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A 220.0-mL sample of helium gas is in a cylinder with a movable piston at 105 kPa and 275 K. The piston is pushed in until the sample has a volume of 95.0 mL. The new temperature of the gas is 310. K. What is the new pressure of the sample?(1) 5l.1 kPa(2) 216 kPa(3) 243 kPa(4) 274 kPa

Verified Answer :

pmills3
  1. This problem is really simple if you realize that you can use the combined gas law:

    P1V1/T1 = P2V2/T2

    Therefore, just substitute in and solve:

    (105 kPa)(220.0 mL)/(275K) = ?kPa(95.0 mL)/(310 K)
    Solve for ?
    Answer is 274. kPa
nuhulawal20

The new pressure of the sample of helium gas in a cylinder with a movable piston is 274 kPa.

Option 4)274 kPa is the correct answer.

What is Combined gas law?

Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.

It is expressed as;

P₁V₁/T₁ = P₂V₂/T₂

Given the data in the question;

  • Initial volume V₁ = 220.0mL = 0.22L
  • Initial pressure P₁ = 105kPa = 1.03627atm
  • Initial temperature T₁ = 275K
  • Final volume V₂ = 95.0mL = 0.095L
  • Final temperature T₂ = 310K
  • Final pressure P₂ = ?

To calculate the new temperature of the gas, we substitute our given values into the equation above.

P₁V₁/T₁ = P₂V₂/T₂

P₁V₁T₂ = P₂V₂T₁

P₂ = (P₁V₁T₂) / V₂T₁

P₂ = ( 1.03627atm × 0.22L × 310K ) / ( 0.095L × 275K )

P₂ = 70.673614LatmK / 26.125LK

P₂ = 2.705atm

P₂ = (2.705 × 101.325)kPa

P₂ = 274 kPa

The new pressure of the sample of helium gas in a cylinder with a movable piston is 274 kPa.

Option 4)274 kPa is the correct answer.

Learn more about the combined gas law here: brainly.com/question/25944795

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