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

Asked by Pragya Singh | 1 year ago |  108

##### Solution :-

Initial pressure, P1 = 1.2bar

Initial volume, V1 = 120mL

Final volume, V2 = 180mL

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

According to the Boyle’s law,

P1V1 = P2V2

P=$$\dfrac{P_{1}V_{1}}{V_{2}}$$

=$$\dfrac{1.2 \; \times \; 120}{180}$$

= 0.8bar

Therefore, the min pressure required is 0.8bar.

Answered by Pragya Singh | 1 year ago

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