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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

Respuesta :

pmills3
4. 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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