What will be the minimum pressure required to compress 500 dm^{3} of air at 1 bar to 200 dm^{3} at 30°C?

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#### Solution 1

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_1V_1 = p_2V_2`

`=>p_2 = (p_1V_1)/V_2`

`= (1xx500)/200 "bar"`

`= 2.5 "bar"`

Therefore, the minimum pressure required is 2.5 bar.

#### Solution 2

P_{1} = 1 bar,P_{2} = ? V_{1}= 500 dm^{3} ,V_{2}=200 dm^{3}

As temperature remains constant at 30°C,

P_{1}V_{1}=P_{2}V_{2}

1 bar x 500 dm3 = P_{2} x 200 dm3 or P_{2}=500/200 bar=2.5 bar

Concept: Intermolecular Forces - Introduction

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