Pressure change with constant volume

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SUMMARY

The discussion focuses on calculating the change in temperature (\(\Delta T\)), work done, and internal energy change for a gas undergoing an isochoric process, starting with an initial pressure of \(P_i = 2 \text{ Atm}\) and volume of \(V_i = 0.001 \text{ m}^3\). The gas is heated to a final pressure of \(2P_i\). The first law of thermodynamics is applied, leading to the conclusion that all heat added to the system translates into internal energy change, as the volume remains constant. The participants discuss the implications of assuming the ideal gas law for deriving temperature changes based on pressure.

PREREQUISITES
  • Understanding of the first law of thermodynamics
  • Familiarity with isochoric processes in thermodynamics
  • Knowledge of internal energy equations, specifically \(U = aK_{B}T\)
  • Basic principles of the ideal gas law
NEXT STEPS
  • Study the derivation of the ideal gas law and its implications for real gases
  • Explore the relationship between pressure, volume, and temperature in isochoric processes
  • Learn about the mathematical formulation of the first law of thermodynamics
  • Investigate specific heat capacities and their role in internal energy calculations
USEFUL FOR

Students and professionals in physics and engineering, particularly those focusing on thermodynamics and gas laws, will benefit from this discussion.

SadStudent
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Homework Statement


Given two moles of some gas with two atoms per molecule which has energy U=aK_{B}T.
Assume the initial pressure is P_{i}= 2 Atm and the initial volume is P_{i}= 0.001\ m^{3}.
Now we heat it in an isochoric process to 2P_{i}.

What would be \Delta T?
What would be the work required to do that?
What would be the change in the gas particles' internal energy?

Homework Equations


U=aK_{B}T
dU=PdS-TdV

The Attempt at a Solution


Well using the first law of thermodynamics, I got that under constant volume dV=0 so \frac{dQ}{dT}=\frac{\partial U}{\partial T} so the entire energy that we spent on heating would be transferred into the gas.
What's a tricky to me is to formulate the change in temperature as a function of pressure because dV=0 so I'd really welcome a hint.

Thanks!
 
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SadStudent said:

Homework Equations



dU=PdS-TdV

Oops, check that.

What's a tricky to me is to formulate the change in temperature as a function of pressure because dV=0 so I'd really welcome a hint.

Can you assume the gas obeys the ideal gas law?
 
TSny said:
Oops, check that.Can you assume the gas obeys the ideal gas law?

And yes of course it was a typo :) it's supposed to be dU=TdS-PdV

I'm really not sure whether I can, is there a way of solving this without this assumption?
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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