Calculate the volume of 3 mole of an ideal gas at STP

At STP

Pressure P=10^{5} Pascal

Temperature T=0° C=273 K

From ideal gas equation

PV=nRT (1)

where n is number of moles of gas. Thus from (1)

V=nRT/P

For 3 moles of an Ideal gas

V=3RT/P

R=8.3 J/mol-K

Now at STP

V=3 x 8.3 x 273/10^{5}

=6.79 x 10^{-2} m^{3}

Calculate the mass of 4 cm

We know that

At STP

Pressure P=10^{5} Pascal

Temperature T=0° C=273 K

From ideal gas equation

PV=nRT

where n is number of moles of gas

n=PV/RT

Substituting the values

n=1.76 x 10^{-4} Moles

We know that number of moles of the gas

N=Mass of gas in gm/Molecular mass in gm/moles

m=nM

=1.76 x 10^{-4 }x 28

=4.94 mg

Equal masses of air are filled and sealed in two container ,one of volume V

Given the mass of air filled in both the containers are equal

For gas in first container

T=200K,V=V_{0},N=n and P=P_{1}

For gas in second container

T=800K,V=4V_{0} N=n since the mass are equal

Ideal gas equation for first gas

P_{1}V_{0}=nRx200---(1)

Ideal gas equation for Second gas

P_{2}x4V_{0}=nRx800---(2)

Dividing equation 1 and 2

P_{1}:P_{2}=1:1

An glass bulb of volume 390 cc was sealed at a pressure of 2*10

g=10 m/s

We know that for any given height h ,Pressure P is

P=ρhg

For 2 x 10^{-3} mm of mercury

P=2 x 10^{-3} x 13600 x 10/1000

=.272Nm^{2}

From Ideal gas equation

PV=nRT

or n=PV/RT

Substituting values

=4.2 x 10^{-8} moles

Number of air molecules=nN_{A}

=2.57 x 10^{16}

A gas container has a walls that can bear maximum pressure of 2.2 x 10

Maximum pressure that the container can bear

P_{m}=2.2 x 10^{6} pa---(1)

Ideal gas equation when the gas is in initial state

PV=nRT

4 x 10^{5} xV=nRx350

Ideal gas equation when the gas reaches Pressure P_{m}

P_{m}V=nRT_{m}

2.2 x 10^{6}xV=nRxT_{m}---(2)

Dividing equation 1 and 2

T_{m}=855.5K

Density of an ideal gas at STP is 3 x 10

Given that

ρ=3 x 10^{-3} gm/cm^{3}

Molecular weight of the gas is given as

M=nN_{a}

and we know that ρ=m/V

Now PV=nRT

or PV=(m/M)RT

M=ρRT/P

At STP

Pressure P=10^{5} Pascal

Temperature T=0° C=273 K

So substituting all the values

M=67.97gm/mole

A vessel of volume V=30 L contains ideal gas at temperature 0° C. After the portion of gas has been let out, the pressure in the vessel decreased by ΔP=.78 atm(Temp remaining constant).Find the mas of the gas released. The gas density under normal condition is ρ=1.3 g/L

Let m_{1} mass of the gas before the gas is released

m_{2} mass of the gas after the gas is released

Thus the mass of gas released

Δm=m_{1}-m_{2}

From ideal gas equation

PV=nRT

Now before release

P_{1}V=n_{1}RT_{0}

Now after release

P_{2}V=n_{2}RT_{0}

where n_{1}=m_{1}/M and n_{2}=m_{2}/M

Number of moles of the gas before and after the release and V and T will remain same

(P_{1}-P_{2})V=(n_{1}-n_{2})RT_{0}

=(m_{1}-m_{2})RT_{0}/M

or Δm=(P_{1}-P_{2})VM/RT_{0}---(1)

Also we know that

PV=(m/M)RT

M/RT_{0}=ρ/P_{0}---(2)

where P_{0}=standard atmospheric pressure and T_{0}=273K

From 1 and 2

Δm=ρVΔP/P_{0}

Substituting all the values

Δm=30gm

A vessel of volume V=20L contains a mixture of hydrogen and helium at T=20° C and Pressure P=2.0 atm.The mass of the mixture is equal to m=5g.Find the ratio of the mass of the hydrogen to that mass of the helium in the mixture. Given R=.082 L.atm.mol

Let the mixture contains:-

m_{1} mass of hydrogen

m_{2} mass of Helium

n_{1} moles of hydrogen

n_{2} moles of Helium

M_{1} Molecular mass of hydrogen

M_{2} Molecular mass of Helium

then m_{1}=n_{1}M_{1} and m_{2}=n_{2}M_{2}

Also m=m_{1}+m_{2}

=n_{1}M_{1} +n_{2}M_{2} (1)

let n be the total no of moles in the mixture then

n=n_{1}+n_{2}

Now putting n_{1}=n-n_{2} in 1 we get

n_{2}=(m-nM_{1})/(M_{2}-M_{1})

Similarly putting n_{2}=n-n_{1} in 1 we get

n_{1}=(nM_{2}-m)/(M_{2}-M_{1})

therefore

m_{1}=M_{1}[(nM_{2}-m)/(M_{2}-M_{1})]

Similarly

m_{2}=M_{2}[(m-nM_{1})/(M_{2}-M_{1})]

so, ratio of m_{1} and m_{2} is

m_{1}/m_{2}=(M_{1}/M_{2}(nM_{2}-m/m-nM_{1})

Now for hydrogen M_{1}=2

for helium M_{2}=4

Given m=5 gm

For mixture

PV=nRT

n=PV/RT

substituting the values

n=1.66

So m_{1}/m_{2}=.50

Find the rms speed of the hydrogen molecules in a sample of hydrogen gas at 250K .Find the temperature at which rms speed is three times the speed at 250K

we know that

V_{rms}=√(3RT/M)---(1)

Now T=25oK

M=2 gm/Moles=.002Kg/mole

R=8.3J/mol-K

V_{rms}=1.76 x 10^{3} m/s

Now we have to calculate the temperature at which rms becomes triple

Squaring equation 1

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

T=V_{rms}^{2}M/3R

=(3 x 1.76 x 10^{3}) x.002/3 x 8.3

=2239K

An ideal gas sample of .203 gm occupies 1000 cm

Mass of sample=.203 gm

V=1000 cm^{3}

At STP

Pressure P=10^{5} Pascal

Temperature T=0°C=273K

V_{rms}=√(3P/ρ)

Now ρ=m/V

So

V_{rms}=√(3PV/m)

Substituting all the values

V_{rms}=1215.6 m/s

The mean speed of the molecules of a hydrogen sample equals the mean speed of the molecules of helium sample. Calculate the ratio of temperature of hydrogen sample to the temperature of the helium sample.

(V_{rms})_{H}=(V_{rms})_{He}

if the T_{H} and T_{He} are respective temperature of the hydrogen and helium samples then

√(3RT_{H}/2)=√(3RT_{He}/4)

or T_{H}:T_{He}=1:2

Find the ratio of the mean speed of the hydrogen molecule to the mean speed of the oxygen molecules in a sample containing mixture of two gases

We know that

V_{rms}=√(3RT/M)

Where M is the molecular weight

(V_{rms})_{H}=√(3RT_{H}/2)

(V_{rms})_{O}=√(3RT_{H}/32)

so (V_{rms})_{H}:(V_{rms})_{O}=4:1

A vessel is partitioned by a fixed diathermic separator. Different ideal gases are filled in the two parts. The rms speed of the molecules in the left part equals the mean speed in the right part. Calculate the ratio of the mass of the molecule in the left part to the mass of the molecule in the right part?

On left side rms speed of the molecule is given as

V_{rms}=√(3RT/m_{1})

On right side

V_{mean}=√(8RT/πm_{2})

Now

(V_{rms})_{left}=(V_{mean})_{right}

or m_{1}:m_{2}=1.18

An air bubble of radius 3.00 mm is formed at the bottom of the 4 m deep river. Calculate the radius of the air bubble as it comes to the surface .Atmospheric pressure=10

Given that radius of bubble formed at bottom=3 mm=3 x 10^{-3} m

Depth of the river=4 m

Pressure on bubble at the bottom is

P_{bottom}=P_{a} + ρgH

P_{a} is atmospheric pressure

So, P_{bottom}=1.4 x 10^{5} pa

Now from ideal gas equation

(PV)_{bottom}=(PV)_{surface} as temperature of water are same

at bottom and surface

1.4 x 10^{5}x(4/3)xπx(3 x 10^{-3})^{3}=10^{5}x(4/3)xπxr^{3}

r=3.35 mm

Air is pumped into the tubes of a cycle at 1.5 atm pressure and volume of tube at this pressure is .002 m

Air is filled in cycle tyres at pressure

P=1.5 atm=1.5 x 10^{5} pa

and volume is V_{0}=.002 m^{3}

When tube get punctured, its volume becomes

V=.00004 m^{3}

T=300K remains constant

Now from ideal gas equation

PV=nRT , where n is the no of moles

If n_{1} is no the moles before puncture

n_{1}=PV_{0}/RT

=.120 moles

After puncture, pressure of tube will become atmospheric pressure. If n_{1} is the no of moles after puncture

n_{2}=PV/RT

=.0160 moles

Number of moles of gas leaked out

=n_{1}-n_{2}

=.120-.016

=.104 moles

A tank attached to an air compressor contains 30L of air at temperature of 40° C and gauge pressure 5*10

Average molecular mas of air which is composed of many gases is M=28.8 gm/mol.

Gauge pressure is the amount of pressure in excess of atmospheric pressure. So total pressure is

P=5 x 10^{5} pa +1 x 10^{5} pa

=6 x 10^{5} pa.

Expressing temperature in kelvin 40° C=313K.

Also express the volume in cubic meters 30L=30 x 10^{-3} m^{3}

To find the number of moles

n=PV/RT

substituting all the values

n=6.91 mole

Mass of gas=nM=.199 kg

Volume at 1 atm and 0° C is

V=nRT/P=71.2 L

Figure below shows two vessels A and B with rigid walls containing ideal gases. The pressure ,temperature and volume are P

P/T=1/2[P

when equilibrium is reached

Ideal gas equation for gases in vessel A and B are

P_{A}V=n_{A}RT_{A}

P_{B}V=n_{B}RT_{B}

where n_{A} and n_{B} are no of molecules of gas in vessel A and B respectively

when equilibrium is reached, after connecting both the vessels gas equation becomes

P(2V)=(n_{A}+n_{B})RT

or

P/T=(R/2V)(n_{A}+n_{B})

P/T=(R/2V)[(P_{A}V/RT_{A})+(P_{B}V/RT_{B})] or

P/T=1/2[P_{A}/T_{A} + P_{B}/T_{B}]

which is the required relation

Figure shows a cylindrical tube of length 30 cm partitioned by a tight fitting separator.The separator is weekly conducting and can freely slide along the tube. Ideal gases are filled in the two parts of the vessel. In the beginning, temp in two parts A and B are 400K and 100 K respectively. The separator slides to a momentary equilibrium position as shown in fig.Find the equilibrium position of the separator reached after the long time

Let, n_{1} and n_{2} are no of molecules of gas in left part and right part respectively

Ideal gas equation for gases in left and right part are

PV_{1}=n_{1}RT_{1}

PV_{2}=n_{2}RT_{2}

As cross-sectional area is same for both the portion

20AP=n_{1}R x 400

10AP=n_{2}R x 100

or n_{1}:n_{2}=1:2

If x is the distance of separator from left end and y from right end, then at common temperature

xAP=n_{1}RT

yAP=n_{2}RT

or x:y=n_{1}:n_{2}=1:2

so 2x=y ---(1)

now we know that

x+y=30 ----(2)

Solving 1 and 2

we get

x=10 and y=20

Assertion and Reason

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

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

STATEMENT 1:The root mean square velocity of molecules of a gas having Maxwellian distribution of velocities is higher than then the most probable speed at any temperature

STATEMENT II:A very small number of molecules of as gas which possess very large velocities,increase the root mean square velocity without effecting the most probable speed.

(a)

STATEMENT 1:Root mean square velocity of the gas does not change in a Isothermal process

STATEMENT 2:Isothermal process may be achieved by immersing the system in large reservoir and performing the process slowly

(b)

STATEMENT 1:In an gas $v_{rms} > v_{mean}$

STATEMENT 2: In an gas $v_{rms} < v_{most probable}$

(c)
V_{rms}=√(3RT/M)

V_{m}=√(8RT/πM)

V=√2RT/M

STATEMENT 1:Thermal conduction is the heat transfer that occurs in the direction of temperature decrease

STATEMENT 2:Thermal conduction is solids occurs due to vibrational energy of the molecules about the equilibrium position

(b) We know that Thermal conduction is the heat transfer that occurs in the direction of temperature decrease Also Thermal conduction is solids occurs due to vibrational energy of the molecules about the equilibrium position Both the statement are true ,but statement II is not a correct explanation for statement I

Three identical cylinder $A_1$,$A_2$,$A_3$ contains equal moles of Helium,Nitrogen,Carbon dioxide at the same temperature

Match the column A to column B

A.Translational Energy is maximum in the cylinder

B.Total energy is minimum in the cylinder

C.Root mean square velocity is maximum in the cylinder

D.Mean speed will be minimum in the cylinder

P. $A_1$

Q. $A_2$

R. $A_3$

S. No appropriate match given

A->S

B->P

C->P

D->R

Molecular Mass

$M_{Helium} < M_{Oxygen} < M_{Carbon-dioxide}$

We know that Translational Energy is same for all molecules at the same temperature

Helium is monatomic gas while oxygen and carbon dioxide are diatomic and polyatomic gases so they posses rotation and vibrational energy also.So total energy is minimum for helium as root mean square velocity is given by $\sqrt {\frac {3RT}{M}}$ as M is least for helium so root mean square velocity will be maximum for Helium

Similarly Mean speed will be minimum for carbon dioxide

Consider an ideal gas with following distribution of speeds

(i) Calculate $V_{rms}$ and hence T. (m= 3.0 × 10

(ii) If all the molecules with speed 1000 m/s escape from the system, calculate new $V_{rms}$ and hence T.

$K_b= 1.38 \times 10^{-23} $

i. $V_{rms}^2= \frac {10 \times (200)^2 + 20 \times (400)^2 + 40 \times (600)^2 + 20 \times (800)^2 + 10 \times (1000)^2}{100}$

$V_{rms}=639 m/s$

Now

$\frac {1}{2} mv_{rms}^2 =\frac {3}{2}K_b T$

or

$T = \frac {1}{3K_b} mv_{rms}^2$

$T=296 K$

ii. $V_{rms}^2= \frac {10 \times (200)^2 + 20 \times (400)^2 + 40 \times (600)^2 + 20 \times (800)^2 }{90}$

$V_{rms}=584 m/s$

Now

$\frac {1}{2} mv_{rms}^2 =\frac {3}{2}K_b T$

or

$T = \frac {1}{3K_b} mv_{rms}^2$

$T=248 K$

An ideal gas sample of .203 gm occupies 1000 cm

Mass of sample=.203 gm

V=1000 cm^{3}

At STP

Pressure P=10^{5} pascal

Temperature T=0°C=273K

V_{rms}=√(3P/ρ)

Now ρ=m/V

So

V_{rms}=√(3PV/m)

Substituting all the values

V_{rms}=1215.6 m/s

How will the rate of collision of a rigid diatomic molecules against the vessel will change ,if the gas is expanded adiabatically η times

Solution