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(B) Given:Initial pressure,$p_{1} = 1 \, bar$Initial volume,$V_{1} = 500 \, dm^{3}$Final volume,$V_{2} = 200 \, dm^{3}$Since the temperature remains c

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(B) Given:Initial pressure,$p_{1} = 1 \, bar$Initial volume,$V_{1} = 500 \, dm^{3}$Final volume,$V_{2} = 200 \, dm^{3}$Since the temperature remains constant,the final pressure $(p_{2})$ can be calculated using Boyle's law.According to Boyle's law,$p_{1} V_{1} = p_{2} V_{2}$$\Rightarrow p_{2} = \frac{p_{1} V_{1}}{V_{2}}$$= \frac{1 	imes 500}{200} \, bar$$= 2.5 \, bar$Therefore,the minimum pressure required is $2.5 \, bar$.

What will be the minimum pressure required to compress $500 \, dm^{3}$ of air at $1 \, bar$ to $200 \, dm^{3}$ at $30^{\circ} C$ (in $, bar$)?

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