Question

Single Correct Answer Type
1. Volume-tempero
PHYSICS
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temperature graph at atmospheric
sure for a monoatomic gas
(V in m?,T in
VA
C) is
Four molecules of a gas have speeds 1,2,3
and 4 kms-1. The value of rms speed of the
gas molecules is
a) T5 kas b tokom 200
1
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C) 2.5 kms
mes
1663
TO
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5 1365 ms 1, The gas is
d) CO2
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5. The root mean square velocity of gas
molecules at 27°C is 1365 ms.
a) 0₂ b) He c) N₂
6. The temperature of a given ma
increased from 27°C to 327"
velocity of the molecule
a) 2 times
c)2V2 times
TO
m 27°C to 327°C. The rms
hy of the molecules increases
2 times
d)4 times
o what temperature should the hydrogen
vany
at 327°C be cooled at constant pressure, so
that the root mean square velocity of its
molecules becomes half of its previous
value
.) -100°C d)0°C
) is expanded
nany times has the gas
t mean
imes?
reduce​

Answer 1

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Answer 2

Given : kinetic energy of oxygen molecules is 8.368 kJ/mol.

To find : the root mean square speed of oxygen molecules and temperature at which kinetic energy of the gas is 8.368 kJ/mol.

solution : kinetic energy of gaseous molecule is given by, K.E = 3/2 nRT

⇒8.368 × 10³ J/mol = 3/2 × 1 × 8.314 J/mol/K × T

⇒T = (8368 × 2)/(3 × 8.314) = 670 K

Therefore the required temperature is 670K

now root mean square speed of oxygen gas is given by, v = √{3RT/M}

Here M is molecular weight i.e., M = 32 × 10¯³ kg/mol [ for oxygen gas ]

T = 670 K, R = 8.314 J/mol/K

so, v = √{3 × 8.314 × 670/32 × 10¯³}

= √{522.223 × 10³}

= √(522223)

= 722.65 ≈ 723 m/s

Therefore the root mean square speed of oxygen molecules is 723 m/s.

Mean square speed of the Oxygen = 130 m/s.

Using the formula,

Mean Square Speed = 

Where R is the Universal constant of gas, and T is the temperature, M is the Molar mass of the Oxygen = 32 × 10⁻³ g.

∴ 130 = √[2 × 8.31T/(32 × 10⁻³)]

130 × √(32 × 10⁻³) = √(8.31 × 2T)

∴ √T = 5.7

∴ T = 32.49 K.

Root mean square speed of Argan atom

( Vrms1 ) = √{3RT1/M1} -----(1)

Root mean square spped of Helium atom ( Vrms2) = √ { 3RT2/M2}-------(2)

Divide equations (1) and (2)

Vrms1/Vrms2 = √{3RT1/M1}/√{3RT2/M2}

= √{ T1 ×M2/T2×M2}

But ,

Vrms1 = Vrms2 [ A/C to question , ]

1 = √{ T1 × M2/M1 × T2}

M1/M2 = T1/T2

So,

T1 = T2 × { M1/M2}

Here,

T2 = -20°C = -20+273 = 253 K

M1 = 40g/mol

M2 = 4 g/mol

T1 = ?

T1 = 253 × { 40 /4 }

= 2530 K:

Explanation: