Non-ideal Gas: Withdrawing gas from bottle with increasing temperature

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Homework Statement

A gas mixture (80.1 mol% CO2, 12.2 mol% CO and balance Ar) is stored in a gas bottle. The volume of the gas bottle is 317.0 liters. Initially the pressure of the gas in the bottle is 14.2 MPa and the temperature is 25.3 °C

Qn: The gas is being withdrawn from the bottle at a constant rate of 310.2 moles/hour. As this is occurring the temperature of the gas in the bottle is increasing at a rate of 13.7 °C/hr. How long does it take for the pressure of the gas in the bottle to reach 3.0 MPa?

Homework Equations

Non-ideal gas equation: PV = znRT

The Attempt at a Solution

I tried doing this by calculating the pressure of the gas in the bottle every one hour. Since the temperature is increasing, the pressure will constantly be changing every moment. Next i plotted a graph of pressure of gas vs time and joined the dots. i could locate the 3.0 MPa point but i think it's not accurate enough. Can anyone help out please? Thanks!

 

[Jhero]

Thank you for your question. In order to accurately calculate the time it takes for the pressure of the gas in the bottle to reach 3.0 MPa, we need to take into account the change in temperature and volume as well.

First, we can use the ideal gas law to calculate the initial volume of the gas in the bottle:

PV = nRT
(14.2 MPa)(317.0 L) = (80.1 mol + 12.2 mol)(8.314 J/mol*K)(298.45 K)
V = 339.3 L

Next, we can use the non-ideal gas equation, PV = znRT, to calculate the new pressure of the gas in the bottle after a certain amount of time has passed:

P2 = (nRT2)/(V2)
P2 = (92.3 mol)(8.314 J/mol*K)(311.15 K)/(339.3 L)
P2 = 8.5 MPa

We can then use the ideal gas law to calculate the new volume of the gas in the bottle at this new pressure and temperature:

V2 = (nRT2)/(P2)
V2 = (92.3 mol)(8.314 J/mol*K)(311.15 K)/(8.5 MPa)
V2 = 339.3 L

This shows that the volume of the gas in the bottle remains constant even as the pressure decreases. Therefore, we can use the ideal gas law to calculate the time it takes for the pressure to decrease to 3.0 MPa:

P = (nRT)/(V)
(3.0 MPa) = (92.3 mol)(8.314 J/mol*K)(T)/(339.3 L)
T = 1.6 hours

So, it takes approximately 1.6 hours for the pressure of the gas in the bottle to reach 3.0 MPa. I hope this helps and please let me know if you have any further questions.