A quantity of heat 'Q' is suppplied to a monotomic ideal gas which ecpands at constant pressure. The fractionm of heat that goes into workdone by the gas is

0%
$2/5$
0%
$3/5$
0%
$2/3$
0%
$1$

Four litre of

{C O}_{2}

${ CO }_{ 2 }$ is kept at

27^{\circ} C

$27^\circ C$ . What will be the volume if the temperature is lowered to 150 K at the same pressure.

0%
3 litre
0%
2 litre
0%
5 litre
0%
4 litre

How much work to be done in decreasing the volume of an ideal gas by an amount of 2.4 x

10^{- 4} m^{3}

$10^{-4} m^3$ at constant normal pressure of 1 x

10^{5} N / m^{2}

$10^5 N/m^2$ ?

0%
28 joule
0%
27 joule
0%
24 joule
0%
25 joule

One mole of an ideal monoatomic gas is heated at a constant pressure of one atmosphere from

0^{o} C

${0}^{o}C$ to

100^{o} C

${100}^{o}C$ . Then the change in the internal energy is

0%
$6.56$ joules
0%
$8.32\times{10}^{2}$ joules
0%
$12.48\times{10}^{2}$ joules
0%
$20.80$ joules

A monoatomic ideal gas undergoes a process given by

2 d U + 3 d W = 0

$2dU+3dW=0$ then the process is:

0%
isobaric
0%
adiabatic
0%
isothermal
0%
none of these

An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time between two successive collisions is

6 \times 10^{- 8} s

$6\times { 10 }^{ -8 }s$ . If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to:

0%
$4\times { 10 }^{ -8 }s$
0%
$3\times { 10 }^{ -6 }s$
0%
$2\times { 10 }^{ -7 }s$
0%
$0.5\times { 10 }^{ -8 }s$

The pressure of a given mass of gas is 100 cm at

127^{o} C

${ 127 }^{ o }C$ .The pressure of the same mass of gas having the same volume at

27^{o} C

${ 27 }^{ o }C$ is

0%
75 cm
0%
76 cm
0%
84 cm
0%
64 cm

What amount of heat must be supplied to 35 g of oxygen at room temperature to raise its temperature by

80^{\circ} C

${ 80 }^{ \circ }C$ at constant volume (molecular mass of oxygen is 32 and R = 8.3 J

{m o I}^{- 1} K^{- 1}

${ moI }^{ -1 }{ K }^{ -1 }$ )

0%
1.52 kJ
0%
3.23 kJ
0%
1.81 kJ
0%
1.62 kJ

A fixed mass of gas is taken through a process

A \to B \to C \to A

$A\rightarrow B\rightarrow C\rightarrow A$ . Here

A \to B

$A\rightarrow B$ is isobaric

B \to C

$B\rightarrow C$ is adiabatic and

C \to A

$C\rightarrow A$ is isothermal

The pressure at

C

$C$ is given by

(γ = 1.5)

$(\gamma=1.5)$

0%
$\frac{10^5}{64}\,N/m^2$
0%
$\frac{10^5}{32}\,N/m^2$
0%
zero
0%
${10}^5\,N/m^2$

An ideal diatomic gas is heated at constant volume, the supplied being Q. The increase in rotational kinetic energy is

0%
Q
0%
Q/2
0%
3Q/5
0%
2Q/5

One gram mole of helium at

60 C

$60C$ is mixed with two gram moles of argon at

30 C

$30C$ . The temperature of the mixture is

0%
45 $C$
0%
40 $C$
0%
38 $C$
0%
35 $C$

An ideal gas is heated at constant volume until its pressure doubles.Which one of the following statements is correct?

0%
The mean speed of the molecules doubles.
0%
Root mean square speed of the molecules doubles.
0%
Mean square speed of the molecules doubles.
0%
Mean square speed of the molecules remains uncharged.

According to kinetic theory of gases kinetic energy of gas molecules:

0%
inversely proportional to absolute temperature
0%
inversely proportional to square of temperature
0%
directly proportional to square of temperature
0%
directly proportional to absolute temperature

A flask breaks when pressure inside is 1.5 atmospheres. To what temperature can it be heated without breaking it if the initial pressure is atmospheric pressure at

0 C

$0C$

0%
$127F$
0%
$400.5C$
0%
$136.5C$
0%
$135.5K$

19 gms of oxygen occupies 5 liters at 2 at. when the pressure is increased to 2.8 at. and the volume is reduced to 4 lit the percentage increase of temperature is .

0%
30
0%
12
0%
21
0%
0

When 2 gms of a gas are introduced into an evacuated flask kept at

25^{0}

$25^0$ C the pressure is found to be one atmosphere. If 3 gms of another gas added to the same flask the pressure becomes 1.5 atmospheres. The ratio of the molecular weights of these gases will be

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

Explanation

When $2$ grams of a gas $A$ is introduced in an evacuated flask kept at $25 C$ ; the pressure was found to be $1atm$ i.e.

The partial pressure of gas $A=1atm$ .

And let moles be $n_a=\dfrac{2}{M_{A}}$

if $3g$ of another gas $B$ is then added to the same flask;

Partial pressure of gas $B=1.5atm-1atm=0.5atm$

moles of gas $B=\dfrac{3}{M_{B}}$

Again the temp and volume are constant therefore,

$\dfrac{P_{1}}{n_{1}}=\dfrac{P_{2}}{n_{2}}$

Hence, $\dfrac{\dfrac{1}{2}}{M_{A}}$ = $\dfrac{\dfrac{0.5}{3}}{M_{B}}$

$M_{A}=\dfrac{0.5M_{B}}{3}\times2$

$\dfrac{M_{A}}{1}=\dfrac{M_{B}}{3}$

$\dfrac{M_{A}}{M_{B}}=\dfrac{1}{3}$

One mole of gas occupies 10 ml at 50 mm pressure. The volume of 3 moles of the gas at 100 mm pressure and same temperature is

0%
15 ml
0%
100 ml
0%
200 ml
0%
500 ml

A diatomic gas at

17^{0} C

$17^0 C$ is compressed to 1/32 of its original volume. Its final temperature is

0%
1100 K
0%
1500 K
0%
1260 K
0%
1160 K

An open mounted bottle contains a gas at

60^{\circ}

${ 60 }^{ \circ }$ C. What is the temperature to which the gas is to be heated so that 1/4 th mass of the gas may leave it.

0%
${ 143 }^{ \circ }C$
0%
${ 17.1 }^{ \circ }C$
0%
${ 414 }^{ \circ }C$
0%
${ 416 }^{ \circ }C$

Velocity of gas molecules is

500 m / s

$500\ m/s$ at

300 K

$300 K$ , at what temperature velocity will be equal to

1500 m / s

$1500 m/s$ ?

0%
$600\ K$
0%
$900\ K$
0%
$1800\ K$
0%
$2700\ K$

Average kinetic energy of a gas molecule is

0%
Inversely proportional to the square of its absolute temperature
0%
Directly proportional to the square root of its absolute temperature
0%
Directly proportional to its absolute temperature
0%
Directly proportional to square of absolute temperature

What is the approximate root mean speed of line smoke particle of mass

10^{- 17}

$10^{-17}$ kg at

27^{0} C

$27^0C$ ?

[k = 1.38 \times 10^{- 23} J K^{- 1}

$[k=1.38\times 10^{-23}JK^{-1}$ ]

0%
$2.0\times 10^{-2}ms^{-1}$
0%
$2.7\times 10^{-2}ms^{-1}$
0%
$3.5\times 10^{-2}ms^{-1}$
0%
$4.4\times 10^{-2}ms^{-1}$

The number of molecules in a gas pressure

1.64 \times 10^{- 3}

$1.64\times 10^{-3}$ atmospheres and temperature

200 K

$200K$ having the volume

1 c c

$1cc$ are-

0%
$6.02\times 10^{16}$
0%
$2.63\times 10^{16}$
0%
$3.01\times 10^{19}$
0%
$12.04\times 10^{19}$

The degree of freedom per molecule of a gas is

3

$3$ . The heat absorbed by the gas at constant pressure is

150 J

$150\,J$ . Then increase in internal energy is

0%
$90\,J$
0%
$50\,J$
0%
$120\,J$
0%
$30\,J$

One mole of distance ideal gas

(γ = 1.4)

$( \gamma = 1.4 )$ is taken through a cyclic process staring from point A. The process

A \to B

$A \rightarrow B$ is an adiabatic compression,

B \to C

$B \rightarrow C$ is isobaric expansion

C \to D

$C \rightarrow D$ an adiabatic expansion and

D \to A

$D \rightarrow A$ is isochoric. The volume ratios are

V_{A} / V_{B} = 16

$V_A / V_B = 16$ and

V_{C} / V_{B} = 2

$V_C / V_B = 2$ and the temperature at A is

T_{A} = 300 K

$T_A = 300 K$ Calculate the temperature of the temperature of the gas at the points B and D and final the efficiency of the cycle.

0%
75 %
0%
61.4 %
0%
52.5%
0%
80.6 %

If the volume of the gas containing

n

$n$ number of molecules is

V

$V$ , then pressure will decrease due to force of intermolecular attraction in the proportion :

0%
$\dfrac {n}{V}$
0%
$\dfrac {n}{V^2}$
0%
$( \dfrac {n}{V} )^2$
0%
$( \dfrac {1}{V^2} )$

In a process

P T = C o n s t a n t

$PT=Constant$ , if molar heat capacity of a gas is

C = 37.35 J / m o l = K

$C=37.35J/mol=K$ , then find the number of degrees of freedom of molecules in the gas.

0%
$n=10$
0%
$n=5$
0%
$n=6$
0%
$n=7$

Four cylinders contain an equal number of moles of argon, hydrogen, nitrogen, and carbon dioxide at the same temperature. The energy is minimum in?

0%
argon
0%
hydrogen
0%
nitrogen
0%
both nitrogen and hydrogen

Explanation

Using the formula

U = m C_{v} p

$U=mC_vp$

We know that with increase in atomicity,

C_{v}

$C_v$ also increases.

The Energy will be minimum for a monoatomic gas i.e. argon.

Option A is corect.

Calculate the total KE of the mixture of 4 gm

H_{2}

$H_{2}$ and 4 gm He at 300 K-`

0%
1800 R
0%
1750 R
0%
1950 R
0%
None of these

At ordinary pressure and temperature, the ratio of interatomic distance of a gas and a liquid is of the order of

0%
$1000$
0%
$10000$
0%
$10$
0%
$100$

2

$2$ mole ideal He gas and

3

$3$ mole ideal

H_{2}

$H_2$ gas at constant volume find out

C_{v}

$C_v$ of mixture

0%
$\dfrac{21 R}{10}$
0%
$\dfrac{11 R}{10}$
0%
$\dfrac{21 R}{5}$
0%
$\dfrac{11 R}{5}$

Explanation

The given mixture of two moles of monoatomic gas mixed with three mole of diatomic gas can be given as

C = \frac{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}}

$C=\dfrac{n_1C_{v_1}+n_2C_{v_2}}{n_1+n_2}$

= \frac{2 \times \frac{3}{2} R + 3 \times \frac{5}{2} R}{2 + 3}

$=\dfrac{2\times \dfrac{3}{2}R+3\times \dfrac{5}{2}R}{2+3}$

= \frac{3 R + 15 \frac{R}{2}}{5} = (\frac{21 R}{10})

$=\dfrac{3R+15\dfrac{R}{2}}{5}=\left(\dfrac{21 R}{10}\right)$

In a jar having a mixture of

H_{2}

$H_{2}$ and He

0%
Hydrogen has more mean KE
0%
Helium has more mean KE
0%
Both have same mean KE
0%
Both have same internal enery

When an ideal diatomic gas is heated at constant pressure. the fraction of the heat energy supplied which increases the internal energy of he gas is

0%
$\dfrac{2}{5}$
0%
$\dfrac{3}{5}$
0%
$\dfrac{3}{7}$
0%
$\dfrac{5}{7}$

A vessel contains mixture of hydrogen and oxygen gases in teh ratio of their masses equal to 1:The ratio of mean kinetic energy of the two gases is

0%
5:16
0%
16:5
0%
1:1
0%
6:1

r.m.s. speed of ideal gas at

127^{o} C

$127^oC$ is

200 m / s

$200m/s$ , the r.m.s. speed of same ideal gas at temperature

227^{o} C

$227^oC$ is:

0%
$100\sqrt{5}$
0%
$200\sqrt{5}$
0%
$100\sqrt{15}$
0%
$100\sqrt{10}$

Explanation

v = \sqrt{\frac{3 R T}{M}}

$v = \sqrt{\dfrac{3RT}{M}}$

\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} = \frac{200}{v_{2}} = \sqrt{\frac{400}{500}}

$\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}} = \dfrac{200}{v_2} = \sqrt{\dfrac{400}{500}}$

v_{2} = 100 \sqrt{5}

$v_2 = 100\sqrt{5}$

The mean kinetic enrgy of one gm-mole of a perfect gas at absoulte temperature T is

0%
$\frac{1}{2}T$
0%
$\frac{1}{2}RT$
0%
$\frac{2}{3}KT$
0%
$\frac{3}{2}RT$

An ideal gas has volume

V

$V$ and pressure

P

$P$ . The total translational kinetic energy of all the molecule of the gas is.

0%
PV only if the gas is monoatomic
0%
PV only if the gas is diatomic
0%
greater than PV if the gas is diatomic
0%
PV in all cases antainar is evacuated to leave just one

For silver ,

C_{p} (J K^{- 1} m o l^{- 1}) = 23 + 0.01 T

$C_p(JK^{-1}mol^{-1})=23+0.01T$ . If the temperature (T) of 3 moles of silver is raised from

300 K

$300K$ to

1000 K

$1000K$ at

1 a t m

$1\ atm$ pressure, the value of

Δ H

$\Delta H$ will be close to:

0%
$21\ kJ$
0%
$16\ kJ$
0%
$13\ kJ$
0%
$62\ kJ$

Explanation

Solution:- (D)

62 k J

$62 \; kJ$

Δ H = n \int_{T_{1}}^{T_{2}} C_{p . m} d T = 3 \times \int_{300}^{1000} (23 + 0.01 T) d T

$\Delta H=n\displaystyle\int _{ T_{ 1 } }^{ T_{ 2 } }{ C } _{ p.m }dT=3\times \int _{ 300 }^{ 1000 }{ (23+0.01T)dT }$

= 3 [23 (1000 - 300) + \frac{0.01}{2} (1000^{2} - 300^{2})]

$=3[23(1000-300)+\frac{0.01}{2}(1000^2-300^2)]$

= 61950 J \approx 62 k J

$=61950J \approx 62kJ$

Which row correctly describes the ordering and motion of the molecules in liquid water and in ice when both are at a temperature of

0^{o} C

${0}^{o}C$

0%
ordering : a regular pattern of molecules in ice but not in water ; motion : molecules in both ice and water have the same average speed
0%
ordering : a regular pattern of molecules in ice but not in water ; motion : molecules in ice travel more slowly than those in water
0%
ordering : a regular pattern of molecules in both ice and water ; motion : molecules in ice travel more slowly than those in water
0%
ordering : a regular pattern of molecules in both ice and water ; motion : molecules in both ice and water have the same average speed

A volume of

2.5 L

$2.5\ L$ of a sample of a gas at

27^{o}

$27^o$ C and

1

$1$ bar pressure is compressed to a volume of

500 m l

$500\ ml$ keeping the temperature constant, the percentage increase in pressure is?

0%
$100\%$
0%
$400\%$
0%
$500\%$
0%
$80\%$

Explanation

V_{1} = 2.5 L

$V_1=2.5\ L$

T_{1} = 27^{o} C = 300 K

$T_1=27^oC=300K$

P_{1} = 1

$P_1=1$ bar

V_{2} = 500

$V_2=500$ ml

= 0.5 L

$=0.5\ L$

T_{2} = 300 K

$T_2=300\ K$

P_{2} = ?

$P_2=?$

P_{1} V_{1} = P_{2} V_{2}

$P_1V_1=P_2V_2$

2.5 \times 1 = P_{2} \times 0.5

$2.5\times 1=P_2\times 0.5$

∴ P_{2} = 5

$\therefore P_2=5$ bar

Now

%

$\%$ increase in pressure

= \frac{P_{2} - P_{1}}{P_{1}}

$=\dfrac{P_2-P_1}{P_1}$

= \frac{5 - 1}{1} \times 100

$=\dfrac{5-1}{1}\times 100$

= 400 %

$=400\%$ .

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 12 - 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 12 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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