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An observation balloon is filled with helium gas to a volume of 55.0 l at a pressure of 2.5 atm. calculate the volume when the pressure is changed to 1.5 atm assuming that the temperature remains constant.

Sagot :

SOLUTION:

Step 1: List the given values.

[tex]\begin{aligned} & P_1 = \text{2.5 atm} \\ & V_1 = \text{55.0 L} \\ & P_2 = \text{1.5 atm} \end{aligned}[/tex]

Step 2: Calculate the final volume by using Boyle's law.

[tex]\begin{aligned} P_1V_1 & = P_2V_2 \\ P_2V_2 & = P_1V_1 \\ \frac{P_2V_2}{P_2} & = \frac{P_1V_1}{P_2} \\ V_2 & = \frac{P_1V_1}{P_2} \\ & = \frac{(\text{2.5 atm})(\text{55.0 L})}{\text{1.5 atm}} \\ & = \boxed{\text{91.7 L}} \end{aligned}[/tex]

Hence, the volume is 91.7 L.

[tex]\\[/tex]

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