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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoKinetic Theory of Gases and Gas LawsMedium2 minQB

PhysicsMediumnumerical

A tube of volume 1500 $c m^{3}$ holds air at $3 \times 1 0^{5}$ N/ $m^{2}$ . After a puncture its volume is 400 $c m^{3}$ in atmospheric conditions ( $1 0^{5}$ N/ $m^{2}$ , $2 7^{\circ}$ C). Find the moles of air that leaked out (two decimals).

Answer:

0.16

Solution:

$n_{1} = \frac{3 \times 1 0 ^{5} \times 1.5 \times 1 0 ^{- 3}}{8.314 \times 300} = 0.180$

mol.

$n_{2} = \frac{1 0 ^{5} \times 4 \times 1 0 ^{- 4}}{8.314 \times 300} = 0.016$

mol. Leaked = $0.16$ mol.

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

⏱ 2mℹ️ Source: QB

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