A vacuum tube of volume 250 c m 3 is sealed at a pressure of 3.0 × 1 0 − 8 mm of mercury and 1 7 ∘ C. Find the number of molecules in the tube ( k = 1.38 × 1 0 − 23 J/K, 1 mm Hg = 133.3 Pa). Give the answer as a multiple of 1 0 10 to the nearest integer.

P = 3.0 × 1 0 − 8 × 133.3 = 4.0 × 1 0 − 6 Pa. N = k T P V = 1.38 × 1 0 − 23 × 290 4.0 × 1 0 − 6 × 2.5 × 1 0 − 4 = 2.5 e + 11 , which is 25 in units of 1 0 10 .

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoKinetic Theory of Gases and Gas LawsMedium2 minQB

PhysicsMediumnumerical

A vacuum tube of volume 250 $c m^{3}$ is sealed at a pressure of $3.0 \times 1 0^{- 8}$ mm of mercury and $1 7^{\circ}$ C. Find the number of molecules in the tube ( $k = 1.38 \times 1 0^{- 23}$ J/K, 1 mm Hg = $133.3$ Pa). Give the answer as a multiple of $1 0^{10}$ to the nearest integer.

Answer:

25.00

Solution:

$P = 3.0 \times 1 0^{- 8} \times 133.3 = 4.0 \times 1 0^{- 6}$

Pa.

$N = \frac{P V}{k T} = \frac{4.0 \times 1 0 ^{- 6} \times 2.5 \times 1 0 ^{- 4}}{1.38 \times 1 0 ^{- 23} \times 290} = 2.5 e + 11$ ,

which is 25 in units of $1 0^{10}$ .

Subject:PhysicsTopic:Heat & ThermoSubtopic:Kinetic Theory of Gases and Gas Laws

⏱ 2mℹ️ Source: QB

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