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.

Question

Junior Johns

Chemistry

9 December, 17:49

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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Answers (1)

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Cristal · Answer 1

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