A gas that has a volume of 28 liters, a temperature of 45C, And an unknown pressure has its volume increased to 34 liters and it's temperature decreased to 35C. If I measure the power after the change to be 2.0 ATM , what was the original pressure of the gas? Don't forget to use the right units in your answer.

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11-05-2020
Chemistry

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A gas that has a volume of 28 liters, a temperature of 45C, And an unknown pressure has its volume increased to 34 liters and it's temperature decreased to 35C. If I measure the power after the change to be 2.0 ATM , what was the original pressure of the gas? Don't forget to use the right units in your answer.

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aabdulsamir aabdulsamir · Answer 1 · 2020-05-16T02:07:33+02:00

aabdulsamir aabdulsamir

16-05-2020

Answer:

P1 = 2.5ATM

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15K

P1 = ?

P2 = 2ATM

applying combined gas equation,

P1V1 / T1 = P2V2 / T2

P1*V1*T2 = P2*V2*T1

Solving for P1

P1 = P2*V2*T1 / V1*T2

P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

P1 = 21634.2 / 8628.2

P1 = 2.5ATM

The initial pressure was 2.5ATM

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throwdolbeau throwdolbeau · Answer 2 · 2022-02-24T13:10:01+01:00

The original pressure of the gas, P₁ = 2.5 atm

Given:

V₁ = 28L

T₁ = 45°C = (45 + 273.15)K = 318.15K

V₂ = 34L

T₂ = 35°C = (35 + 273.15)K = 308.15K

To find:

P₁ = ?

P₂ = 2 atm

Combined gas equation:

P₁*V₁ / T₁ = P₂*V₂ / T₂

P₁*V₁*T₂ = P₂* V₂*T₁

Solving for P₁:

P₁ = P₂*V₂*T₁ / V₁*T₂

P₁= (2.0 * 34 * 318.15) / (28 * 308.15)

P₁ = 21634.2 / 8628.2

P₁= 2.5atm

The initial pressure was 2.5 atm.

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