How to Calculate Gas Usage in Pressure Drop Situation?

AI Thread Summary
The discussion revolves around calculating gas usage during a pressure drop in a welder's oxygen tank, where the pressure decreases from 150 atm to 120 atm. The ideal gas law is referenced, emphasizing the need to consider temperature and volume constants. Participants agree that the tank's volume remains unchanged, as it is not elastic. However, the challenge lies in the unknown temperatures before and after the pressure change, leading to confusion about how to proceed with calculations. Ultimately, it is suggested to assume that the tank is not insulated, allowing for equal temperatures at room temperature for the calculations.
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Homework Statement


Pressure drops in welders tank of oxygen gas from p1=150atm to p2=120atm .
How much of the gas will be used ?

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The Attempt at a Solution


In my mind the simplest way of looking at this problem is to consider the process isothermal.
But then the volume of the gas increases when pressure drops (Ideal gas law). This contradicts the question.

Since I do not know the temperatures before and after, I don't know what to do. I am looking for different view at this problem, some help?
 
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If you want to treat the gas as an ideal gas then state the ideal gas law.
Looking at the law, which of the variables is constant and which variable? List them.
i.e. does the volume of the tank change?
 
Volume of the gas does not change if the tank is not elastic (I think not in this question) since the gas will always fill the whole tank.
 
Simon Bridge said:
If you want to treat the gas as an ideal gas then state the ideal gas law.
Looking at the law, which of the variables is constant and which variable? List them.
i.e. does the volume of the tank change?
Guneykan Ozgul said:
Volume of the gas does not change if the tank is not elastic (I think not in this question) since the gas will always fill the whole tank.

OK, I Think it is logical to assume that Volume of tank does not change. Now I can rethink my solution.
##P_1V=n_1RT_1## and ##P_2V=n_2RT_2##
## \frac {n_1RT_1} {P_1} = \frac {n_2RT_2} {P_2}##
## \frac {n_2} {n_1}=\frac {T_2P_1} {T_1P_2}##

But It seems that I need the temperature before and after, which i do not have. And to assume that T=const seems to be to only way, but it also seems not logical.
 
prehisto said:
OK, I Think it is logical to assume that Volume of tank does not change. Now I can rethink my solution.
##P_1V=n_1RT_1## and ##P_2V=n_2RT_2##
## \frac {n_1RT_1} {P_1} = \frac {n_2RT_2} {P_2}##
## \frac {n_2} {n_1}=\frac {T_2P_1} {T_1P_2}##

But It seems that I need the temperature before and after, which i do not have. And to assume that T=const seems to be to only way, but it also seems not logical.
You're supposed to assume that the tank is not insulated, so that the final and initial temperatures are equal to room temperature.
 
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