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The gas in a 250. mL piston experiences a change in pressure from 1.00 atm to 4.45 atm. What is the new volume (in mL) assuming the moles of gas and temperature are held constant

Respuesta :

Answer:

56.2 mL

Explanation:

Given data

  • Initial volume (V₁): 250 mL
  • Initial pressure (P₁): 1.00 atm
  • Final volume (V₂): ?
  • Final pressure (P₂): 4.45 atm

Assuming the gas has an ideal behavior, we can find the final volume using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 1.00 atm × 250 mL/ 4.45 atm

V₂ = 56.2 mL

The new volume will be "56.2 mL".

According to the question,

Volume,

  • [tex]V_1 = 250 \ mL[/tex]
  • [tex]V_2 = \ ?[/tex]

Pressure,

  • [tex]P_1 = 1.00 \ atm[/tex]
  • [tex]P_2 = 4.45 \ atm[/tex]

By using the Boyle's Law, we get

→ [tex]P_1 V_1 = P_2 V_2[/tex]

or,

→     [tex]V_2 = \frac{P_1V_1}{P_2}[/tex]

By putting the values, we get

           [tex]= \frac{1.00\times 250}{4.45}[/tex]

           [tex]= \frac{250}{4.45}[/tex]

           [tex]= 56.2 \ mL[/tex]

Thus the above answer is right.

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