How Does Gas Behavior on a PV Diagram Indicate Changes in Internal Energy?

AI Thread Summary
The discussion focuses on how gas behavior on a PV diagram reflects changes in internal energy (ΔEint). When the gas moves along the isotherm, ΔEint remains zero since temperature does not change. Moving to region 1 results in a decrease in temperature, leading to a negative ΔEint, while moving to region 2 results in an increase in temperature, resulting in a positive ΔEint. The analysis assumes the gas behaves ideally, confirming that the answers for each scenario are accurate. Understanding these relationships is crucial for interpreting gas behavior in thermodynamics.
Any Help
Messages
79
Reaction score
2

Homework Statement

figure: http://www.chegg.com/homework-help/questions-and-answers/dot-figure-represents-initial-state-gas-isotherm-dot-divides-p-v-diagram-regions-1-2-follo-q5905441

The dot in Figure (b) represents the initial state of a gas, and the isotherm through the dot divides the p-V diagram into regions 1 and 2. For the following processes, determine whether the change ΔEint in the internal energy of the gas is positive, negative, or zero.

(a) the gas moves up along the isotherm(b) the gas moves down along the isotherm(c) the gas moves to anywhere in region 1(d) the gas moves to anywhere in region 21.ΔEint is positive
2.ΔEint is negative
3.ΔEint is zero

Homework Equations



curve of an isotherm is a hyperbola so as the hyperbola becomes closer to the center (0,0) at its inflection point the temperature decreases

deltaE internal= 3/2 nRdelta T means as temperature increases deltaE positive and Vise Versa

The Attempt at a Solution



when moving on isotherm the delta Einternal will be zero since Einternal depends on the change of temperature the a and b will be zero (a,b =3)

but while moving under the curve then we are getting a lesser temperature. the temperature decreases so delta E negative so c is negative (c= 2

similarly for moving above the curve then d is positive (d=1)

Is my answer correct?
 
Last edited:
Physics news on Phys.org
Yes, provided in a and b you are dealing with an ideal gas.
 
  • Like
Likes Any Help
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top