Understanding Pressure Equations in Linear and Abstract Algebra

  • Thread starter Auto Engineer
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In summary: Apparently you did not understand the distributive law:ax+ay=a(x+y). If it is "P" you want to get on its own, why don't you take P out rather than A?
  • #1
Auto Engineer
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Hi everyone, could someone please advise;

The equation;

PA - P atm A = K, change in X

If I wish to factor this equation to get "p" on its own, then;

PA - P atm A = K, change in X
= A ( P - P atm)

so the conclusion I draw is;

(P - P atm)A = K, change in X

My confusion is that I cannot work out how or where the second "A" went?

Could somebody put some light on the topic for me:confused:

David:smile:
 
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  • #2


Auto Engineer said:
Hi everyone, could someone please advise;

The equation;

PA - P atm A = K, change in X

If I wish to factor this equation to get "p" on its own, then;

PA - P atm A = K, change in X
= A ( P - P atm)
If it is "P" you want to get on its own, why don' you take P out rather than A?

so the conclusion I draw is;

(P - P atm)A = K, change in X

My confusion is that I cannot work out how or where the second "A" went?

Could somebody put some light on the topic for me:confused:

David:smile:[/QUOTE]
PA- P atm A= K
P(A- atm A)= K

P= k/(A- atm A)= k/(A(1- atm)).

12 - 6= 3(4)- 3(2)= 3(4- 2). Where did the second "3" go? That is the distributive law: ax+ ay= a(x+ y).
 
  • #3


HallsofIvy said:
If it is "P" you want to get on its own, why don' you take P out rather than A?

so the conclusion I draw is;

(P - P atm)A = K, change in X

My confusion is that I cannot work out how or where the second "A" went?

Could somebody put some light on the topic for me:confused:

David:smile:
PA- P atm A= K
P(A- atm A)= K

P= k/(A- atm A)= k/(A(1- atm)).

12 - 6= 3(4)- 3(2)= 3(4- 2). Where did the second "3" go? That is the distributive law: ax+ ay= a(x+ y).[/QUOTE]

The above distributive law is ok, but my example I feel you have misunderstood it:frown:

If "P" referred to "pressure" and "A" referred to "area", and "atm" was atmosphere, then one cannot remove "P" in that way and leave "atm" because "P" is the pressure of the atmosphere?
If using your method we said; PA - P atm A = K, and then you said; P(A - atm A) = K, the atmosphere cannot exist without "P" pressure, so A - A = 0, so P = K, the rest does not make sense to me, sorry:frown:

When the equation ends up with; (P - P atm)A = K, change in X, then I could work out from that the change in length of X?, this is because P - P atm means that say something (600 000 - 100 0000 = 500 000 x A = K, change in X, if you follow me with what I mean?

The equation is being factored to show the result obtained, I just need to understand how A seems to go missing?:frown:

David:smile:
 
  • #4


Apparently I did misunderstand. Perhaps if you told me what all this (particularly "the atmosphere cannot exist without "P" pressure") has to do with "Linear and Abstract Algebra", it would help.
 
  • #5


HallsofIvy said:
Apparently I did misunderstand. Perhaps if you told me what all this (particularly "the atmosphere cannot exist without "P" pressure") has to do with "Linear and Abstract Algebra", it would help.

Sorry that is my fault:eek:

P = pressure, but the equation is using more than one pressure which is why there are two "P"'s present, so (P - P atm) is saying that one pressure could be say liquid and then the other P atm is the atmospheric pressure. Also "A" is considered in this equation to be the area of a property, so a cylinder would be an example. K is understood to be a constant, and X is the change in length of a property like a spring, but the right hand side of the equation can be ignored when re aranging for "P". I am to understand the method is to factorise it, but I can only do so much of it.

So to recap;

PA - P atm A = K, change in X

= (P - P atm)A

This is all I know, what confuses me is where the second "A" went?

David:smile:
 

FAQ: Understanding Pressure Equations in Linear and Abstract Algebra

What does "PA" stand for in the equation PA - P atm A = K?

PA stands for partial pressure of gas A.

Why is the atmospheric pressure (P atm) subtracted from the partial pressure in the equation PA - P atm A = K?

This is because the partial pressure of a gas is the pressure it would exert if it were the only gas present in a container. By subtracting the atmospheric pressure, we can isolate the pressure exerted by gas A alone.

What is the significance of the constant "K" in the equation PA - P atm A = K?

The constant "K" represents the equilibrium constant for the reaction involving gas A. It is a measure of the extent to which the reaction has reached equilibrium.

How is the equation PA - P atm A = K used in scientific research?

This equation is commonly used in the field of physical chemistry to calculate the partial pressure of a gas in a reaction mixture. It is also used to determine the equilibrium constant for a reaction involving gases.

Can the equation PA - P atm A = K be applied to all gases?

Yes, this equation can be applied to any gas present in a reaction mixture. However, care must be taken to ensure that the units for pressure are consistent throughout the equation.

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