# Homework Help: Thermodynamics cycle work question

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1. Jun 20, 2017

### Vitalius6189

1. The problem statement, all variables and given/known data
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

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

3. 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?

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2. Jun 20, 2017

### Vitalius6189

Anyone?

3. Jun 20, 2017

### 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 $V_{3}=0,5 m^3$.

4. Jun 20, 2017

### Vitalius6189

Can you please explain this to me?

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

Last edited: Jun 20, 2017
5. Jun 20, 2017

### 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

6. Jun 20, 2017

### Vitalius6189

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

7. Jun 21, 2017

### Vitalius6189

bump for visibility

8. Jun 23, 2017

### 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