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A 250 Liter Helium gas at a temperature of 29o Celsius and a pressure of 2.00 atm was transferred to a tank with a volume of 68.0 Liters. What is the final pressure of the tank if the temperature is maintained?

Sagot :

SOLUTION:

Step 1: List the given values.

[tex]\begin{aligned} & P_1 = \text{2.00 atm} \\ & V_1 = \text{250 L} \\ & V_2 = \text{68.0 L} \end{aligned}[/tex]

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

[tex]\begin{aligned} P_1V_1 & = P_2V_2 \\ P_2V_2 & = P_1V_1 \\ \frac{P_2V_2}{V_2} & = \frac{P_1V_1}{V_2} \\ P_2 & = \frac{P_1V_1}{V_2} \\ & = \frac{(\text{2.00 atm})(\text{250 L})}{\text{68.0 L}} \\ & = \boxed{\text{7.35 atm}} \end{aligned}[/tex]

Hence, the final pressure of the tank is 7.35 atm.

[tex]\\[/tex]

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