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A gas has a volume of 250.0 liters at 17.0 °C and 3.00 atm of pressure. what volume must the gas be increased for the gas to be at STP conditions?

Sagot :

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Answer:

0.706 L (Liters)

Explanation:

P1V1 = P2 V2 3 atm 250.0 mL = 1 atm X

T1 T2 290 K 273 K

706 mL = X

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The increased volume will be "706 L".

According to the question,

Pressure,

  • Initial, [tex]P_1 = 3 \ atm[/tex]
  • Final, [tex]P_2 = 1 \ atm[/tex]

Temperature,

  • Initial, [tex]T_1 = 17^{\circ} C[/tex] or, [tex]290 \ K[/tex]
  • Final, [tex]T_2 = 273 \ K[/tex]

Volume,

  • Initial, [tex]V_1 = 250 \ L[/tex]
  • Final, [tex]V_2 = \ ?[/tex]

By using the Boyl's and Charl's combination law, we get

→ [tex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}[/tex]

or,

→ [tex]V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}[/tex]

By putting the values, we get

→      [tex]= \frac{3\times 250\times 273}{290\times 1}[/tex]

→      [tex]= 706 \ L[/tex]

Thus the aforementioned answer is right.

Learn more about pressure here:

https://brainly.com/question/19156022

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