Find the number of moles in 2.0 L of an ideal gas at 1.013 × 1 0 5 Pa and 300 K (three decimals). R = 8.314 J/(mol K).

n = RT P V = 8.314 × 300 ( 1.013 × 1 0 5 ) ( 2.0 × 1 0 − 3 ) = 2494.2 202.6 = 0.0812 mol, so 0.081 mol.

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PhysicsHeat & ThermoKinetic Theory of Gases and Gas LawsMedium2 minQB

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Find the number of moles in 2.0 L of an ideal gas at $1.013 \times 1 0^{5} Pa$ and 300 K (three decimals). $R =$ 8.314 J/(mol K).

Answer:

0.08

Solution:

$n = \frac{P V}{RT} = \frac{( 1.013 \times 1 0 ^{5} ) ( 2.0 \times 1 0 ^{- 3} )}{8.314 \times 300} = \frac{202.6}{2494.2} = 0.0812$

mol, so 0.081 mol.

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

⏱ 2mℹ️ Source: QB

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