How Does Pressure Change Affect Internal Energy in an Ideal Gas?

  • Thread starter Thread starter hockeyfan123
  • Start date Start date
  • Tags Tags
    Gas Ideal gas
AI Thread Summary
The discussion focuses on calculating the total change in internal energy for a monoatomic ideal gas, given initial and final pressures of 1.9 atm and 4.8 atm. The user attempts to apply an incorrect formula for internal energy, leading to confusion about the path dependency of energy changes. It is emphasized that the change in internal energy should be calculated as the difference between two states, not as an absolute value. Additionally, inconsistencies in the provided pressure values suggest a potential error in the problem setup, particularly regarding temperature and volume relationships. The correct approach involves using the equations ΔU = nC_vΔT and Δ(PV) = nRΔT to find the change in internal energy accurately.
hockeyfan123
Messages
1
Reaction score
0
Homework Statement
Monoatomic ideal gas question
Relevant Equations
Internal Energy
Screen Shot 2022-12-04 at 4.14.28 PM.png


P1=1.9 and P2=4.8.

Question: what is the total change in internal energy

This is what I have so far but it is still incorrect I believe:

U= (3/2)(1/2)(2.9)(1.01x10^3)(8x10^-3)

Where am I going wrong?
 
Physics news on Phys.org
hockeyfan123 said:
Homework Statement:: Monoatomic ideal gas question
Relevant Equations:: Internal Energy

View attachment 318195

P1=1.9 and P2=4.8.

Question: what is the total change in internal energy

This is what I have so far but it is still incorrect I believe:

U= (3/2)(1/2)(2.9)(1.01x10^3)(8x10^-3)

Where am I going wrong?
Hello and welcome to PF!

It's not clear what formula you are using. You did not list the formula in the "Relevant Equations" section. Also, it would be good to include the units with your numerical values.

It appears to me that the data given in the problem is inconsistent. The change in internal energy should not depend on the particular path going from state A to state D. But I find different changes in energy for different paths.
 
I don't quite understand this equation
hockeyfan123 said:
U= (3/2)(1/2)(2.9)(1.01x10^3)(8x10^-3)
Please write symbolically first then show the numbers that you substitute. You can post equations in LateX. Click on the link "LaTeX guide", lower left, to learn how.

One wrong thing is that your answer is an internal energy at some point, not clear which. You are supposed to find the change in internal energy from A to D. This means the difference ##U_D-U_A##.
 
Another way to see that the data is inconsistent is to consider states E and D. They are at the same temperature. State E has half the volume of state D. Therefore, from Boyles's law (or from PV = nRT), state E should have twice the pressure of state D. But ##P_D## is given to be 1.9 atm and ##P_E## is given to be 4.8 atm. Check to see if the 4.8 atm was actually given as 3.8 atm.
 
The two equations you should be working with are $$\Delta U=nC_v\Delta T$$ and $$\Delta (PV)=nR\Delta T$$
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top