Kinetic Theory

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New answer posted•a year ago
S
a year ago

Frequency of collisions = 1/T = Vavg/λ = 600/ (3 × 10? ) = 2 × 10? sec? ¹

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New answer posted•a year ago
S
a year ago

V_rms = √ (3RT/M)
V_N? = V_H?
√ (3RT_N? / M_N? ) = √ (3RT_H? / M_H? )
573/28 = T_H? /2
⇒ T_H? = 40.928

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

λ = kT / (√2πd²P)
= (1.38×10? ²³ × 300) / (√2 × 3.14 × (0.3×10? )² × 1.01×10? )
≈ 102 nm

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New answer posted•a year ago
S
a year ago

1 litre, T = 300K, P = 2 atm, KE = 2×10? J/molecule, no of molecule =?
No. of molecules = (no of moles) × NA = nNA
Also, n = PV/RT = PV/ (NAkT)
KE = (3/2)kT = 2×10? J [Given]
kT = (4/3)×10?
P = 2 atm = 2 × 1.013 × 10? N/m²
vol = 1 lit = 10? ³ m³
No. of molecules = PV/kT = (2×1.013×10? × 10? ³)/ (4/3)×10?

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New answer posted•a year ago
S
a year ago

T = 2 7 ° C , P = 1 a t m = 1 0 5 N / m 2 , V r m s = 2 0 0 m / s

As we know

v r m s = 3 R T M ⇒ v r m s v ' r m s = 3 R T M 3 R T ' M

V ' r m s = 2 0 0 × 2 3 = x 3

x = 400

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New answer posted•a year ago
S
a year ago

According to KTG, the gas exerts pressure because its molecule :

suffer change in momentum when impinge on the walls of container.

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New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
S
a year ago

At same temperature, curve with higher volume corresponds to lower pressure.

V 3 > V 2 > V 1

⇒ P 1 > P 2 > P 3

(We draw a straight line parallel to volume axis to get this)

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