Thermodynamics cycle work question

[Vitalius6189]

Homework Statement

A mass of gas occupying volume V1 = 2 m3 at the pressure P1 = 4*10^5 Pa performs the cycle represented in the Figure that i have uploaded.What is the work of gas in this cycle, knowing that the pressure P2 = 10^5 Pa

Homework Equations

Work=1/2 * (P1 - P2) * (V1 - V3)

The Attempt at a Solution

The work is the area within the triangle.
1/2 base * height
= 1/2 * (P1 - P2) * (V1 - V3) The question that i have is how do i find V3?[/B]

 

[Vitalius6189]

Anyone?

 

[Ssnow]

The cartesian equation of the line from the step ##3## to the step ##2## is ##P(V)=2\cdot 10^5 \cdot V## so ## V_{3}=0,5 m^3##.

 

[Vitalius6189]

The cartesian equation of the line from the step ##3## to the step ##2## is ##P(V)=2\cdot 10^5 \cdot V## so ## V_{3}=0,5 m^3##.

Can you please explain this to me?

For example how did you get 2*10^5?

 

[Ssnow]

The number ##2\cdot 10^5## is the angular coefficient of the line from the process ##3## to ##2##. The line pass through the origin so the line has equation of the form ##P=m\cdot V## in order to find ##m## you can use the point ##(V_{1},P_{1})=(2,4\cdot 10^5)##, so ##4\cdot 10^5=m\cdot 2## and ##m=2\cdot 10^5##...
Ssnow

 

[Vitalius6189]

The number ##2\cdot 10^5## is the angular coefficient of the line from the process ##3## to ##2##. The line pass through the origin so the line has equation of the form ##P=m\cdot V## in order to find ##m## you can use the point ##(V_{1},P_{1})=(2,4\cdot 10^5)##, so ##4\cdot 10^5=m\cdot 2## and ##m=2\cdot 10^5##...
Ssnow

What is m? Is this some kinda new thing like volume or pressure or not?
also now that i know m how do i find V3

and also i apologize for being obnoxious and wasting your time but... why is the line passing through the origin means that the equation is of the form P=m*V?
again sorry for so many questions but i am genuinly curios and really want to learn before school starts

 

[Vitalius6189]

bump for visibility

 

[Ssnow]

In a Cartesian plane, with coordinates ##x## and ##y## you can rapresent a line (a direct proportionality law between ##x## and ##y##) using the equation ##y=mx+q## where ##m## is called the angular coefficient and ##q## is the pont where the line intersect the ##y## axis. In your case instead ##y## and ##x## there are ##P## and ##V## so the equation of your line is ##P=mV+q##. Now ##q## must be ##0## because the line pass in the origin, so from ##0=m\cdot 0 + q## you find ##q=0##, and ##P=mV## . In order to find the line equation you must determine the coefficient ##m##. You will use the other point on the line so as I said in the previous post ##P=2\cdot 10^5\cdot V##. Now you have the general equation of the line and using the point ##(P_{2},V_{3})=(10^5,V_{3})## you can find ##V_{3}##...
Ssnow