At STP
Pressure P=105 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 273/105
=6.79 x 10-2 m3
We know that
At STP
Pressure P=105 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
Given the mass of air filled in both the containers are equal
For gas in first container
T=200K,V=V0,N=n and P=P1
For gas in second container
T=800K,V=4V0 N=n since the mass are equal
Ideal gas equation for first gas
P1V0=nRx200---(1)
Ideal gas equation for Second gas
P2x4V0=nRx800---(2)
Dividing equation 1 and 2
P1:P2=1:1
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
=.272Nm2
From Ideal gas equation
PV=nRT
or n=PV/RT
Substituting values
=4.2 x 10-8 moles
Number of air molecules=nNA
=2.57 x 1016
Maximum pressure that the container can bear
Pm=2.2 x 106 pa---(1)
Ideal gas equation when the gas is in initial state
PV=nRT
4 x 105 xV=nRx350
Ideal gas equation when the gas reaches Pressure Pm
PmV=nRTm
2.2 x 106xV=nRxTm---(2)
Dividing equation 1 and 2
Tm=855.5K
Given that
ρ=3 x 10-3 gm/cm3
Molecular weight of the gas is given as
M=nNa
and we know that ρ=m/V
Now PV=nRT
or PV=(m/M)RT
M=ρRT/P
At STP
Pressure P=105 Pascal
Temperature T=0° C=273 K
So substituting all the values
M=67.97gm/mole
Let m1 mass of the gas before the gas is released
m2 mass of the gas after the gas is released
Thus the mass of gas released
Δm=m1-m2
From ideal gas equation
PV=nRT
Now before release
P1V=n1RT0
Now after release
P2V=n2RT0
where n1=m1/M and n2=m2/M
Number of moles of the gas before and after the release and V and T will remain same
(P1-P2)V=(n1-n2)RT0
=(m1-m2)RT0/M
or Δm=(P1-P2)VM/RT0---(1)
Also we know that
PV=(m/M)RT
M/RT0=ρ/P0---(2)
where P0=standard atmospheric pressure and T0=273K
From 1 and 2
Δm=ρVΔP/P0
Substituting all the values
Δm=30gm
Let the mixture contains:-
m1 mass of hydrogen
m2 mass of Helium
n1 moles of hydrogen
n2 moles of Helium
M1 Molecular mass of hydrogen
M2 Molecular mass of Helium
then m1=n1M1 and m2=n2M2
Also m=m1+m2
=n1M1 +n2M2 (1)
let n be the total no of moles in the mixture then
n=n1+n2
Now putting n1=n-n2 in 1 we get
n2=(m-nM1)/(M2-M1)
Similarly putting n2=n-n1 in 1 we get
n1=(nM2-m)/(M2-M1)
therefore
m1=M1[(nM2-m)/(M2-M1)]
Similarly
m2=M2[(m-nM1)/(M2-M1)]
so, ratio of m1 and m2 is
m1/m2=(M1/M2(nM2-m/m-nM1)
Now for hydrogen M1=2
for helium M2=4
Given m=5 gm
For mixture
PV=nRT
n=PV/RT
substituting the values
n=1.66
So m1/m2=.50
we know that
Vrms=√(3RT/M)---(1)
Now T=25oK
M=2 gm/Moles=.002Kg/mole
R=8.3J/mol-K
Vrms=1.76 x 103 m/s
Now we have to calculate the temperature at which rms becomes triple
Squaring equation 1
(Vrms)2=3RT/M
T=Vrms2M/3R
=(3 x 1.76 x 103) x.002/3 x 8.3
=2239K
Mass of sample=.203 gm
V=1000 cm3
At STP
Pressure P=105 Pascal
Temperature T=0°C=273K
Vrms=√(3P/ρ)
Now ρ=m/V
So
Vrms=√(3PV/m)
Substituting all the values
Vrms=1215.6 m/s
(Vrms)H=(Vrms)He
if the TH and THe are respective temperature of the hydrogen and helium samples then
√(3RTH/2)=√(3RTHe/4)
or TH:THe=1:2
We know that
Vrms=√(3RT/M)
Where M is the molecular weight
(Vrms)H=√(3RTH/2)
(Vrms)O=√(3RTH/32)
so (Vrms)H:(Vrms)O=4:1
On left side rms speed of the molecule is given as
Vrms=√(3RT/m1)
On right side
Vmean=√(8RT/πm2)
Now
(Vrms)left=(Vmean)right
or m1:m2=1.18
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
Pbottom=Pa + ρgH
Pa is atmospheric pressure
So, Pbottom=1.4 x 105 pa
Now from ideal gas equation
(PV)bottom=(PV)surface as temperature of water are same
at bottom and surface
1.4 x 105x(4/3)xπx(3 x 10-3)3=105x(4/3)xπxr3
r=3.35 mm
Air is filled in cycle tyres at pressure
P=1.5 atm=1.5 x 105 pa
and volume is V0=.002 m3
When tube get punctured, its volume becomes
V=.00004 m3
T=300K remains constant
Now from ideal gas equation
PV=nRT , where n is the no of moles
If n1 is no the moles before puncture
n1=PV0/RT
=.120 moles
After puncture, pressure of tube will become atmospheric pressure. If n1 is the no of moles after puncture
n2=PV/RT
=.0160 moles
Number of moles of gas leaked out
=n1-n2
=.120-.016
=.104 moles
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 105 pa +1 x 105 pa
=6 x 105 pa.
Expressing temperature in kelvin 40° C=313K.
Also express the volume in cubic meters 30L=30 x 10-3 m3
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
Ideal gas equation for gases in vessel A and B are
PAV=nARTA
PBV=nBRTB
where nA and nB 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)=(nA+nB)RT
or
P/T=(R/2V)(nA+nB)
P/T=(R/2V)[(PAV/RTA)+(PBV/RTB)] or
P/T=1/2[PA/TA + PB/TB]
which is the required relation
Let, n1 and n2 are no of molecules of gas in left part and right part respectively
Ideal gas equation for gases in left and right part are
PV1=n1RT1
PV2=n2RT2
As cross-sectional area is same for both the portion
20AP=n1R x 400
10AP=n2R x 100
or n1:n2=1:2
If x is the distance of separator from left end and y from right end, then at common temperature
xAP=n1RT
yAP=n2RT
or x:y=n1:n2=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
(a)
(b)
(c)
Vrms=√(3RT/M)
Vm=√(8RT/πM)
V=√2RT/M
(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
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
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
$T = \frac {1}{3K_b} mv_{rms}^2$
$T=248 K$
Mass of sample=.203 gm
V=1000 cm3
At STP
Pressure P=105 pascal
Temperature T=0°C=273K
Vrms=√(3P/ρ)
Now ρ=m/V
So
Vrms=√(3PV/m)
Substituting all the values
Vrms=1215.6 m/s
