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Eduardwww [97]
3 years ago
8

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.​
Chemistry
1 answer:
patriot [66]3 years ago
3 0

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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Explanation:

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Replacing in Kp formula:

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A sample of gas occupies 10 L at STP. What
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Pressure is 5.7 atm

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