An electric bulb of volume 250 c m 3 was sealed off during manufacture at pressure 1 0 − 3 mm of Hg at 2 7 ∘ C. Compute the number of air molecules in the bulb (in units of 1 0 15 , to two decimals).

N = k T P V = 1.38 × 1 0 − 23 × 300 1.333 × 1 0 − 1 × 2.5 × 1 0 − 4 = 8.05 × 1 0 15 .

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoKinetic Theory of Gases and Gas LawsMedium2 minQB

PhysicsMediumnumerical

An electric bulb of volume 250 $c m^{3}$ was sealed off during manufacture at pressure $1 0^{- 3}$ mm of Hg at $2 7^{\circ}$ C. Compute the number of air molecules in the bulb (in units of $1 0^{15}$ , to two decimals).

Answer:

8.05

Solution:

$N = \frac{P V}{k T} = \frac{1.333 \times 1 0 ^{- 1} \times 2.5 \times 1 0 ^{- 4}}{1.38 \times 1 0 ^{- 23} \times 300} = 8.05 \times 1 0^{15}$ .

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

⏱ 2mℹ️ Source: QB

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