Question

A vessel of 120 mL capacity contains a certain amount of gas at 35°c and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°c. What would be its pressure?

Answer 1

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solutions..

According to Boyle’s Law P1 V1 = P2 V2

Here the temperature is constant. Therefore

1.2 X 120 = P2 X 180

OR

P2 = 1.2 X 120 /180 = 0.8 bar

Therefore, the pressure would be 0.8 bar

 

Answer 2

Hello!!

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$\Huge{\mathfrak{\blue{\underline{ANSWER}}}}$

Initial pressure, P1 = 1.2 bar

Initial volume, V1 = 120 mL

Final volume, V2 = 180 mL

As the temperature remains same, final pressure (P2) can be calculated with the help of Boyle’s law.

According to the Boyle’s law,

P1V1 = P2V2

$p2 = \frac{p1v1}{v2}$

= 1.2×120/180

= 0.8 bar

Therefore, the min pressure required is 0.8 bar.

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Thanks!!