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Fluorine exerts a pressure of 900. torr. When the pressure is changed to 1.5 atm, its
volume is 250. mL. What was the original volume?

Respuesta :

sebassandin

Answer:

[tex]V_1=312.5mL[/tex]

Explanation:

Hello!

In this case, since this is a problem about the Boyle's law as we are given the initial and final pressure and the final volume, we can write:

[tex]P_1V_1=P_2V_2[/tex]

Whereas we can solve for the initial volume V1 as shown below:

[tex]V_1=\frac{P_2V_2}{P_1}[/tex]

However, both pressures must be in the same units, therefore, we plug in the values as shown below:

[tex]V_1=\frac{1.5atm*250mL}{900torr*\frac{1atm}{760torr} }\\\\V_1=312.5mL[/tex]

Best regards!

asuwafohjames

The original volume of the Fluorine gas is 316.67 mL.

To calculate the original volume of the fluorine gas, we apply Boyles law.

Formula:

  • PV = P'V'.............. Equation 1

Where:

  • P = Original pressure of the gas
  • V = Original volume of the gas
  • P' = Final pressure of the gas
  • V' = Final volume of the gas.

Make V the subject of the equation

  • V = P'V'/P................. Equation 2

From the question,

Given:

  • P = 900 torr
  • P' = 1.5 atm = (1.5×760) = 1140 torr.
  • V' = 250 mL

Substitute these values into equation 2

  • V = (1140×250)/900
  • V = 316.67 mL

Hence, the original volume of the Fluorine gas is 316.67 mL

Learn more about Boyles law here: https://brainly.com/question/24938688