Question

A flask of volume 300.0 x 10^-6 m^3 contains air at a pressure of 5.00 x 10^5 Pa and a temperature of 27.0°C. The flask loses molecules at a rate of 3.00 x 10^19 per second. Estimate how long it takes for the pressure in the flask to fall to half its original value. (Assume that the temperature of the air remains constant during this time.)

Name: A flask of volume 300.0 x 10^-6 m^3 contains air at a pressure of 5.00 x 10^5 Pa and a temperature of 27.0°C. The flask loses molecules at a rate of 3.00 x 10^19 per second. Estimate how long it takes for the pressure in the flask to fall to half its original value. (Assume that the temperature of the air remains constant during this time.)
Uploaded: 2022-10-03T17:34:24-08:00
Duration: 1 min 27 s
Channel: Natalie Belcher
Description: A flask of volume 300.0 x 10^-6 m^3 contains air at a pressure of 5.00 x 10^5 Pa and a temperature of 27.0°C. The flask loses molecules at a rate of 3.00 x 10^19 per second. Estimate how long it takes for the pressure in the flask to fall to half its original value. (Assume that the temperature of the air remains constant during this time.)

          A flask of volume 300.0 x 10^-6 m^3 contains air at a pressure of 5.00 x 10^5 Pa and a temperature of 27.0°C. The flask loses molecules at a rate of 3.00 x 10^19 per second. Estimate how long it takes for the pressure in the flask to fall to half its original value. (Assume that the temperature of the air remains constant during this time.)

A flask of volume 300.0 x 10^-6 m^3 contains air at a pressure of 5.00 x 10^5 Pa and a temperature of 27.0°C. The flask loses molecules at a rate of 3.00 x 10^19 per second. Estimate how long it takes for the pressure in the flask to fall to half its original value. (Assume that the temperature of the air remains constant during this time.)

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University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Natalie Belcher

10/03/2022

Step 1

0 \times 10^5 \, Pa \] \[ V = 300 \times 10^{-6} \, m^3 \] \[ R = 8.31 \, J/(mol \cdot K) \] \[ T = 300 \, K \] \[ n = \frac{5.0 \times 10^5 \times 300 \times 10^{-6}}{8.31 \times 300} \] \[ n = 0.06 \, mol \] Show more…

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