Solving a Physics Problem: Work Done by OR on a Gas

[kent davidge]

Homework Statement

Homework Equations

no

The Attempt at a Solution

no

Since the problem asks how much work was done by OR on the gas, I did not understand why the book's answer is 162 J instead ±81 J that I've found. (sorry my bad english)

Sorry, the correct question on the problem is how much work was done from b to c instead from a to b as it's in the image.

 

[Isaac0427]

Show us how you got your answer, then we can point out what you did wrong. Also, be careful with your units. I don't know all the unit conversions but to use Joules the units for P and V are pascals and meters cubed, not atmospheres and liters.

 

[kent davidge]

Okay.

dW = dV p

In this case we have the initial and final values of V and p. So, W = (V_c - V_b) x 10^-3m³ x (P_c - P_b) x 1.013 x 10⁵Pa, which gives W = 81.04 J.

 

[haruspex]

Okay.

dW = dV p

In this case we have the initial and final values of V and p. So, W = (V_c - V_b) x 10^-3m³ x (P_c - P_b) x 1.013 x 10⁵Pa, which gives W = 81.04 J.

You have calculated $\Delta V\Delta p$. That is not the same as $\int p.dV$.

 

[kent davidge]

and how can I solve the integral for T?

 

[haruspex]

and how can I solve the integral for T?

In the graph, p is the y ordinate and V the x ordinate. So the integral is equivalent to $\int y.dx$. what's a geometric interpretation of that integral?

 

[kent davidge]

oh yes, I see that and I solve the problem by this way. But I wonder if there's anyway to solve this integral for T using only calculations without the graph. Is there a way?

 

[haruspex]

oh yes, I see that and I solve the problem by this way. But I wonder if there's anyway to solve this integral for T using only calculations without the graph. Is there a way?

Yes, but you first have to turn the graph into an equation relating p to V. Then plug that function into $\int p.dV$.

Edit: When you say you solved the problem that way, are you referring to your solution in post #3? That solution was wrong.

 

[kent davidge]

Ok. Thank you.

 

[Herman Trivilino]

Since the problem asks how much work was done by OR on the gas,

They do not want an answer with a ± sign in front of it. They want a positive number and they want you to determine whether it's "on" or "by".