How Does Gas Escape Affect Volume and Mass in Thermodynamics?

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Homework Help Overview

The discussion revolves around a thermodynamics problem involving a container of oxygen gas, where participants analyze the effects of gas escape on volume and mass. The initial conditions include a specific mass of gas at a defined pressure and temperature, followed by changes in these parameters due to gas escaping.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the ideal gas law (PV=NRT) to determine the volume of the container and the mass of gas that has escaped. There are questions regarding the calculations of volume and the resulting number of moles, with some participants expressing confusion over the results obtained.

Discussion Status

Some participants have provided calculations and corrections to earlier mistakes, indicating a productive exchange of ideas. There is acknowledgment of errors in initial calculations, and a clearer understanding of the volume is emerging, though inconsistencies in the number of moles remain a point of discussion.

Contextual Notes

Participants are working under the constraints of the ideal gas law and are addressing the implications of changes in pressure and temperature on the system. There is a focus on ensuring accurate calculations and understanding the relationships between the variables involved.

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Homework Statement


A container contains 1g of O_2 at a pressure of 1 atm and temperature T_i=47°C. At a certain time, due to an escape of gas, the pressure is 5/8 atm and the temperature decreases to 27°C.
1)What is the volume of the container?
2)How much mass of O_2 has escaped?


Homework Equations


PV=NRT.


The Attempt at a Solution


1g of O_2 is worth 1/32 mol of O_2. Thus V=\frac{N_iRT_i}{P_i}. I took care of converting degrees celsius to kelvin, took 1atm =10⁵ pascal. This gave me V=8.31m^3.
While N_f = \frac{P_fV}{RT_f} \approx 208.25 mol which makes absolutely no sense. I should get less than 1/32, yet I get a number greater than 200. I don't know what I did wrong, I tried another approach leading to another inconsistency: N_f=1.5 N_i.
 
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Let's see...

$$V_i={N_i R T_i \over P_i}={\frac 1 {32} \cdot 8.31 \cdot 320 \over 10^5}=8.31 \cdot 10^{-4} m^3$$
How did you get ##V_i=8.31 m^3##?
 
I like Serena said:
Let's see...

$$V_i={N_i R T_i \over P_i}={\frac 1 {32} \cdot 8.31 \cdot 320 \over 10^5}=8.31 \cdot 10^{-4} m^3$$
How did you get ##V_i=8.31 m^3##?

Hi! Thanks :biggrin: Misuse of calculator apparently. I must be extremely careful to put all parenthesis I think. I now get your result.
Number of moles in the end: 0.005 or about 1/200. Thus about 0.16 g of O_2.
 
Yeah, I thought as much.
And this is a problem with numbers that are "nice", so you can easily do it without calculator! ;)
 

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