A tyre having volume $0.02 m^{3}$ and pressure $2.5 a t m$ suddenly bursts. What is the new volume occupied by air present?

0.036 m^{3}

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0.063 m^{3}

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0.072 m^{3}

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none

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Solution

The correct option is D $0.036 m^{3}$
$P_{1} V_{1}^{r} = P_{2} V_{2}^{r}$
or
$V_{2} = (\frac{P_{1} V_{1}^{r}}{P_{2}})^{\frac{1}{r}}$
$= 0.02 \times (\frac{2.5}{1})^{(\frac{1}{1.4})} = 0.036 m^{3}$

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A tyre having volume 0.02 m3 and pressure 2.5 atm suddenly bursts. What is the new volume occupied by air present?

The correct option is D 0.036 m3P1Vr1=P2Vr2orV2=(P1Vr1P2)1r=0.02×(2.51)(11.4)=0.036 m3

A tyre having volume $0.02 m^{3}$ and pressure $2.5 a t m$ suddenly bursts. What is the new volume occupied by air present?

The correct option is D $0.036 m^{3}$
$P_{1} V_{1}^{r} = P_{2} V_{2}^{r}$
or
$V_{2} = (\frac{P_{1} V_{1}^{r}}{P_{2}})^{\frac{1}{r}}$
$= 0.02 \times (\frac{2.5}{1})^{(\frac{1}{1.4})} = 0.036 m^{3}$