Let A & B are two sample of ideal gases of equal mole .let T be the temperature of both the gas Let $E_A$ and $E_B$ are there total energy respectively .Let $M_A$ and $M_B$ are these respective Molecular Mass .which of these is true

a. $E_A > E_B$

b. $E_A < E_B$

c. $E_A =E_B$

d. none of these

E_{A} = 3/2 nRT

E_{B} = 3/2 nRT

So E_{A} =E_{B}

An Ideal gas undergoes an state change according to PV diagram. what is the value $V_x$

a. $\frac {V_0}{2}$

b. $V_0$

c. $2V_0$

d. $\frac {V_0}{4}$

As the temperature is constant PV = constant

2P_{0} V_{x} = P_{0}V_{0}

V_{x} = V_{0}/ 2

A container contains $N_2$ gas at T K. The no of moles of gas is $n_0$. Consider it behaves like ideal gas. It rms speed is $v_0$

What is the total translational kinetic energy

a. $\frac {3}{2} n_0RT$

b. $\frac {3}{2} RT$

c. $\frac {1}{2} RT$

d. $\frac {1}{2} n_0RT$

KE = (3/2) nRT

= (3/2) n_{0}RT

Suppose the temperature of gas is tripled and $N_2$ molecules dissociate into atom. Then what will be the rms speed of atom.

a. $v_0 \sqrt {6}$

b. $ \sqrt {6 v_0}$

c. $v_0 \sqrt {3}$

d. $\sqrt {3 v_0}$

v_{0} = √3KT/M

v_{t} = √9KT/(M/2)

=v_{0}√6

if $V_p$ denotes most probable velocity of N

a. $V_p : V_0= \sqrt {2} : \sqrt {3}$

b. $V_p : V_0= 1 : 1$

c. $V_p : V_0= 2 : 3$

d. $V_p : V_0= \sqrt {2 }: 3$

V_{p} = √2kT/M

= √2/3 x 3kT/M

= V_{0}√(2/3)

An ideal gas undergoes the process describe by equation

$P = P_0 - aV^2$

Where $P_0$ , a are positive constant and V is the volume of one mole of gas

Maximum temperature attainable by gas

a. $\frac {2}{3} (\frac {P_0}{R}) \sqrt { \frac {P_0}{3a}}$

b. $3 P_0 \sqrt {\frac {P_0}{2Ra}}$

c. $(\frac {1}{3R}) \sqrt {\frac {P_0}{3a}}$

d. $\frac {4}{3} (\frac {P_0}{R}) \sqrt { \frac {P_0}{2}}$

P = P_{0} - aV^{2}

= P_{0}-a(RT/P)^{2}

PV = RT for one mole of gas

T = (1/R√a )P√(P_{0} -P)

= (1/R√a )√(P_{0}P^{2} - P^{3})

For T_{max}

d (P_{0}P^{2} - P^{3}) /dP= 0

P = 2P_{0}/3

Putting this value

T_{max} = 2/3 (P_{0}/R) √(P_{0}/3a)

Let V, V

is m then

a. No molecule can have a speed greater the (2)

b. No molecule can have a speed less the V

c. V

d. The average kinetic charge of a molecule (3/4 )mV

V_{rms}= (3kT/M)^{1/2}

V = (8KT/πM)^{1/2}

V_{p}= (2KT/M)^{1/2}

So it is clear

V_{p} < V < V_{rms}

Also

Average kinetic energy

=(½)mV_{rms}^{2}

V_{rms}^{2} = 3KT/M

V_{p}^{2} = 2KT/M

V_{rms}^{2} = 3/2 V_{p}^{2}

So E=3/4 V_{p}^{2}

So Ans. C&D

A flask contains Oxygen, Hydrogen & chlorine in the ration of 3:2:1 mixture at 27 °C

Molecular mass of Oxygen = 32

Molecular mass of Hydrogen = 2

Molecular mass of Chlorine = 70.9

Find the ratio of average kinetic energy per molecule of Oxygen & Hydrogen

a. 1:2

b. 1:1

c. 1:16

d. 2:1

find the ratio of average Kinetic energy per molecule of Hydrogen & chlorine

a. 1:2

b. 1:1

c. 1:16

d. 1:1

All the molecule have same kinetic energy at same temperature

find the ratio of mean speed of Oxygen, chlorine, Hydrogen

a. 4:1 : (.45)

b. 5:2 :(5)

c. 1: (.45)

d. 2 :(.45)

V= (8KT/πM)^{1/2}

V_{H}= (8KT/πM_{H})^{1/2}

V_{O}= (8KT/πM_{O})^{1/2}

V_{CL}= (8KT/πM_{CL})^{1/2}

V_{O} : V_{CL}: V_{H} = (8KT/πM_{O})^{1/2} : (8KT/πM_{CL})^{1/2}: (8KT/πM_{H})^{1/2}

= 1/(M_{O}) ^{1/2}: 1/(M_{CL})^{1/2} : 1/(M_{H}) ^{1/2}

= 1: (.45)^{1/2}:4

Suppose a container is evacuated to have just one molecule of a gas in it. Let V

a. V

b. V

c. V

d. V

Average speed is define

V_{a} = ∑V / N

= V/1 = V

mean square speed = ∑ V^{2} / N

= V^{2}

root mean square = (∑ V^{2} / N)^{1/2}= V

So V_{a} = V_{rms}

The velocities of the molecules are v, 2v, 3v, 4v & 5v. The rms speed will be

a. 11v

b. v(11)

c. v

d. 3.3v

Vrms= (∑ V^{2} / N)^{1/2}

= [(V^{2} + 4V^{2} + 9V^{2} + 16V^{2} + 25V^{2})/5]^{1/2}

=v(11)^{1/2}

Equal Number of molecules of hydrogen & Oxygen are contained in a vessel at one atmosphere

pressure. The ratio of the collision frequency of hydrogen molecules to the of Oxygen molecules on the

container

a. 1:4

b. 4:1

c. 1:16

d. 16:1

n_{1} = Vrms / 2L ---Hydrogen

n_{2} = Vrms/2L ---Oxygen

n_{1}/n_{2} = Vrms of Hydrogen/ Vrms of oxygen

Vrms of Hydrogen/ Vrms of oxygen = (M_{O}/M_{H})^{1/2}

So n_{1}/n_{2} = 4:1

The prefect gases A, B & C having masses m

is the final temperature of mixture.

a. [(m

b. [(M

c. [(m

d. [(M

Internal kinetic energy

= (3/2) nRT

so for A KE_{1} = (3/2 )(m_{1}/M_{1}) RT_{1},

for B KE_{2} = (3/2 )(m_{2}/M_{2}) RT_{2},

for C KE_{3} = (3/2 )(m_{3}/M_{3}) RT_{3},

Now final Kinetic energy

KE = 3/2 (m_{1}/M_{1}+m_{2}/M_{2}+m_{3}/M_{3}) RT

Now

KE = KE_{1} + KE_{3}+KE_{3}

T = [(m_{1}/M_{1})T_{1} + (m_{2}/M_{2})T_{2} + (m_{3}/M_{3})T_{3}]/(m_{1}/M_{1}+m_{2}/M_{2}+m_{3}/M_{3})

M moles of a ideal polyatomic gas(C

(a) $3Q=4MRT$

(b) 2Q=3MRT

(c) $Q=4MRT$

(d) $7Q=4MRT$

Internal energy before supplying the heat

$= \frac {7}{2} MRT$

Total no of polyatomic moles after the heat is supplied

=2M/3

Total no of monatomic moles after the heat is supplied

=5M/3

Total internal after the split

$= \frac {7}{2} (\frac {2M}{3})RT + \frac {3}{2} (\frac {5M}{3}) RT$

$= \frac {29}{6} MRT$

So

$Q= \frac {29}{6} MRT - \frac {7}{2} MRT$

3Q=4MRT

Ans is a

A gas mixture consist of molecules of type A, B, C, D with molecular

masses M

Two statement are drawn from it

E

V

which one of following is true

a. Only A correct

b. Only B correct

c. A & B both are correct

d. A & B both are wrong

Average kinetic energy is same for all the gases at the same temperature

rms speed is inversely proportional to the Molecular mass of the gas

since M_{a} > M_{b} > M_{c} >M_{d}

V_{D} > V_{C} > V_{C} > V_{A}

a. Statement I is true ,statement II is true ,statement II is correct explanation for statement I

b. Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I

c. Statement I is true, Statement II is false

d. Statement I is False, Statement II is True

There are two statement

which one of the following is correct

a. A and B both

b. A only

c B only

d. A and B both are incorrect

Statement A is Avogadro law. So true

Avogadro no is 6.0255 * 10^{23 } So second one is true

There are two statement about Ideal gases

which one of the following is correct

a. A and B both

b. A only

c. B only

d. A and B both are incorrect

v_{rms} depends on mass and temperature

So A is correct . Ans B is incorrect

The molar specific heat capacity of the ideal gas in isothermal process is infinity as there is no change in temperature

Heat transfer is non zero in the Isothermal process as dU=0 so dQ=dW so dQ not equal to zero

So both the statement is true but statement II is not a correct explanation for statement I

Equation of state for a real gas is (P+a/V^{2})(V-b)=nRT.This statement is true

Molecular attraction is not negligible and the size of molecules are not negligible in comparison to average separation between them. This is also true

Equation of state for a ideal gas is PV=nRT

but Equation of state for a real gas is (P+a/V^{2})(V-b)=nRT because Molecular attraction is not negligible and the size of molecules are not negligible in comparison to average separation between them

A container has a mixture of 1 mole of oxygen and 2 moles of nitrogen at 330K.The ratio of average rotational kinetic energy per O

a. 2:1

b. 1:2

c. 1:1

d. None of these

Rotational kinetic energy depends only on degree of freedom associated with which is same for both diatomic molecule ie 1:1

So c is correct

1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it.

At any given time

(a) the pressure on EFGH would be zero.

(b) the pressure on all the faces will the equal.

(c) the pressure of EFGH would be double the pressure on ABCD.

(d) the pressure on EFGH would be half that on ABCD.

(d) In the ideal case that we normally consider,each collision transfers twice the magnitude of its normal momentum. On the face EFGH, it transfers only half of that.

According the Maxwell's Speed distribution law

$A(v) = 4 \pi (\frac {M}{2\pi RT})^{3/2} v^2 e^{-Mv^2/2RT}$

Where V is molecular speed

M = molar mass of gas

R = gas constant

T = Temperature

A (v) = Probability distribution function

A (v) dv = fraction of the molecules whose speed lie in the internal of width dv speed center on v

What is most probable speed

a. $\sqrt {\frac {2RT}{M}}$

b. $\sqrt {\frac {RT}{M}}$

c. $\sqrt {\frac {3RT}{M}}$

d.$\sqrt {\frac {8RT}{M}}$

(a)

This is derived from Maxwell equation.Derivation is beyond the scope of the course

When is of these is true with respect to above paragraph where the integral is from 0 to infinity

a. $\int_{0}^{\infty}A (v) dv = 1$

b. $\int_{0}^{\infty}A (v) dv = 2$

c. $\int_{0}^{\infty}A (v) dv = 1/2$

d. $\int_{0}^{\infty}A (v) dv = 0$

(a)

A(v)dv represent the fraction of molecules...so when all the fractions are considered...this should be equal to 1

let A

a. A

b. A

c. A

d. None of these

(a)

A rectangular strip in this curve is represent by =A(v)dv

Area of the curve =∫ A(v)dv

A(v)dv represent the fraction of molecules...so when all the fractions are considered...this should be equal to 1

Area of the curve =1

So

Area under the curve =1 and it is true for all variation.

What is the formula for V

a. $V_{avg} = \int_{0}^{\infty} vA (v) dv $

b. $V_{avg} = \int_{0}^{\infty} A (v) dv $

c. $V_{avg} = \int_{-\infty}^{\infty} vA (v) dv $

d. $V_{avg} = \int_{-\infty}^{\infty} A (v) dv $

(a)

A(v)dv represent the fraction of molecules

v is the speed.

Now average speed formula is

=(n1v1+n2v2+n3v3...)/(n1+n2+n3)

or

average speed formula =(fraction of molecules have speed v1*v1)+(fraction of molecules have speed v2*v2)+(fraction of molecules have speed v3*v3)

Similarly we find average speed of the molecules in a gas with the following procedure

We weight each value of v in the distribution,that is we multiply it by the fraction of molecules A(v)v of the molecules

Then we add us all these values of vA(v)dv to get the average speed

So $V_{avg} = \int_{0}^{\infty} vA (v) dv $

The below figure show a hypothetical speed distribution for a sample of N gas particles.

What is the relation but a, b & v

a. $v_0 (2a+3b) = 2$

b. $v_0 (2a- 3b) = 2$

c. $v_0 (a+ b) = 1$

d. $v_0 (a- b) = 1$

(a)

Area of curve should be one

so

$ \frac {1}{2}av_0 +av_0 + \frac {1}{2}v_0(b-a)+bv_0=1$

or $v_0 (2a+3b) = 2$

How much fraction of molecules are in speed range v

a. $(\frac {v_0}{2}) (3b+a)$

b. $v_0(3b+c)$

c. $(\frac {v_0}{4}) (3b+a)$

d. $(\frac {v_0}{4}) (3b-a)$

(a)

Fraction of molecules are in speed range v0 & 3v0 = Area of the curve between the speed range

$=bv_0+av_0+(1/2)v_0(b-a)$

$=(\frac {v_0}{2}) (3b+a)$

What is the average speed of the molecules

a. $\frac {6av_0^2 + 20bv_0^2}{6}$

b.$\frac {6av_0^2 - 38bv_0^2}{6}$

c. $3av_0^2 + 2bv_0^2$

d. $3av_0^2 - 2bv_0^2$

(a)
V_{av} = ∫[vA(v)]dv where limits are 0 to infinity

= ∫[vA(v)]dv where limits are 0 to v_{0}

+ ∫[vA(v)]dv where limits are v_{0} to 2v_{0}

+ ∫[vA(v)]dv where limits are 2v_{0} to 3v_{0}

= ∫[v(a/v_{0})vdv where limits are 0 to v_{0}

+ ∫v[{(b-a)/v_{0}}v+(2a-b)]dv where limits are v_{0} to 2v_{0}

+ ∫[v b]dv where limits are 2v_{0} to 3v_{0}

= (6av_{0}^{2} + 20bv_{0}^{2})/6

A. Average energy of monatomic molecule

B. Average energy of Diatomic molecule

C. Average energy of Polyatomic molecule with no vibration

d. Average energy of Diatomic molecule which vibrate

U. $3K_BT$

V. $\frac {7}{2} K_BT$

X. $\frac {5}{2} K_BT$

Y. $\frac {3}{2} K_BT$

a. A - V,B - X,C - U,D - Y

b. A - Y,B - X,C - U,D - V

c. A -X,B - Y,C - U,D - V

d. A - Y,B - X,C - V,D - U

(b)

Average energy of monatomic molecule =(degree of Freedom)*(KT/2)=3KT/2 as degree of Freedom=3

Average energy of Diatomic molecule =(degree of Freedom)*(KT/2)=5KT/2 as degree of Freedom=5

Average energy of Polyatomic molecule =(degree of Freedom)*(KT/2)=3KT as degree of Freedom=6

Average energy of Diatomic molecule which vibrate =(degree of Freedom)*(KT/2)=7KT/2 as degree of Freedom=7

A vessel of volume V contains a mixture of 1 mole of Hydrogen and 1 mole of Oxygen (both considered as ideal). Let $f_1(v)dv$, denote the fraction of molecules with speed between v and (v + dv) with $f_2 (v)dv$, similarly for oxygen. Then

(a) $f_1(v) + f_2(v) =f(v)$ obeys the Maxwell's distribution law.

(b) $f_1 (v)$, $f_2(v)$ will obey the Maxwell's distribution law separately.

(c) Neither $f_1 (v)$, nor $f_2 (v)$ will obey the Maxwell's distribution law.

(d) $f_2 (v)$ and $f_1 (v)$ will be the same.

(b)

A gas has five molecules having velocities v,2v,3v,4v,5v.Match the column A to column B

Here V

P. (V

Q. (V

R (V

S. (V

W. 9v

X. 11v

Y. 10v

Z. 0

a.P-W,Q-X,R-Y,S-Z

b.P-X,Q-W,R-Z,S-Y

c.P-X,Q-W,R-Y,S-Z

d.P-X,Q-W,R-X,S-Z

V_{mean}=(∑ v/N)=[v+(2v)+(3v)+(4v)+(5v)]/5=3v

V_{rms}=√(∑ v^{2}/N)=√[v^{2}+(2v)^{2}+(3v)^{2}+(4v)^{2}+(5v)^{2}]/5=v(11)^{1/2}

A gas has $v_0$ as rms at temperature $T_0$ and pressure $p_0$.which of the following is correct

a.If the pressure is doubled keeping temperature constant T0 ,rms will remain same

b.if the mass of the gas molecules is tripled,rms will become .58v0

c. if the temperature is increased such that T=4T0,rms will become 2v0

d. None of the above

(a),(b) and (c)

V_{rms}=v_{0}

V_{rms}=(3RT/M)^{1/2}

Now since V_{rms} depends on temperature and mass

so when pressure is doubled keeping temperature constant, rms remains same

Now .if the mass of the gas molecules is tripled

then

V_{rms}=(3RT/3M)^{1/2}=v_{0}(1/3)^{1/2}

=.58v_{0}

Now temperature is increased such that T=4T_{0},

V_{rms}=(3RX4T/M)^{1/2}=2v_{0}

V_{mean}=(∑ v/N)=[v+(2v)+(3v)+(4v)+(5v)]/5=3v

V_{rms}=√(∑ v^{2}/N)=√[v^{2}+(2v)^{2}+(3v)^{2}+(4v)^{2}+(5v)^{2}]/5=v(11)^{1/2}

A cubic vessel (with faces horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of 200 m/s in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground

(a) remains the same because 200 m/s is very much smaller than $v_{rms}$ of the gas.

(b) remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls.

(c) will increase by a factor equal to $(\frac { v_{rms}^2 + 2500)^2}{v_{rms}^2})$ where $v_{rms}$ was the original mean square velocity of the gas.

(d) will be different on the top wall and bottom wall of the vessel.

(b)

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